Mathematics 1174 Calculus I - with Economics & Business Applications First Custom Edition 2023 This Book is Adapted from the Open Textbook Titled APEX Calculus by Gregory Hartman Modified for Langara College by Pichmony Anhaouy panhaouy@langara.ca Copyrights: © 2023 • Designed Cover belongs to Pichmony Anhaouy, All Rights Reserved. • The Langara Logo belongs to the Langara College, All Rights Reserved. • The APEX Calculus, version 3.0, by Gregory Hartman, and this custom book are licensed under the Creative Commons Attribution- NonCommercialShare-Alike 4.0 International License. A license can be found here https://creativecommons.org/licenses/by-nc-sa/4.0/. Contents Table of Contents iii Preface v 1 Limits 1.1 An Introduction To Limits . . . . . . . . . . . . . . . . . . . 1.2 Epsilon-Delta Definition of a Limit . . . . . . . . . . . . . . 1.3 Finding Limits Analytically . . . . . . . . . . . . . . . . . . 1.4 One Sided Limits . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Limits Involving Infinity . . . . . . . . . . . . . . . . . . . . 1.6 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 10 19 32 39 52 2 Derivatives 65 2.1 Introduction to Derivative . . . . . . . . . . . . . . . . . . . 65 2.2 Interpretations of the Derivative . . . . . . . . . . . . . . . 79 2.3 Basic Differentiation Rules . . . . . . . . . . . . . . . . . . . 86 2.4 The Product and Quotient Rules . . . . . . . . . . . . . . . 92 2.5 Derivative as Rates of Change . . . . . . . . . . . . . . . . . 100 2.5.1 Linear Motion . . . . . . . . . . . . . . . . . . . . . 100 2.5.2 Marginal Analysis . . . . . . . . . . . . . . . . . . . 102 2.6 Derivatives of Trigonometric Functions . . . . . . . . . . . . 107 2.7 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . 112 2.8 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . 122 2.9 Derivatives of Exponential and Logarithmic Functions . . . 132 2.10 Derivatives of Inverse Trigonometric Functions . . . . . . . 142 3 The Graphical Behaviour of Functions 147 3.1 Extreme Values . . . . . . . . . . . . . . . . . . . . . . . . . 147 3.2 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . 155 3.3 Increasing and Decreasing Functions . . . . . . . . . . . . . 160 3.4 Concavity and the Second Derivative . . . . . . . . . . . . . 169 Contents 3.5 Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . 177 4 Applications of the Derivative 185 4.1 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . 185 4.2 Applied Optimization Problems . . . . . . . . . . . . . . . . 192 4.3 Economics and Business Applications . . . . . . . . . . . . 199 4.3.1 Elasticity of Demand Problems . . . . . . . . . . . . 199 4.3.2 Optimal Harvest Problems . . . . . . . . . . . . . . 204 4.3.3 Inventory Costs Control . . . . . . . . . . . . . . . . 207 4.4 Linear Approximation and Differentials . . . . . . . . . . . 212 4.4.1 Linear Approximation (Tangent Line Approximation Revisited) . . . . . . . . . . . . . . . . . . . . . 212 4.4.2 Differentials . . . . . . . . . . . . . . . . . . . . . . . 215 4.5 L’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . 223 4.6 Antiderivatives and Indefinite Integration . . . . . . . . . . 232 A Solutions To Selected Problems A.1 Index A.11 Notes: Page iv Custom Made for Math 1174 at Langara College by P. Anhaouy Preface This custom book is created in response to the ever increasing price of calculus textbooks available out there through many publication companies. It is created specifically for Mathematics 1174 at Langara College. This book is mainly adapted from the first four chapters of the open source calculus textbook titled APEX Calculus, Version 3.0, by Gregory Hartman, department of applied mathematics, Virginia Military Institute. This complete calculus book is under the Creative Commons Attribution- NonCommercial-Share-Alike 4.0 International License. It is available for free at www.apexcalculus.com. Chapter 2 is greatly modified in such a way that exponential, logarithmic, and trigonometric functions and their derivatives have their own sections. Also, the book is partly taken from the Math 1174 Lecture Notes, by Pichmony Anhaouy, department of mathematics and statistics, Langara College. The main topics from the notes are economics and business applications in chapter 4, section 4.3, linear motion and marginal analysis in chapter 2, section 2.5. In addition, topic such as linear approximation is added to section 4.4. We add two examples to section 1.5, three to section 1.4, one to section 2.3, and two examples showing how the chain rule is used in business problems to section 2.7. These examples can be found at the end of each section. Few more examples and exercises were added to new derivatives of trigonometric and inverse trigonometric functions. I would like to thank my dear colleagues, particularly Ken Collins, Bruce Aubertin, Nora Franzova, and Sonoko Nakano, and my former students in the Mathematics and Statistic Department for their encouragements and contributions that lead to the existence of this book. For more information regarding this book, whether it is about contents or typos, please contact: Pichmony Anhaouy at panhaouy@langara.ca. 1: Limits Calculus means “a method of calculation or reasoning.” When one computes the sales tax on a purchase, one employs a simple calculus. When one finds the area of a polygonal shape by breaking it up into a set of triangles, one is using another calculus. Proving a theorem in geometry employs yet another calculus. Despite the wonderful advances in mathematics that had taken place into the first half of the 17th century, mathematicians and scientists were keenly aware of what they could not do. In particular, two important concepts eluded mastery by the great thinkers of that time: area and rates of change. Area seems innocuous enough; areas of circles, rectangles, parallelograms, etc., are standard topics of study for students today just as they were then. However, the areas of arbitrary shapes could not be computed, even if the boundary of the shape could be described exactly. Rates of change were also important. When an object moves at a constant rate of change, then “distance = rate × time.” But what if the rate is not constant – can distance still be computed? Or, if distance is known, can we discover the rate of change? It turns out that these two concepts were related. Two mathematicians, Sir Isaac Newton and Gottfried Leibniz, are credited with independently formulating a system of computing that solved the above problems and showed how they were connected. Their system of reasoning was “a” calculus. However, as the power and importance of their discovery took hold, it became known to many as “the” calculus. Today, we generally shorten this to discuss “calculus.” The foundation of “the calculus” is the limit. It is a tool to describe a particular behaviour of a function. This chapter begins our study of the limit by approximating its value graphically and numerically. After a formal definition of the limit, properties are established that make “finding limits” tractable. Once the limit is understood, then the problems of area and rates of change can be approached. 1.1 An Introduction To Limits We begin our study of limits by considering examples that demonstrate key concepts that will be explained as we progress. Consider the function y = sin x . When x is near the value 1, what x Chapter 1 Limits y 1 0.8 0.6 . x 1 0.5 Figure 1.1: 1.5 sin(x) near x = 1. x y 1 0.9 value (if any) is y near? While our question is not precisely formed (what constitutes “near the value 1”?), the answer does not seem difficult to find. One might think first to look at a graph of this function to approximate the appropriate y sin x values. Consider Figure 1.1, where y = is graphed. For values of x x near 1, it seems that y takes on values near 0.85. In fact, when x = 1, sin 1 then y = ≈ 0.84, so it makes sense that when x is “near” 1, y will 1 be “near” 0.84. Consider this again at a different value for x. When x is near 0, what value (if any) is y near? By considering Figure 1.2, one can see that it seems that y takes on values near 1. But what happens when x = 0? We have “ 0 ” sin 0 → . y→ 0 0 The expression “0/0” has no value; it is indeterminate. Such an expression gives no information about what is going on with the function nearby. We cannot find out how y behaves near x = 0 for this function simply by letting x = 0. Finding a limit entails understanding how a function behaves near a particular value of x. Before continuing, it will be useful to establish some notation. Let y = f (x); that is, let y be a function of x for some function f . The expression “the limit of y as x approaches 1” describes a number, often referred to as L, that y nears as x nears 1. We write all this as 0.8 lim y = lim f (x) = L. . x→1 x −1 1 Figure 1.2: sin(x) near x = 0. x This is not a complete definition (that will come in the next section); this is a pseudo-definition that will allow us to explore the idea of a limit. Above, where f (x) = sin(x)/x, we approximated lim x→1 x 0.9 0.99 0.999 1 1.001 1.01 1.1 sin(x) x 0.870363 0.844471 0.841772 0.841471 0.84117 0.838447 0.810189 Figure 1.3: Values of near 1. x→1 sin x ≈ 0.84 x and lim x→0 sin x ≈ 1. x (We approximated these limits, hence used the “≈” symbol, since we are working with the pseudo-definition of a limit, not the actual definition.) Once we have the true definition of a limit, we will find limits analytically; that is, exactly using a variety of mathematical tools. For now, we will approximate limits both graphically and numerically. Graphing a function can provide a good approximation, though often not very precise. Numerical methods can provide a more accurate approximation. We have already approximated limits graphically, so we now turn our attention to numerical approximations. sin(x) with x x Notes: Page 2 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.1 sin(x) . To approximate this limit numerically, we x→1 x can create a table of x and f (x) values where x is “near” 1. This is done in Figure 1.3. sin(x) near 0.841. The Notice that for values of x near 1, we have x x = 1 row is in bold to highlight the fact that when considering limits, we are not concerned with the value of the function at that particular x value; we are only concerned with the values of the function when x is near 1. sin(x) numerically. We already approximated Now approximate lim x→0 x the value of this limit as 1 graphically in Figure 1.2. The table in Figure 1.4 shows the value of sin(x)/x for values of x near 0. Ten places after the decimal point are shown to highlight how close to 1 the value of sin(x)/x gets as x takes on values very near 0. We include the x = 0 row in bold again to stress that we are not concerned with the value of our function at x = 0, only on the behavior of the function near 0. This numerical method gives confidence to say that 1 is a good apsin(x) ; that is, proximation of lim x→0 x An Introduction To Limits Consider again lim sin(x) ≈ 1. x→0 x lim Later we will be able to prove that the limit is exactly 1. Example 1 Approximating the value of a limit Use graphical and numerical methods to approximate 2 x −x−6 lim 2 . x→3 6x − 19x + 3 Solution x -0.1 -0.01 -0.001 0 0.001 0.01 0.1 sin(x) x 0.9983341665 0.9999833334 0.9999998333 not defined 0.9999998333 0.9999833334 0.9983341665 Figure 1.4: Values of near 0. sin(x) with x x y 0.34 0.32 0.3 0.28 0.26 . x 2.5 3 3.5 Figure 1.5: Graphically approximating a limit in Example 1. To graphically approximate the limit, graph x2 − x − 6 y= 2 6x − 19x + 3 on a small interval that contains 3. To numerically approximate the limit, create a table of values where the x values are near 3. This is done in Figures 1.5 and 1.6, respectively. The graph shows that when x is near 3, the value of y is very near 0.3. By considering values of x near 3, we see that y = 0.294 is a better approximation. The graph and the table imply that x2 − x − 6 ≈ 0.294. lim 2 x→3 6x − 19x + 3 x 2.9 2.99 2.999 3 3.001 3.01 3.1 x2 − x − 6 6x2 − 19x + 3 0.29878 0.294569 0.294163 not defined 0.294073 0.293669 0.289773 Figure 1.6: Numerically approximating a limit in Example 1. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 3 Chapter 1 Limits This example may bring up a few questions about approximating limits (and the nature of limits themselves). 1. If a graph does not produce as good an approximation as a table, why bother with it? 2. How many values of x in a table are “enough?” In the previous example, could we have just used x = 3.001 and found a fine approximation? y 1 0.5 .−1 x −0.5 0.5 1 Figure 1.7: Graphically approximating a limit in Example 2. x -0.1 -0.01 -0.001 0.001 0.01 0.1 f (x) 0.9 0.99 0.999 0.999999 0.9999 0.99 Figure 1.8: Numerically approximating a limit in Example 2. Graphs are useful since they give a visual understanding concerning the behaviour of a function. Sometimes a function may act “erratically” near certain x values which is hard to discern numerically but very plain graphically. Since graphing utilities are very accessible, it makes sense to make proper use of them. Since tables and graphs are used only to approximate the value of a limit, there is not a firm answer to how many data points are “enough.” Include enough so that a trend is clear, and use values (when possible) both less than and greater than the value in question. In Example 1, we used both values less than and greater than 3. Had we used just x = 3.001, we might have been tempted to conclude that the limit had a value of 0.3. While this is not far off, we could do better. Using values “on both sides of 3” helps us identify trends. Example 2 Approximating the value of a limit Graphically and numerically approximate the limit of f (x) as x approaches 0, where  x+1 x<0 f (x) = . −x2 + 1 x > 0 Again we graph f (x) and create a table of its values Solution near x = 0 to approximate the limit. Note that this is a piecewise defined function, so it behaves differently on either side of 0. Figure 1.7 shows a graph of f (x), and on either side of 0 it seems the y values approach 1. Note that f (0) is not actually defined, as indicated in the graph with the open circle. The table shown in Figure 1.8 shows values of f (x) for values of x near 0. It is clear that as x takes on values very near 0, f (x) takes on values very near 1. It turns out that if we let x = 0 for either “piece” of f (x), 1 is returned; this is significant and we’ll return to this idea later. The graph and table allow us to say that limx→0 f (x) ≈ 1; in fact, we are probably very sure it equals 1. Notes: Page 4 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.1 Identifying When Limits Do Not Exist An Introduction To Limits y 3 A function may not have a limit for all values of x. That is, we cannot say limx→c f (x) = L for some numbers L for all values of c, for there may not be a number that f (x) is approaching. There are three ways in which a limit may fail to exist. 2 1. The function f (x) may approach different values on either side of c. 1 2. The function may grow without upper or lower bound as x approaches c. . 3. The function may oscillate as x approaches c. We’ll explore each of these in turn. x 0.5 x 0.9 0.99 0.999 1.001 1.01 1.1 x→1 f (x) = x2 − 2x + 3 x ≤ 1 . x x>1 A graph of f (x) around x = 1 and a table are given Figures 1.9 and 1.10, respectively. It is clear that as x approaches 1, f (x) does not seem to approach a single number. Instead, it seems as though f (x) approaches two different numbers. When considering values of x less than 1 (approaching 1 from the left), it seems that f (x) is approaching 2; when considering values of x greater than 1 (approaching 1 from the right), it seems that f (x) is approaching 1. Recognizing this behavior is important; we’ll study this in greater depth later. Right now, it suffices to say that the limit does not exist since f (x) is not approaching one value as x approaches 1. Solution The Function Grows Without Bound 1 Explore why lim does not exist. x→1 (x − 1)2 1.5 2 Figure 1.9: Observing no limit as x → 1 in Example 3. Example 3 Different Values Approached From Left and Right Explore why lim f (x) does not exist, where  1 f (x) 2.01 2.0001 2.000001 1.001 1.01 1.1 Figure 1.10: Values of f (x) near x = 1 in Example 3. y 100 50 Example 4 1 are given in Fig(x − 1)2 ures 1.11 and 1.12, respectively. Both show that as x approaches 1, f (x) grows larger and larger. Solution Notes: A graph and table of f (x) = . x 0.5 1 1.5 2 Figure 1.11: Observing no limit as x → 1 in Example 4. x 0.9 0.99 0.999 1.001 1.01 1.1 f (x) 100. 10000. 1. × 106 1. × 106 10000. 100. Figure 1.12: Values of f (x) near x = 1 in Example 4. Custom Made for Math 1174 at Langara College by P. Anhaouy Page 5 Chapter 1 Limits We can deduce this on our own, without the aid of the graph and table. If x is near 1, then (x − 1)2 is very small, and: 1 = very large number. very small number Since f (x) is not approaching a single number, we conclude that 1 x→1 (x − 1)2 lim does not exist. Example 5 The Function Oscillates 1 does not exist. Explore why lim sin x→0 x Two graphs of f (x) = sin(1/x) are given in Figures 1.13. Solution Figure 1.13(a) shows f (x) on the interval [−1, 1]; notice how f (x) seems to oscillate near x = 0. One might think that despite the oscillation, as x approaches 0, f (x) approaches 0. However, Figure 1.13(b) zooms in on sin(1/x), on the interval [−0.1, 0.1]. Here the oscillation is even more pronounced. Finally, in the table in Figure 1.13(c), we see sin(x)/x evaluated for values of x near 0. As x approaches 0, f (x) does not appear to approach any value. It can be shown that in reality, as x approaches 0, sin(1/x) takes on all values between −1 and 1 infinite times! Because of this oscillation, 1 lim sin does not exist. x→0 x y y 1 1 0.5 0.5 x −1 −0.5 0.5 1 x −0.1 −5 · 10−2 −0.5 0.1 −0.5 −1 . 5 · 10−2 . (a) −1 (b) f Figure 1.13: Observing that f (x) = sin . 1 x has no limit as x → 0 in Example 5. Notes: x 2 4 sin(1/x) −0.544021 −0.506366 0.82688 −0.305614 0.0357488 −0.349994 0.420548 (c) 20 10 x 0.1 0.01 0.001 0.0001 1. × 10−5 1. × 10−6 1. × 10−7 6 Figure 1.14: Interpreting a difference quotient as the slope of a secant line. Page 6 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.1 Limits of Difference Quotients f We have approximated limits of functions as x approached a particular number. We will consider another important kind of limit after explaining a few key ideas. Let f (x) represent the position function, in feet, of some particle that is moving in a straight line, where x is measured in seconds. Let’s say that when x = 1, the particle is at position 10 ft., and when x = 5, the particle is at 20 ft. Another way of expressing this is to say f (1) = 10 and An Introduction To Limits f (5) = 20. 20 10 x . 2 4 6 4 6 4 6 (a) Since the particle traveled 10 feet in 4 seconds, we can say the particle’s average velocity was 2.5 ft/s. We write this calculation using a “quotient of differences,” or, a difference quotient: f 20 10 f (5) − f (1) = = 2.5ft/s. 5−1 4 This difference quotient can be thought of as the familiar “rise over run” used to compute the slopes of lines. In fact, that is essentially what we are doing: given two points on the graph of f , we are finding the slope of the secant line through those two points. See Figure 1.14 on the previous page. Now consider finding the average speed on another time interval. We again start at x = 1, but consider the position of the particle h seconds later. That is, consider the positions of the particle when x = 1 and when x = 1 + h. The difference quotient is now 10 lim h→0 2 (b) f 20 f (1 + h) − f (1) f (1 + h) − f (1) = . (1 + h) − 1 h Let f (x) = −1.5x2 + 11.5x; note that f (1) = 10 and f (5) = 20, as in our discussion. We can compute this difference quotient for all values of h (even negative values!) except h = 0, for then we get “0/0,” the indeterminate form introduced earlier. For all values h 6= 0, the difference quotient computes the average velocity of the particle over an interval of time of length h starting at x = 1. For small values of h, i.e., values of h close to 0, we get average velocities over very short time periods and compute secant lines over small intervals. See Figure 1.15. This leads us to wonder what the limit of the difference quotient is as h approaches 0. That is, x . 10 . x 2 (c) Figure 1.15: Secant lines of f (x) at x = 1 and x = 1 + h, for shrinking values of h (i.e., h → 0). f (1 + h) − f (1) = ? h Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 7 Chapter 1 h −0.5 −0.1 −0.01 0.01 0.1 0.5 Limits f (1 + h) − f (1) h 9.25 8.65 8.515 8.485 8.35 7.75 Figure 1.16: The difference quotient evaluated at values of h near 0. As we do not yet have a true definition of a limit nor an exact method for computing it, we settle for approximating the value. While we could graph the difference quotient (where the x-axis would represent h values and the y-axis would represent values of the difference quotient) we settle for making a table. See Figure 1.16. The table gives us reason to assume the value of the limit is about 8.5. Proper understanding of limits is key to understanding calculus. With limits, we can accomplish seemingly impossible mathematical things, like adding up an infinite number of numbers (and not get infinity) and finding the slope of a line between two points, where the “two points” are actually the same point. These are not just mathematical curiosities; they allow us to link position, velocity and acceleration together, connect cross-sectional areas to volume, find the work done by a variable force, and much more. In the next section we give the formal definition of the limit and begin our study of finding limits analytically. In the following exercises, we continue our introduction and approximate the value of limits. Notes: Page 8 Custom Made for Math 1174 at Langara College by P. Anhaouy Exercises 1.1 Terms and Concepts 1. In your own words, what does it mean to “find the limit of f (x) as x approaches 3”? 2. An expression of the form 00 is called . 3. T/F: The limit of f (x) as x approaches 5 is f (5). 13. lim f (x), where x→3  2 x −x+1 f (x) = 2x + 1 x≤3 . x>3 14. lim f (x), where x→0  cos x f (x) = x2 + 3x + 1 x≤0 . x>0 4. Describe three situations where lim f (x) does not exist. x→c 15. 5. In your own words, what is a difference quotient? lim f (x), where x→π/2  f (x) = sin x cos x x ≤ π/2 . x > π/2 Problems In Exercises 6 – 16, approximate the given limits both numerically and graphically. 6. lim (x2 + 3x − 5) In Exercises 16 – 24, a function f and a value a are given. Approximate the limit of the difference quotient, f (a + h) − f (a) lim , using h = ±0.1, ±0.01. h→0 h x→1 7. lim (x3 − 3x2 + x − 5) 16. f (x) = −7x + 2, a=3 17. f (x) = 9x + 0.06, a = −1 18. f (x) = x2 + 3x − 7, a=1 x→0 x+1 8. lim 2 x→0 x + 3x 2 9. lim x − 2x − 3 x→3 x2 − 4x + 3 19. f (x) = 1 , x+1 a=2 2 10. lim x + 8x + 7 x→−1 x2 + 6x + 5 20. f (x) = −4x2 + 5x − 1, 11. lim x2 + 7x + 10 x→2 x2 − 4x + 4 21. f (x) = ln x, a=5 12. lim f (x), where x→2  x+2 f (x) = 3x − 5 22. f (x) = sin x, a=π 23. f (x) = cos x, a=π x≤2 . x>2 a = −3 Custom Made for Math 1174 at Langara College by P. Anhaouy Page 9 Chapter 1 Limits 1.2 Epsilon-Delta Definition of a Limit This section introduces the formal definition of a limit. Many refer to this as “the epsilon–delta,” definition, referring to the letters  and δ of the Greek alphabet. Before we give the actual definition, let’s consider a few informal ways of describing a limit. Given a function y = f (x) and an x-value, c, we say that “the limit of the function f , as x approaches c, is a value L”: 1. if “y tends to L” as “x tends to c.” 2. if “y approaches L” as “x approaches c.” 3. if “y is near L” whenever “x is near c.” The problem with these definitions is that the words “tends,” “approach,” and especially “near” are not exact. In what way does the variable x tend to, or approach, c? How near do x and y have to be to c and L, respectively? The definition we describe in this section comes from formalizing 3. A quick restatement gets us closer to what we want: 30 . If x is within a certain tolerance level of c, then the corresponding value y = f (x) is within a certain tolerance level of L. The traditional notation for the x-tolerance is the lowercase Greek letter delta, or δ, and the y-tolerance is denoted by lowercase epsilon, or . One more rephrasing of 30 nearly gets us to the actual definition: 300 . If x is within δ units of c, then the corresponding value of y is within  units of L. We can write “x is within δ units of c” mathematically as |x − c| < δ, which is equivalent to c − δ < x < c + δ. Letting the symbol “−→” represent the word “implies,” we can rewrite 300 as |x−c| < δ −→ |y−L| <  or c−δ < x < c+δ −→ L− < y < L+. The point is that δ and , being tolerances, can be any positive (but typically small) values. Finally, we have the formal definition of the limit with the notation seen in the previous section. Notes: Page 10 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.2 Definition 1 Epsilon-Delta Definition of a Limit The Limit of a Function f Let I be an open interval containing c, and let f be a function defined on I, except possibly at c. The limit of f (x), as x approaches c, is L, denoted by lim f (x) = L, x→c means that given any  > 0, there exists δ > 0 such that for all x 6= c, if |x − c| < δ, then |f (x) − L| < . (Mathematicians often enjoy writing ideas without using any words. Here is the wordless definition of the limit: lim f (x) = L ⇐⇒ ∀  > 0, ∃ δ > 0 s.t. 0 < |x−c| < δ −→ |f (x)−L| < x→c .) Note the order in which  and δ are given. In the definition, the ytolerance  is given first and then the limit will exist if we can find an x-tolerance δ that works. An example will help us understand this definition. Note that the explanation is long, but it will take one through all steps necessary to understand the ideas. Example 6 √ Evaluating a limit using the definition Show that lim x = 2. x→4 Before we use the formal definition, let’s try some nuSolution merical tolerances. What if the y tolerance is 0.5, or  = 0.5? How close to 4 does x have to be so that y is within 0.5 units of 2, i.e., 1.5 < y < 2.5? In this case, we can proceed as follows: 1.5 < y < 2.5 √ 1.5 < x < 2.5 1.52 < x < 2.52 2.25 < x < 6.25. So, what is the desired x tolerance? Remember, we want to find a symmetric interval of x values, namely 4 − δ < x < 4 + δ. The lower bound of 2.25 is 1.75 units from 4; the upper bound of 6.25 is 2.25 units from 4. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 11 Chapter 1 Limits We need the smaller of these two distances; we must have δ ≤ 1.75. See Figure 1.17. Given the y tolerance  = 0.5, we have found an x tolerance, δ ≤ 1.75, such that whenever x is within δ units of 4, then y is within  units of 2. That’s what we were trying to find. Let’s try another value of . What if the y tolerance is 0.01, i.e.,  = 0.01? How close to 4 does x have to be in order for y to be within 0.01 units of 2 (or 1.99 < y < 2.01)? Again, we just square these values to get 1.992 < x < 2.012 , or 3.9601 < x < 4.0401. What is the desired x tolerance? In this case we must have δ ≤ 0.0399, which is the minimum distance from 4 of the two bounds given above. What we have so far: if  = 0.5, then δ ≤ 1.75 and if  = 0.01, then δ ≤ 0.0399. A pattern is not easy to see, so we switch√to general  try to determine δ symbolically. We start by assuming y = x is within  units of 2: y } 2 } |y − 2| <  ε = .5 − < y − 2 <  √ − < x − 2 <  √ 2−< x<2+ ε = .5 Choose ε > 0. Then ... 1 2 (Square all) 4 − 4 +  < x < 4 + 4 +  2 (Expand) (2 − ) < x < (2 + ) x 4 2 6 y } 2 } (Add 2) 2 2 . (Definition of absolute value) √ (y = x) 2 2 4 − (4 −  ) < x < 4 + (4 +  ). (Rewrite in the desired form) The “desired form” in the last step is “4 − something < x < 4 + something.” Since we want this last interval to describe an x tolerance around 4, we have that either δ ≤ 4 − 2 or δ ≤ 4 + 2 , whichever is smaller: δ ≤ min{4 − 2 , 4 + 2 }. ε = .5 ε = .5 ... choose δ smaller than each of these With  = 0.5, we pick any δ < 1.75. Since  > 0, the minimum is δ ≤ 4 − 2 . That’s the formula: given an , set δ ≤ 4 − 2 . We can check this for our previous values. If  = 0.5, the formula gives δ ≤ 4(0.5) − (0.5)2 = 1.75 and when  = 0.01, the formula gives δ ≤ 4(0.01) − (0.01)2 = 0.399. Figure 1.17: Illustrating the −δ process. Notes: 1 . 2 z width = 1.75 }| {z 4 width = 2.25 }| { x 6 Page 12 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.2 Epsilon-Delta Definition of a Limit So given any  > 0, set δ ≤ 4 − 2 . Then if |x − 4| < δ (and x 6= 4), then |f (x) − 2| < , satisfying the √ definition of the limit. We have shown formally (and finally!) that lim x = 2. x→4 The previous example was a little long in that we sampled a few specific cases of  before handling the general case. Normally √ this is not done. The previous example is also a bit unsatisfying in that 4 = 2; why work so hard to prove something so obvious? Many -δ proofs are long and difficult to do. In this section, we will focus on examples where the answer is, frankly, obvious, because the non–obvious examples are even harder. In the next section we will learn some theorems that allow us to evaluate limits analytically, that is, without using the -δ definition. Example 7 Evaluating a limit using the definition Show that lim x2 = 4. x→2 Let’s do this example symbolically from the start. Let Solution  > 0 be given; we want |y − 4| < , i.e., |x2 − 4| < . How do we find δ such that when |x − 2| < δ, we are guaranteed that |x2 − 4| < ? This is a bit trickier than the previous example, but let’s start by noticing that |x2 − 4| = |x − 2| · |x + 2|. Consider: |x2 − 4| <  −→ |x − 2| · |x + 2| <  −→ |x − 2| <  . |x + 2| (1.1)  ? |x + 2| We are close to an answer, but the catch is that δ must be a constant value (so it can’t contain x). There is a way to work around this, but we do have to make an assumption. Remember that  is supposed to be a small number, which implies that δ will also be a small value. In particular, we can (probably) assume that δ < 1. If this is true, then |x − 2| < δ would imply that |x − 2| < 1, giving 1 < x < 3.  Now, back to the fraction . If 1 < x < 3, then 3 < x + 2 < 5 |x + 2| (add 2 to all terms in the inequality). Taking reciprocals, we have Could we not set δ = 1 1 1 < < 5 |x + 2| 3 1 1 < 5 |x + 2|   < . 5 |x + 2| which implies which implies (1.2) Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 13 Chapter 1 Limits This suggests that we set δ ≤ y when we assume |x − 2| < δ: length of ε |x − 2| < δ  |x − 2| < 5 length of δ = ε/5 4  . To see why, let consider what follows 5  5  |x2 − 4| < |x + 2| · 5   = |x2 − 4| < |x + 2| · < |x + 2| · 5 |x + 2| |x − 2| · |x + 2| < |x + 2| · . δ z }| { x 2 Figure 1.18: Choosing δ = /5 in Example 7. (Our choice of δ) (Multiply by |x + 2|) (Combine left side) (Using (1.2) as long as δ < 1) We have arrived at |x2 −4| <  as desired. Note again, in order to make this happen we needed δ to first be less than 1. That is a safe assumption; we want  to be arbitrarily small, forcing δ to also be small. We have also picked δ to be smaller than “necessary.” We could get by with a slightly larger δ, as shown in Figure 1.18. The dashed outer lines show the boundaries defined by our choice of . The dotted inner lines show the boundaries defined by setting δ = /5. Note how these dotted lines are within the dashed lines. That is perfectly fine; by choosing x within the dotted lines we are guaranteed that f (x) will be within  of 4. In summary, given  > 0, set δ =≤ /5. Then |x − 2| < δ implies |x2 − 4| <  (i.e. |y − 4| < ) as desired. This shows that lim x2 = 4. x→2 Figure 1.18 gives a visualization of this; by restricting x to values within δ = /5 of 2, we see that f (x) is within  of 4. Make note of the general pattern exhibited in these last two examples. In some sense, each starts out “backwards.” That is, while we want to 1. start with |x − c| < δ and conclude that 2. |f (x) − L| < , we actually start by assuming 1. |f (x) − L| < , then perform some algebraic manipulations to give an inequality of the form 2. |x − c| < something. When we have properly done this, the something on the “greater than” side of the inequality becomes our δ. We can refer to this as the “scratch– work” phase of our proof. Once we have δ, we can formally start with Notes: Page 14 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.2 Epsilon-Delta Definition of a Limit |x−c| < δ and use algebraic manipulations to conclude that |f (x)−L| < , usually by using the same steps of our “scratch–work” in reverse order. We highlight this process in the following example. Example 8 Evaluating a limit using the definition Prove that lim (x3 − 2x) = −1. x→1 Solution : We start our scratch–work by considering |f (x)−(−1)| < |f (x) − (−1)| <  |x3 − 2x + 1| <  (Now factor) 2 |(x − 1)(x + x − 1)| <  |x − 1| <  . |x2 + x − 1| (1.3) We are at the phase of saying that |x − 1| < something, where something= /|x2 + x − 1|. We want to turn that something into δ. Since x is approaching 1, we are safe to assume that x is between 0 and 2. So 0 ln(1 + ) and ln(1 − ) < 0, we know ln(1 − ) < − ln(1 + ). 1 −  < ex < 1 +  (Exponentiate) − < ex − 1 <  (Subtract 1) In summary, given  > 0, let δ = ln(1 + ). Then |x − 0| < δ implies |ex − 1| <  as desired. We have shown that lim ex = 1. x→0 We note that we could actually show that lim ex = ec for any constant x→c c. We do this by factoring out ec from both sides, leaving us to show lim ex−c = 1 instead. By using the substitution u = x − c, this reduces to x→c showing lim eu = 1 which we just did in the last example. As an added u→0 benefit, this shows that in fact the function f (x) = ex is continuous at all values of x, an important concept we will define in Section 1.6. This formal definition of the limit is not an easy concept grasp. Our examples are actually “easy” examples, using “simple” functions like polynomials, square–roots and exponentials. It is very difficult to prove, using sin(x) = 1, as we approximated in the techniques given above, that lim x→0 x the previous section. There is hope. The next section shows how one can evaluate complicated limits using certain basic limits as building blocks. While limits are an incredibly important part of calculus (and hence much of higher mathematics), rarely are limits evaluated using the definition. Rather, the techniques of the following section are employed. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 17 Exercises 1.2 Terms and Concepts Problems In Exercises 5 – 11, prove the given limit using an  − δ proof. 1. What is wrong with the following “definition” of a limit? 5. lim (3 − x) = −2 x→5 “The limit of f (x), as x approaches a, is K” means that given any δ > 0 there exists  > 0 such that whenever |f (x) − K| < , we have |x − a| < δ. 6. lim (x2 − 3) = 6 x→3 7. lim (x2 + x − 5) = 15 x→4 8. lim (x3 − 1) = 7 x→2 2. Which is given first in establishing a limit, the x– tolerance or the y–tolerance? 3. T/F:  must always be positive. 9. lim 5 = 5 x→2 10. lim (e2x − 1) = 0 x→0 11. lim sin x = 0 (Hint: use the fact that | sin x| ≤ |x|, x→0 4. T/F: δ must always be positive. with equality only when x = 0.) Page 18 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.3 1.3 Finding Limits Analytically Finding Limits Analytically In Section 1.1 we explored the concept of the limit without a strict definition, meaning we could only make approximations. In the previous section we gave the definition of the limit and demonstrated how to use it to verify our approximations were correct. Thus far, our method of finding a limit is 1) make a really good approximation either graphically or numerically, and 2) verify our approximation is correct using a -δ proof. This section gives a series of theorems which allow us to find limits much more quickly and intuitively. Suppose that lim f (x) = 2 and x→2 lim g(x) = 3. What is lim (f (x) + g(x))? Intuition tells us that the limit x→2 x→2 should be 5, as we expect limits to behave in a nice way. The following theorem states that already established limits do behave nicely. Theorem 1 Basic Limit Properties Let b, c, L and K be real numbers, let n be a positive integer, and let f and g be functions with the following limits: lim f (x) = L and lim g(x) = K. x→c x→c The following limits hold. 1. Constants: 2. Identity 3. Sums/Differences: 4. Scalar Multiples: 5. Products: 6. Quotients: 7. Powers: 8. Roots: 9. Compositions: lim b = b x→c lim x = c x→c lim (f (x) ± g(x)) = L ± K x→c lim b · f (x) = bL x→c lim f (x) · g(x) = LK x→c lim f (x) x→c g(x) = L , (K 6= 0) K lim f (x)n = Ln p √ n lim n f (x) = L x→c x→c Adjust our previously given limit situation to: lim f (x) = L, lim g(x) = K and g(L) = K. x→c x→L Then lim g(f (x)) = K. x→c Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 19 Chapter 1 Limits We make a note about Property #8: when n is even, L must be greater than 0. If n is odd, then the statement is true for all L. We apply the theorem to an example. Example 10 Let Using basic limit properties lim f (x) = 2, x→2 lim g(x) = 3 and x→2 p(x) = 3x2 − 5x + 7. Find the following limits:  1. lim f (x) + g(x) 3. lim p(x) x→2 x→2 2. lim 5f (x) + g(x) x→2 Solution  2  1. Using the Sum/Difference rule, we know that lim f (x) + g(x) = x→2 2 + 3 = 5. 2. Using the Scalar Multiple and Sum/Difference rules, we find that  lim 5f (x) + g(x)2 = 5 · 2 + 32 = 19. x→2 3. Here we combine the Power, Scalar Multiple, Sum/Difference and Constant Rules. We show quite a few steps, but in general these can be omitted: lim p(x) = lim (3x2 − 5x + 7) x→2 x→2 = lim 3x2 − lim 5x + lim 7 x→2 x→2 x→2 2 =3·2 −5·2+7 =9 Part 3 of the previous example demonstrates how the limit of a quadratic polynomial can be determined using the properties of Theorem 1. Not only that, recognize that lim p(x) = 9 = p(2); x→2 i.e., the limit at 2 was found just by plugging 2 into the function. This holds true for all polynomials, and also for rational functions (which are quotients of polynomials), as stated in the following theorem. Notes: Page 20 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.3 Theorem 2 tions Finding Limits Analytically Limits of Polynomial and Rational Func- Let p(x) and q(x) be polynomials and c a real number. Then: 1. lim p(x) = p(c) x→c 2. lim p(x) x→c q(x) = p(c) , where q(c) 6= 0. q(c) Example 11 Finding a limit of a rational function Using Theorem 2, find 3x2 − 5x + 1 . x→−1 x4 − x2 + 3 lim Solution Using Theorem 2, we can quickly state that 3x2 − 5x + 1 3(−1)2 − 5(−1) + 1 = x→−1 x4 − x2 + 3 (−1)4 − (−1)2 + 3 9 = = 3. 3 lim It was likely frustrating in Section 1.2 to do a lot of work to prove that lim x2 = 4 x→2 as it seemed fairly obvious. The previous theorems state that many functions behave in such an “obvious” fashion, as demonstrated by the rational function in Example 11. Polynomial and rational functions are not the only functions to behave in such a predictable way. The following theorem gives a list of functions whose behavior is particularly “nice” in terms of limits. In the next section, we will give a formal name to these functions that behave “nicely.” Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 21 Chapter 1 Limits Theorem 3 Special Limits Let c be a real number in the domain of the given function and let n be a positive integer. The following limits hold: 1. lim sin x = sin c 4. lim csc x = csc c 7. lim ax = ac (a > 0) 2. lim cos x = cos c 5. lim sec x = sec c 8. lim ln x = ln c 3. lim tan x = tan c 6. lim cot x = cot c √ √ 9. lim n x = n c x→c x→c x→c x→c x→c x→c x→c x→c x→c Example 12 Evaluating limits analytically Evaluate the following limits. 1. lim cos x 4. lim eln x x→π x→1 2 2 2. lim (sec x − tan x) sin x x→0 x 5. lim x→3 3. lim cos x sin x x→π/2 Solution 1. This is a straightforward application of Theorem 3. cos π = −1. lim cos x = x→π 2. We can approach this in at least two ways. First, by directly applying Theorem 3, we have: lim (sec2 x − tan2 x) = sec2 3 − tan2 3. x→3 Using the Pythagorean Theorem, this last expression is 1; therefore lim (sec2 x − tan2 x) = 1. x→3 We can also use the Pythagorean Theorem from the start. lim (sec2 x − tan2 x) = lim 1 = 1, x→3 x→3 using the Constant limit rule. Either way, we find the limit is 1. Notes: Page 22 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.3 Finding Limits Analytically 3. Applying the Product limit rule of Theorem 1 and Theorem 3 gives lim cos x sin x = cos(π/2) sin(π/2) = 0 · 1 = 0. x→π/2 4. Again, we can approach this in two ways. First, we can use the exponential/logarithmic identity that eln x = x and evaluate lim eln x = x→1 lim x = 1. x→1 We can also use the limit Composition Rule of Theorem 1. Using Theorem 3, we have lim ln x = ln 1 = 0 and lim ex = e0 = 1, x→1 x→0 satisfying the conditions of the Composition Rule. Applying this rule, lim eln x = lim ex = e0 = 1. x→1 x→0 Both approaches are valid, giving the same result. 5. We encountered this limit in Section 1.1. Applying our theorems, we attempt to find the limit as “ 0 ” sin x sin 0 → → . x→0 x 0 0 lim This, of course, violates a condition of Theorem 1, as the limit of the denominator is not allowed to be 0. Therefore, we are still unable to evaluate this limit with tools we currently have at hand. The section could have been titled “Using Known Limits to Find Unknown Limits.” By knowing certain limits of functions, we can find limits involving sums, products, powers, etc., of these functions. We further the development of such comparative tools with the Squeeze Theorem, a clever and intuitive way to find the value of some limits. Before stating this theorem formally, suppose we have functions f , g and h where g always takes on values between f and h; that is, for all x in an interval, f (x) ≤ g(x) ≤ h(x). If f and h have the same limit at c, and g is always “squeezed” between them, then g must have the same limit as well. That is what the Squeeze Theorem states. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 23 Chapter 1 Limits Theorem 4 Squeeze Theorem Let f , g and h be functions on an open interval I containing c such that for all x in I, f (x) ≤ g(x) ≤ h(x). If lim f (x) = L = lim h(x), (1, tan θ) x→c x→c then (cos θ, sin θ) lim g(x) = L. x→c . θ (1, 0) Figure 1.19: The unit circle and related triangles. It can take some work to figure out appropriate functions by which to “squeeze” the given function of which you are trying to evaluate a limit. However, that is generally the only place work is necessary; the theorem makes the “evaluating the limit part” very simple. We use the Squeeze Theorem in the following example to finally prove sin x = 1. that lim x→0 x Example 13 Using the Squeeze Theorem Use the Squeeze Theorem to show that sin x = 1. x→0 x lim We begin by considering the unit circle. Each point on Solution the unit circle has coordinates (cos θ, sin θ) for some angle θ as shown in Figure 1.19. Using similar triangles, we can extend the line from the origin through the point to the point (1, tan θ), as shown. (Here we are assuming that 0 ≤ θ ≤ π/2. Later we will show that we can also consider θ ≤ 0.) Figure 1.19 shows three regions have been constructed in the first quadrant, two triangles and a sector of a circle, which are also drawn below. The area of the large triangle is 21 tan θ; the area of the sector is θ/2; the area of the triangle contained inside the sector is 21 sin θ. It is then clear from the diagram that Notes: Page 24 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.3 Finding Limits Analytically tan θ sin θ . θ . θ . θ 1 1 tan θ 2 θ 2 ≥ Multiply all terms by 1 ≥ sin θ 2 2 , giving sin θ 1 θ ≥ ≥ 1. cos θ sin θ Taking reciprocals reverses the inequalities, giving cos θ ≤ sin θ ≤ 1. θ (These inequalities hold for all values of θ near 0, even negative values, since cos(−θ) = cos θ and sin(−θ) = − sin θ.) Now take limits. lim cos θ ≤ lim θ→0 θ→0 sin θ ≤ lim 1 θ→0 θ sin θ ≤1 θ→0 θ cos 0 ≤ lim 1 ≤ lim θ→0 Clearly this means that lim θ→0 sin θ ≤1 θ sin θ = 1. θ Two notes about the previous example are worth mentioning. First, one might be discouraged by this application, thinking “I would never have come up with that on my own. This is too hard!” Don’t be discouraged; within this text we will guide you in your use of the Squeeze Theorem. As one gains mathematical maturity, clever proofs like this are easier and easier to create. Second, this limit tells us more than just that as x approaches 0, sin(x)/x approaches 1. Both x and sin x are approaching 0, but the ratio Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 25 Chapter 1 Limits of x and sin x approaches 1, meaning that they are approaching 0 in essentially the same way. Another way of viewing this is: for small x, the functions y = x and y = sin x are essentially indistinguishable. We include this special limit, along with three others, in the following theorem. Theorem 5 Special Limits sin x =1 x→0 x 3. lim (1 + x) x = e cos x − 1 =0 x→0 x 4. lim 1 1. lim x→0 ex − 1 =1 x→0 x 2. lim A short word on how to interpret the latter three limits. We know that as x goes to 0, cos x goes to 1. So, in the second limit, both the numerator and denominator are approaching 0. However, since the limit is 0, we can interpret this as saying that “cos x is approaching 1 faster than x is approaching 0.” In the third limit, inside the parentheses we have an expression that is approaching 1 (though never equaling 1), and we know that 1 raised to any power is still 1. At the same time, the power is growing toward infinity. What happens to a number near 1 raised to a very large power? In this particular case, the result approaches Euler’s number, e, approximately 2.718. In the fourth limit, we see that as x → 0, ex approaches 1 “just as fast” as x → 0, resulting in a limit of 1. Our final theorem for this section will be motivated by the following example. Example 14 Using algebra to evaluate a limit Evaluate the following limit: x2 − 1 . x→1 x − 1 lim Solution We begin by attempting to apply Theorem 3 and sub- Notes: Page 26 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.3 Finding Limits Analytically stituting 1 for x in the quotient. This gives: x2 − 1 12 − 1 “ 0 ” = = , x→1 x − 1 1−1 0 lim and indeterminate form. We cannot apply the theorem. By graphing the function, as in Figure 1.20, we see that the function seems to be linear, implying that the limit should be easy to evaluate. Recognize that the numerator of our quotient can be factored: (x − 1)(x + 1) x2 − 1 = . x−1 x−1 The function is not defined when x = 1, but for all other x, x2 − 1 (x − 1)(x + 1) (x − 1)(x + 1) = = = x + 1. x−1 x−1 x−1 Clearly lim x + 1 = 2. Recall that when considering limits, we are not x→1 concerned with the value of the function at 1, only the value the function x2 − 1 and x + 1 are the same at all approaches as x approaches 1. Since x−1 points except x = 1, they both approach the same value as x approaches 1. Therefore we can conclude that x2 − 1 lim = 2. x→1 x − 1 x2 − 1 and x−1 y = x + 1 are identical except at x = 1. Since limits describe a value the function is approaching, not the value the function actually attains, the limits of the two functions are always equal. y 3 2 1 . x 1 2 Figure 1.20: Graphing f in Example 14 to understand a limit. The key to the above example is that the functions y = Theorem 6 Point Limits of Functions Equal At All But One Let g(x) = f (x) for all x in an open interval, except possibly at c, and let lim g(x) = L for some real number L. Then x→c lim f (x) = L. x→c Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 27 Chapter 1 Limits The Fundamental Theorem of Algebra tells us that when dealing with g(x) a rational function of the form and directly evaluating the limit f (x) g(x) lim returns “0/0”, then (x − c) is a factor of both g(x) and f (x). x→c f (x) One can then use algebra to factor this term out, cancel, then apply Theorem 6. We demonstrate this once more. Example 15 Evaluating a limit using Theorem 6 x3 − 2x2 − 5x + 6 Evaluate lim 3 . x→3 2x + 3x2 − 32x + 15 We begin by applying Theorem 3 and substituting 3 Solution for x. This returns the familiar indeterminate form of “0/0”. Since the numerator and denominator are each polynomials, we know that (x − 3) is factor of each. Using whatever method is most comfortable to you, factor out (x − 3) from each (using polynomial division, synthetic division, a computer algebra system, etc.). We find that (x − 3)(x2 + x − 2) x3 − 2x2 − 5x + 6 = . 3 2 2x + 3x − 32x + 15 (x − 3)(2x2 + 9x − 5) We can cancel the (x − 3) terms as long as x 6= 3. Using Theorem 6 we conclude: x3 − 2x2 − 5x + 6 (x − 3)(x2 + x − 2) = lim x→3 2x3 + 3x2 − 32x + 15 x→3 (x − 3)(2x2 + 9x − 5) (x2 + x − 2) = lim x→3 (2x2 + 9x − 5) 10 1 = = . 40 4 lim We end this section by revisiting a limit first seen in Section 1.1, a limit of a difference quotient. Let f (x) = −1.5x2 + 11.5x; we approximated the f (1 + h) − f (1) limit lim ≈ 8.5. We formally evaluate this limit in the h→0 h following example. Example 16 Evaluating the limit of a difference quotient f (1 + h) − f (1) . Let f (x) = −1.5x2 + 11.5x; find lim h→0 h Since f is a polynomial, our first attempt should be Solution to employ Theorem 3 and substitute 0 for h. However, we see that this Notes: Page 28 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.3 Finding Limits Analytically gives us “0/0.” Knowing that we have a rational function hints that some algebra will help. Consider the following steps:  −1.5(1 + h)2 + 11.5(1 + h) − −1.5(1)2 + 11.5(1) f (1 + h) − f (1) lim = lim h→0 h→0 h h −1.5(1 + 2h + h2 ) + 11.5 + 11.5h − 10 = lim h→0 h 2 −1.5h + 8.5h = lim h→0 h h(−1.5h + 8.5) = lim h→0 h = lim (−1.5h + 8.5) (using Theorem 6, as h 6= 0) h→0 = 8.5 (using Theorem 3) This matches our previous approximation. This section contains several valuable tools for evaluating limits. One of the main results of this section is Theorem 3; it states that many functions that we use regularly behave in a very nice, predictable way. In the next section we give a name to this nice behavior; we label such functions as continuous. Defining that term will require us to look again at what a limit is and what causes limits to not exist. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 29 Exercises 1.3 Terms and Concepts 1. Explain in your own words, without using -δ formality, why lim b = b. In Exercises 14 – 17, use the following information to evaluate the given limit, when possible. If it is not possible to determine the limit, state why not. • lim f (x) = 2, x→c 2. Explain in your own words, without using -δ formality, why lim x = c. x→c x→1 • lim g(x) = 0, x→1 lim f (x) = 1, f (1) = 1/5 lim g(x) = π, g(10) = π x→10 x→10 14. lim f (x)g(x) x→1 3. What does the text mean when it says that certain functions’ “behavior is ‘nice’ in terms of limits”? What, in particular, is “nice”? 15. lim cos g(x)  x→10 16. lim f (x)g(x) x→1 4. Sketch a graph that visually demonstrates the Squeeze Theorem. 17. lim g 5f (x)  x→1 5. You are given the following information: (a) lim f (x) = 0 In Exercises 18 – 32, evaluate the given limit. 18. lim x2 − 3x + 7 x→1 x→3 (b) lim g(x) = 0 x→1  f (x) (c) lim =2 x→1 g(x) 19. lim What can be said about the relative sizes of f (x) and g(x) as x approaches 1? 20. x→π x−3 x−5 7 lim cos x sin x x→π/4 21. lim ln x x→0 Problems 3 22. lim 4x −8x x→3 In Exercises 6 – 13, use the following information to evaluate the given limit, when possible. If it is not possible to determine the limit, state why not. • lim f (x) = 6, x→9 • lim g(x) = 3, x→9 lim f (x) = 9, x→6 lim g(x) = 3, x→6 f (9) = 6 23. lim csc x x→π/6 24. lim ln(1 + x) x→0 g(6) = 9 x2 + 3x + 5 x→π 5x2 − 2x − 3 25. lim 6. lim (f (x) + g(x)) x→9 26. lim x→π 7. lim (3f (x)/g(x)) x→9  8. lim x→9 f (x) − 2g(x) g(x)  27. lim 3x + 1 1−x x2 − 4x − 12 x→6 x2 − 13x + 42 x2 + 2x x→0 x2 − 2x 28. lim  9. lim x→6 f (x) 3 − g(x)  x2 + 6x − 16 x→2 x2 − 3x + 2 29. lim 10. lim g f (x)  11. lim f g(x)  x→9 x2 − 10x + 16 x→2 x2 − x − 2 30. lim x→6  12. lim g f (f (x)) 31. h i 13. lim f (x)g(x) − f 2 (x) + g 2 (x) 32. x→6 x→6 x2 − 5x − 14 x→−2 x2 + 10x + 16 lim lim x2 + 9x + 8 x→−1 x2 − 6x − 7 Page 30 Custom Made for Math 1174 at Langara College by P. Anhaouy Use the Squeeze Theorem in Exercises 33 – 36, where appropriate, to evaluate the given limit. 33. lim x sin x→0   1 x  34. lim sin x cos x→0 38. lim sin 5x 8x 39. lim ln(1 + x) x 40. lim sin x , where x is measured in degrees, not radians. x x→0 x→0 1 x2  x→0 35. lim f (x), where 3x − 2 ≤ f (x) ≤ x3 . 41. Let f (x) = 0 and g(x) = x→1 (a) Show why lim f (x) = 0. 36. lim f (x), where 6x − 9 ≤ f (x) ≤ x2 . x→2 x→3 Exercises 37 – 41 challenge your understanding of limits but can be evaluated using the knowledge gained in this section. sin 3x x→0 x 37. lim x . x (b) Show why lim g(x) = 1. x→0  (c) Show why lim g f (x) does not exist. x→2 (d) Show why the answer to part (c) does not violate the Composition Rule of Theorem 1. Custom Made for Math 1174 at Langara College by P. Anhaouy Page 31 Chapter 1 Limits 1.4 One Sided Limits In Section 1.1 we explored the three ways in which limits of functions failed to exist: 1. The function approached different values from the left and right, 2. The function grows without bound, and 3. The function oscillates. In this section we explore in depth the concepts behind #1 by introducing the one-sided limit. We begin with formal definitions that are very similar to the definition of the limit given in Section 1.2, but the notation is slightly different and “x 6= c” is replaced with either “x < c” or “x > c.” Definition 2 One Sided Limits Left-Hand Limit Let I be an open interval containing c, and let f be a function defined on I, except possibly at c. The limit of f (x), as x approaches c from the left, is L, or, the left–hand limit of f at c is L, denoted by lim f (x) = L, x→c− means that given any  > 0, there exists δ > 0 such that for all x < c, if |x − c| < δ, then |f (x) − L| < . Right-Hand Limit Let I be an open interval containing c, and let f be a function defined on I, except possibly at c. The limit of f (x), as x approaches c from the right, is L, or, the right–hand limit of f at c is L, denoted by lim f (x) = L, x→c+ means that given any  > 0, there exists δ > 0 such that for all x > c, if |x − c| < δ, then |f (x) − L| < . Practically speaking, when evaluating a left-hand limit, we consider only values of x “to the left of c,” i.e., where x < c. The admittedly imperfect notation x → c− is used to imply that we look at values of x Notes: Page 32 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.4 One Sided Limits to the left of c. The notation has nothing to do with positive or negative values of either x or c. A similar statement holds for evaluating righthand limits; there we consider only values of x to the right of c, i.e., x > c. We can use the theorems from previous sections to help us evaluate these limits; we just restrict our view to one side of c. We practice evaluating left and right-hand limits through a series of examples. Example 17  Let f (x) = the following: Evaluating one sided limits x 0≤x≤1 , as shown in Figure 1.21. Find each of 3−x 11 lim f (x) (c) lim f (x) x→1− x→1 (b) lim f (x) (d) f (1)   x2 x+1 19. f (x) =  −x2 + 2x + 4 x<2 x=2 x>2 lim f (x) (c) lim f (x) x→1+ (a) lim f (x) (c) lim f (x) lim f (x) (d) f (a) x→a x→a+ (c) lim f (x) 4 −2 x→a− lim f (x) x→a−   x+1 1 18. f (x) =  x−1 y (b) lim f (x) x→π −  x→−2 (a) x<π x≥π 17. f (x) = x→−2− . cos x sin x x→2− x→2 (d) f (2) (b) lim f (x) x→2+ a(x − b)2 + c x1 x<0 x≥0 x 0 x 6= 0 x=0 lim f (x) (c) lim f (x) x→0− x→0 (d) f (0) (b) lim f (x) x→0+ (c) lim f (x) x→0 (d) f (0) Review x2 + 5x + 4 . x→−1 x2 − 3x − 4 22. Evaluate the limit: lim  2  x −1 x3 + 1 15. f (x) =  2 x +1 (a) (b) (c) x < −1 −1 ≤ x ≤ 1 x>1 23. Evaluate the limit: lim lim f (x) (e) lim f (x) (f) lim f (x) lim f (x) (g) lim f (x) x→−1− x→−1+ x2 − 16 x→−4 x2 − 4x − 32 . lim f (x) x→1− x2 − 15x + 54 . x→−6 x2 − 6x 24. Evaluate the limit: lim x→1+ x→−1 x→1 (d) f (−1) (h) f (1) x2 − 4.4x + 1.6 . x→0.4 x2 − 0.4x 25. Approximate the limit numerically: lim Page 38 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.5 1.5 Limits Involving Infinity Limits Involving Infinity y 100 In Definition 1 we stated that in the equation lim f (x) = L, both c and L x→c were numbers. In this section we relax that definition a bit by considering situations when it makes sense to let c and/or L be “infinity.” As a motivating example, consider f (x) = 1/x2 , as shown in Figure 1.25. Note how, as x approaches 0, f (x) grows very, very large. It seems appropriate, and descriptive, to state that lim 1 x→0 x2 = ∞. Also note that as x gets very large, f (x) gets very, very small. We could represent this concept with notation such as lim 1 x→∞ x2 50 . x −1 −0.5 0.5 1 Figure 1.25: Graphing f (x) = 1/x2 for values of x near 0. = 0. We explore both types of use of ∞ in turn. Definition 3 Limit of Infinity, ∞ We say lim f (x) = ∞ if for every M > 0 there exists δ > 0 such x→c that for all x 6= c, if |x − c| < δ, then f (x) ≥ M . This is just like the –δ definition from Section 1.2. In that definition, given any (small) value , if we let x get close enough to c (within δ units of c) then f (x) is guaranteed to be within  of f (c). Here, given any (large) value M , if we let x get close enough to c (within δ units of c), then f (x) will be at least as large as M . In other words, if we get close enough to c, then we can make f (x) as large as we want. We can define limits equal to −∞ in a similar way. It is important to note that by saying lim f (x) = ∞ we are implicitly y x→c stating that the limit of f (x), as x approaches c, does not exist. A limit only exists when f (x) approaches an actual numeric value. We use the concept of limits that approach infinity because it is helpful and descriptive. 100 50 Example 21 Evaluating limits involving infinity 1 Find lim as shown in Figure 1.26. x→1 (x − 1)2 . x 0.5 Notes: 1 1.5 2 Figure 1.26: Observing infinite limit as x → 1 in Example 21. Custom Made for Math 1174 at Langara College by P. Anhaouy Page 39 Chapter 1 Limits In Example 4 of Section 1.1, by inspecting values of x Solution close to 1 we concluded that this limit does not exist. That is, it cannot equal any real number. But the limit could be infinite. And in fact, we see that the function does appear to be growing larger and larger, as f (.99) = 104 , f (.999) = 106 , f (.9999) = 108 . A similar thing happens on the√other side of 1. In general, let a “large” value √ M be given. Let δ = 1/ M . If x is within δ of 1, i.e., if |x − 1| < 1/ M , then: 1 |x − 1| < √ M 1 2 (x − 1) < M 1 > M, (x − 1)2 y 50 x −1 . −0.5 1 0.5 which is what we wanted to show. So we may say lim 1/(x − 1)2 = ∞. x→1 Example 22 Evaluating limits involving infinity 1 Find lim , as shown in Figure 1.27. x→0 x −50 1 . x→0 x Figure 1.27: Evaluating lim It is easy to see that the function grows without bound Solution near 0, but it does so in different ways on different sides of 0. Since its 1 behaviour is not consistent, we cannot say that lim = ∞. However, x→0 x we can make a statement about one–sided limits. We can state that 1 1 lim+ = ∞ and lim− = −∞. x→0 x x→0 x Vertical asymptotes If the limit of f (x) as x approaches c from either the left or right (or both) is ∞ or −∞, we say the function has a vertical asymptote at c. Example 23 Finding vertical asymptotes 3x Find the vertical asymptotes of f (x) = 2 . x −4 Vertical asymptotes occur where the function grows withSolution out bound; this can occur at values of c where the denominator is 0. When x is near c, the denominator is small, which in turn can make the function take on large values. In the case of the given function, the denominator is 0 at x = ±2. Substituting in values of x close to 2 and −2 seems to Notes: Page 40 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.5 y indicate that the function tends toward ∞ or −∞ at those points. We can graphically confirm this by looking at Figure 1.28. Thus the vertical asymptotes are at x = ±2. When a rational function has a vertical asymptote at x = c, we can conclude that the denominator is 0 at x = c. However, just because the denominator is 0 at a certain point does not mean there is a vertical asymptote there. For instance, f (x) = (x2 − 1)/(x − 1) does not have a vertical asymptote at x = 1, as shown in Figure 1.29. While the denominator does get small near x = 1, the numerator gets small too, matching the denominator step for step. In fact, factoring the numerator, we get f (x) = (x − 1)(x + 1) . x−1 Limits Involving Infinity 3 2 1 x .−1 1 2 Figure 1.29: Graphically showing x2 − 1 that f (x) = does not have an x−1 asymptote at x = 1. Canceling the common term, we get that f (x) = x + 1 for x 6= 1. So there is clearly no asymptote, rather a hole exists in the graph at x = 1. The above example may seem a little contrived. Another example demonstrating this important concept is f (x) = (sin x)/x. We have considered this function several times in the previous sections. We found that sin x = 1; i.e., there is no vertical asymptote. No simple algebraic lim x→0 x cancellation makes this fact obvious; we used the Squeeze Theorem in Section 1.3 to prove this. If the denominator is 0 at a certain point but the numerator is not, then there will usually be a vertical asymptote at that point. On the other hand, if the numerator and denominator are both zero at that point, then there may or may not be a vertical asymptote at that point. This case where the numerator and denominator are both zero returns us to an important topic. y 10 x −5 Indeterminate Forms −10 We have seen how the limits sin x lim x→0 x 5 and x2 − 1 lim x→1 x − 1 each return the indeterminate form “0/0” when we blindly plug in x = 0 and x = 1, respectively. However, 0/0 is not a valid arithmetical expression. It gives no indication that the respective limits are 1 and 2. . Figure 1.28: 3x . x2 − 4 Graphing f (x) = Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 41 Chapter 1 Limits With a little cleverness, one can come up 0/0 expressions which have a limit of ∞, 0, or any other real number. That is why this expression is called indeterminate. A key concept to understand is that such limits do not really return 0/0. Rather, keep in mind that we are taking limits. What is really happening is that the numerator is shrinking to 0 while the denominator is also shrinking to 0. The respective rates at which they do this are very important and determine the actual value of the limit. Some other common indeterminate forms are ∞ − ∞, ∞ · 0, ∞/∞, 00 , ∞0 and 1∞ . Again, keep in mind that these are the “blind” results of evaluating a limit, and each, in and of itself, has no meaning. The expression ∞ − ∞ does not really mean “subtract infinity from infinity.” Rather, it means “One quantity is subtracted from the other, but both are growing without bound.” What is the result? It is possible to get every value between −∞ and ∞ Note that 1/0 and ∞/0 are not indeterminate forms, though they are not exactly valid mathematical expressions, either. In each, the function is growing without bound, indicating that the limit will be ∞, −∞, or simply not exist if the left- and right-hand limits do not match. Limits at Infinity and Horizontal Asymptotes At the beginning of this section we briefly considered what happens to f (x) = 1/x2 as x grew very large. Graphically, it concerns the behavior of the function to the “far right” of the graph. We make this notion more explicit in the following definition. Definition 4 tote Limits at Infinity and Horizontal Asymp- 1. We say lim f (x) = L if for every  > 0 there exists M > 0 x→∞ such that if x ≥ M , then |f (x) − L| < . 2. We say lim f (x) = L if for every  > 0 there exists M < 0 x→−∞ such that if x ≤ M , then |f (x) − L| < . 3. If lim f (x) = L or x→∞ lim f (x) = L, we say that y = L is a x→−∞ horizontal asymptote of f . Notes: Page 42 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.5 Limits Involving Infinity y Example 24 Approximating horizontal asymptotes x2 Approximate the horizontal asymptote(s) of f (x) = 2 . x +4 1 0.5 We will approximate the horizontal asymptotes by approximating the limits Solution x2 x→−∞ x2 + 4 x2 . x→∞ x2 + 4 and lim x lim −20 −10 . Horizontal asymptotes can take on a variety of forms. Figure 1.31(a) shows that f (x) = x/(x2 + 1) has a horizontal asymptote of y = 0, where 0 is approached from both above and below. √ Figure 1.31(b) shows that f (x) = x/ x2 + 1 has two horizontal asymptotes; one at y = 1 and the other at y = −1. Figure 1.31(c) shows that f (x) = (sin x)/x has even more interesting behavior than at just x = 0; as x approaches ±∞, f (x) approaches 0, but oscillates as it does this. y 0.5 0.5 −10 10 20 −0.5 . 10 100 10000 −10 −100 −10000 0.9615 0.9996 0.999996 0.9615 0.9996 0.999996 (b) 1 0.5 x −20 f (x) y x −20 x Figure 1.30: Using a graph and a table to approximate a horizontal asymptote in Example 24. y 1 −10 10 20 −0.5 −1 . (a) 20 (a) Figure 1.30(a) shows a sketch of f , and part (b) gives values of f (x) for large magnitude values of x. It seems reasonable to conclude from both of these sources that f has a horizontal asymptote at y = 1. Later, we will show how to determine this analytically. 1 10 x −20 −1 . (b) −10 10 20 (c) Figure 1.31: Considering different types of horizontal asymptotes. We can analytically evaluate limits at infinity for rational functions once we understand lim 1/x. As x gets larger and larger, the 1/x gets x→∞ smaller and smaller, approaching 0. We can, in fact, make 1/x as small Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 43 Chapter 1 Limits as we want by choosing a large enough value of x. Given , we can make 1/x <  by choosing x > 1/. Thus we have limx→∞ 1/x = 0. It is now not much of a jump to conclude the following: lim 1 x→∞ xn =0 and lim 1 x→−∞ xn =0 Now suppose we need to compute the following limit: x3 + 2x + 1 . x→∞ 4x3 − 2x2 + 9 lim A good way of approaching this is to divide through the numerator and denominator by x3 (hence dividing by 1), which is the largest power of x to appear in the function. Doing this, we get x3 + 2x + 1 1/x3 x3 + 2x + 1 = lim · 3 2 x→∞ 4x − 2x + 9 x→∞ 1/x3 4x3 − 2x2 + 9 x3 /x3 + 2x/x3 + 1/x3 = lim x→∞ 4x3 /x3 − 2x2 /x3 + 9/x3 1 + 2/x2 + 1/x3 = lim . x→∞ 4 − 2/x + 9/x3 lim Then using the rules for limits (which also hold for limits at infinity), as well as the fact about limits of 1/xn , we see that the limit becomes 1 1+0+0 = . 4−0+0 4 This procedure works for any rational function. In fact, it gives us the following theorem. Notes: Page 44 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.5 Theorem 8 Limits Involving Infinity Limits of Rational Functions at Infinity Let f (x) be a rational function of the following form: f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 , bm xm + bm−1 xm−1 + · · · + b1 x + b0 where any of the coefficients may be 0 except for an and bm . 1. If n = m, then lim f (x) = lim f (x) = x→∞ x→−∞ an . bm 2. If n < m, then lim f (x) = lim f (x) = 0. x→∞ x→−∞ 3. If n > m, then lim f (x) and lim f (x) are both infinite. x→∞ x→−∞ We can see why this is true. If the highest power of x is the same in both the numerator and denominator (i.e. n = m), we will be in a situation like the example above, where we will divide by xn and in the limit all the terms will approach 0 except for an xn /xn and bm xm /xn . Since n = m, this will leave us with the limit an /bm . If n < m, then after dividing through by xm , all the terms in the numerator will approach 0 in the limit, leaving us with 0/bm or 0. If n > m, and we try dividing through by xn , we end up with all the terms in the denominator tending toward 0, while the xn term in the numerator does not approach 0. This is indicative of some sort of infinite limit. Intuitively, as x gets very large, all the terms in the numerator are small in comparison to an xn , and likewise all the terms in the denominator are small compared to bn xm . If n = m, looking only at these two important terms, we have (an xn )/(bn xm ). This reduces to an /bm . If n < m, the function behaves like an /(bm xm−n ), which tends toward 0. If n > m, the function behaves like an xn−m /bm , which will tend to either ∞ or −∞ depending on the values of n, m, an , bm and whether you are looking for lim f (x) or lim f (x). x→∞ x→−∞ With care, we can quickly evaluate limits at infinity for a large number of functions by considering the largest powers of x. For instance, consider x again lim √ , graphed in Figure 1.31(b). When x is very large, x→±∞ x2 + 1 x2 + 1 ≈ x2 . Thus p x2 + 1 ≈ √ x2 = |x|, and √ x x ≈ . |x| x2 + 1 Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 45 Chapter 1 Limits This expression is 1 when x is positive and −1 when x is negative. Hence we get asymptotes of y = 1 and y = −1, respectively. Example 25 Finding a limit of a rational function Confirm analytically that y = 1 is the horizontal asymptote of f (x) = x2 , as approximated in Example 24. x2 + 4 y 0.5 Before using Theorem 8, let’s use the technique of evalSolution uating limits at infinity of rational functions that led to that theorem. The largest power of x in f is 2, so divide the numerator and denominator of f by x2 , then take limits. x −40 −30 −20 −10 −0.5 . x2 /x2 x2 = lim x→∞ x2 /x2 + 4/x2 x→∞ x2 + 4 1 = lim x→∞ 1 + 4/x2 1 = 1+0 = 1. lim (a) y 0.5 x 10 20 30 40 −0.5 . We can also use Theorem 8 directly; in this case n = m so the limit is the ratio of the leading coefficients of the numerator and denominator, i.e., 1/1 = 1. Example 26 Finding limits of rational functions Use Theorem 8 to evaluate each of the following limits. (b) .y 1. x 20 40 x2 + 2x − 1 x→−∞ x3 + 1 lim x2 + 2x − 1 x→∞ 1 − x − 3x2 x2 − 1 x→∞ 3 − x 3. lim 2. lim −20 Solution 1. The highest power of x is in the denominator. Therefore, the limit is 0; see Figure 1.32(a). −40 . (c) Figure 1.32: Visualizing the functions in Example 26. 2. The highest power of x is x2 , which occurs in both the numerator and denominator. The limit is therefore the ratio of the coefficients of x2 , which is −1/3. See Figure 1.32(b). Notes: Page 46 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.5 Limits Involving Infinity 3. The highest power of x is in the numerator so the limit will be ∞ or −∞. To see which, consider only the dominant terms from the numerator and denominator, which are x2 and −x. The expression in the limit will behave like x2 /(−x) = −x for large values of x. Therefore, the limit is −∞. See Figure 1.32(c). Example 27 A company manufacturing tables has a fixed cost of $300 per day and total costs of $5, 100 per day for a daily output of 20 tables. 1. Suppose that the total cost per day C(x) is linearly related to the total output per day x. Find the average cost function, C̄(x) = C(x) . x 2. What does the average cost per table tend to as production increases? Solution 1. Since C(x) is a linear function of x, it can be written in the form C(x) = mx + b. Since the fixed costs are $300, b = 300. Also, C(20) = 5100, so 5100 = m(20) + 300 ⇒ 20m = 4800 ⇒ m = 240. 240x + 300 Therefore, C(x) = 240x + 300, and C̄(x) = . x 300 240x + 300 = 240 + . As x increases, C̄(x) tends to x x $240 per table. 2. C̄(x) = Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 47 Chapter 1 Limits Example 28 Suppose the demand curve is given by p = q is the quantity. 2400 , where p is the price and q+2 1. Find the revenue function, R(q) = p · q. 2. Find lim+ R(q), lim R(q), and lim R(q). q→0 q→∞ q→10 3. How large can R(q) become? Solution 1. R(q) = p · q = 2400q 2400 ·q = . q+2 q+2 2. lim+ R(q) = 0, lim R(q) = q→0 q→10 3. It will be 2400 as q → ∞. Notes: Page 48 Custom Made for Math 1174 at Langara College by P. Anhaouy 2400(10) = 2000 and lim R(q) = 2400. q→∞ 10 + 2 1.5 Limits Involving Infinity Chapter Summary In this chapter we: • defined the limit, • found accessible ways to approximate their values numerically and graphically, • developed a not–so–easy method of proving the value of a limit (-δ proofs), • explored when limits do not exist, • defined continuity and explored properties of continuous functions, and • considered limits that involved infinity. Why? Mathematics is famous for building on itself and calculus proves to be no exception. In the next chapter we will be interested in “dividing by 0.” That is, we will want to divide a quantity by a smaller and smaller number and see what value the quotient approaches. In other words, we will want to find a limit. These limits will enable us to, among other things, determine exactly how fast something is moving when we are only given position information. Later, we will want to add up an infinite list of numbers. We will do so by first adding up a finite list of numbers, then take a limit as the number of things we are adding approaches infinity. Surprisingly, this sum often is finite; that is, we can add up an infinite list of numbers and get, for instance, 42. These are just two quick examples of why we are interested in limits. Many students dislike this topic when they are first introduced to it, but over time an appreciation is often formed based on the scope of its applicability. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 49 Exercises 1.5 Terms and Concepts 1 . (x − 3)(x − 5)2 10. f (x) = (a) 1. T/F: If lim f (x) = ∞, then we are implicitly stating lim f (x) (d) lim f (x) x→5− x→5 x→3− that the limit exists. (b) lim f (x) (e) lim f (x) (c) lim f (x) (f) lim f (x) x→5+ x→3+ 2. T/F: If lim f (x) = 5, then we are implicitly stating x→5 x→3 x→∞ that the limit exists. y 50 3. T/F: If lim f (x) = −∞, then lim f (x) = ∞ x→1− x→1+ x 4 2 4. T/F: If lim f (x) = ∞, then f has a vertical asymptote x→5 at x = 5. −50 6 . 5. T/F: ∞/0 is not an indeterminate form. 11. f (x) = 6. List 5 indeterminate forms. (a) 7. Construct a function with a vertical asymptote at x = 5 and a horizontal asymptote at y = 5. 1 ex + 1 lim f (x) (c) x→−∞ (b) lim f (x) (d) lim f (x) x→∞ x→0+ y 8. Let lim f (x) = ∞. Explain how we know that f is/is 1 x→7 not continuous at x = 7. 0.5 . x −10 Problems −5 5 10 5 10 −0.5 In Exercises 9 – 14, evaluate the given limits using the graph of the function. −1 . 12. f (x) = x2 sin(πx) 9. f (x) = 1 (x + 1)2 (a) lim f (x) (a) (b) lim f (x) x→−∞ x→−1− (b) lim f (x) x→∞ lim f (x) x→−1+ y y 100 100 50 50 x −10 −5 −50 . x −2 −1 .. Page 50 Custom Made for Math 1174 at Langara College by P. Anhaouy −100 lim f (x) x→0− 13. f (x) = cos(x) (a) In Exercises 19 – 24, identify the horizontal and vertical asymptotes, if any, of the given function. lim f (x) 19. f (x) = 2x2 − 2x − 4 x2 + x − 20 20. f (x) = −3x2 − 9x − 6 5x2 − 10x − 15 21. f (x) = x2 + x − 12 7x3 − 14x2 − 21x 22. f (x) = x2 − 9 9x − 9 23. f (x) = x2 − 9 9x + 27 24. f (x) = x2 − 1 −x2 − 1 x→−∞ (b) lim f (x) x→∞ y 1 0.5 x π −4π −3π −2π −π 2π 3π 4π −0.5 −1 . 14. f (x) = 2x + 10 (a) lim f (x) x→−∞ (b) lim f (x) In Exercises 25 – 28, evaluate the given limit. x→∞ x3 + 2x2 + 1 x→∞ x−5 y 25. lim 150 x3 + 2x2 + 1 x→∞ 5−x 100 26. lim 50 . 27. x3 + 2x2 + 1 x→−∞ x2 − 5 28. x3 + 2x2 + 1 x→−∞ 5 − x2 lim x −5 −10 5 lim In Exercises 15 – 18, numerically approximate the following limits: (a) lim f (x) Review x→3− (b) lim f (x) x→3+ 29. Use an  − δ proof to show that lim 5x − 2 = 3. x→1 (c) lim f (x) x→3 30. Let lim f (x) = 3 and lim g(x) = −1. Evaluate the x→2 x2 − 1 15. f (x) = 2 x −x−6 x→2 following limits. (a) lim (f + g)(x) (c) lim (f /g)(x) (b) lim (f g)(x) (d) lim f (x)g(x) x→2 x2 + 5x − 36 16. f (x) = 3 x − 5x2 + 3x + 9 x2 − 11x + 30 17. f (x) = 3 x − 4x2 − 3x + 18 18. f (x) = x2 − 9x + 18 x2 − x − 6 x→2 x→2 x→2 x2 − 1 x<3 . x+5 x≥3 Is f continuous everywhere?  31. Let f (x) = 32. Evaluate the limit: lim ln x. x→e Custom Made for Math 1174 at Langara College by P. Anhaouy Page 51 Chapter 1 Limits 1.6 Continuity As we have studied limits, we have gained the intuition that limits measure “where a function is heading.” That is, if lim f (x) = 3, then as x is close x→1 to 1, f (x) is close to 3. We have seen, though, that this is not necessarily a good indicator of what f (1) actually is. This can be problematic; functions can tend to one value but attain another. This section focuses on functions that do not exhibit such behavior. Definition 5 Continuous Function Let f be a function defined on an open interval I containing c. 1. f is continuous at c if lim f (x) = f (c). x→c 2. f is continuous on I if f is continuous at c for all values of c in I. If f is continuous on (−∞, ∞), we say f is continuous everywhere. A useful way to establish whether or not a function f is continuous at c is to verify the following three things: 1. lim f (x) exists, x→c 2. f (c) is defined, and 3. lim f (x) = f (c). x→c y Example 29 Finding intervals of continuity Let f be defined as shown in Figure 1.33. Give the interval(s) on which f is continuous. 1.5 1 Solution 0.5 We proceed by examining the three criteria for continu- ity. x 1 2 3 1. The limits lim f (x) exists for all c between 0 and 3. x→c . Figure 1.33: A graph of f in Example 29. 2. f (c) is defined for all c between 0 and 3, except for c = 1. We know immediately that f cannot be continuous at x = 1. 3. The limit lim f (x) = f (c) for all c between 0 and 3, except, of course, x→c for c = 1. Notes: Page 52 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.6 Continuity We conclude that f is continuous at every point of (0, 3) except at x = 1. Therefore f is continuous on (0, 1) ∪ (1, 3). Example 30 Finding intervals of continuity The floor function, f (x) = bxc, returns the largest integer smaller than the input x. (For example, f (π) = bπc = 3.) The graph of f in Figure 1.34 demonstrates why this is often called a “step function.” y Give the intervals on which f is continuous. Solution 2 We examine the three criteria for continuity. 1. The limits limx→c f (x) do not exist at the jumps from one “step” to the next, which occur at all integer values of c. Therefore the limits exist for all c except when c is an integer. x −2 2 2. The function is defined for all values of c. 3. The limit lim f (x) = f (c) for all values of c where the limit exist, x→c since each step consists of just a line. We conclude that f is continuous everywhere except at integer values of c. So the intervals on which f is continuous are . −2 Figure 1.34: A graph of the step function in Example 30. . . . , (−2, −1), (−1, 0), (0, 1), (1, 2), . . . . We can extend the definition of continuity to closed intervals by considering the appropriate one-sided limits at the endpoints. Definition 6 Continuity on Closed Intervals Let f be defined on the closed interval [a, b] for some real numbers a, b. f is continuous on [a, b] if: 1. f is continuous on (a, b), 2. lim f (x) = f (a) and x→a+ 3. lim− f (x) = f (b). x→b We can make the appropriate adjustments to talk about continuity on half–open intervals such as [a, b) or (a, b] if necessary. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 53 Chapter 1 Limits Example 31 Determining intervals on which a function is continuous For each of the following functions, give the domain of the function and the interval(s) on which it is continuous. 1. f (x) = 1 x 4. f (x) = 1 − x2 5. f (x) = |x| 2. f (x) = sin x √ 3. f (x) = x Solution √ We examine each in turn. 1 1. The domain of f (x) = is (−∞, 0) ∪ (0, ∞). As it is a rational x function, we apply Theorem 2 to recognize that f is continuous on all of its domain. 2. The domain of f (x) = sin x is all real numbers, or (−∞, ∞). Applying Theorem 3 shows that sin x is continuous everywhere. √ 3. The domain √of f (x) = x is [0, ∞). Applying Theorem 3 shows that f (x) = x is continuous on its domain of [0, ∞). √ 4. The domain of f (x) = 1 − x2 is [−1, 1]. Applying Theorems 1 and 3 shows that f is continuous on all of its domain, [−1, 1]. 5. The domain of f (x) = |x| is (−∞, ∞). We can define the abso−x x < 0 lute value function as f (x) = . Each “piece” of this x x≥0 piecewise defined function is continuous on all of its domain, giving that f is continuous on (−∞, 0) and [0, ∞). We cannot assume this implies that f is continuous on (−∞, ∞); we need to check that lim f (x) = f (0), as x = 0 is the point where f transitions from x→0 one “piece” of its definition to the other. It is easy to verify that this is indeed true, hence we conclude that f (x) = |x| is continuous everywhere. Continuity is inherently tied to the properties of limits. Because of this, the properties of limits found in Theorems 1 and 2 apply to continuity as well. Further, now knowing the definition of continuity we can re– read Theorem 3 as giving a list of functions that are continuous on their domains. The following theorem states how continuous functions can be Notes: Page 54 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.6 Continuity combined to form other continuous functions, followed by a theorem which formally lists functions that we know are continuous on their domains. Theorem 9 Properties of Continuous Functions Let f and g be continuous functions on an interval I, let c be a real number and let n be a positive integer. The following functions are continuous on I. 1. Sums/Differences: f ± g 2. Constant ples: Multi- c·f 3. Products: f ·g 4. Quotients: f /g 5. Powers: fn √ n f 6. Roots: 7. Compositions: Theorem 10 (as long as g 6= 0 on I) (if n is even then f ≥ 0 on I; if n is odd, then true for all values of f on I.) Adjust the definitions of f and g to: Let f be continuous on I, where the range of f on I is J, and let g be continuous on J. Then g ◦ f , i.e., g(f (x)), is continuous on I. Continuous Functions The following functions are continuous on their domains. 1. f (x) = sin x 2. f (x) = cos x 3. f (x) = tan x 4. f (x) = cot x 5. f (x) = sec x 6. f (x) = csc x √ 8. f (x) = n x, 7. f (x) = ln x 9. f (x) = ax (a > 0) (where n is a positive integer) Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 55 Chapter 1 Limits We apply these theorems in the following Example. Example 32 Determining intervals on which a function is continuous State the interval(s) on which each of the following functions is continuous. √ √ 3. f (x) = tan x 1. f (x) = x − 1 + 5 − x √ 4. f (x) = ln x 2. f (x) = x sin x y 3 Solution 2 We examine each in turn, applying Theorems 9 and 10 as appropriate. 1 x . 2 4 Figure 1.35: A graph of f in Example 32(1). 1. The square–root terms are continuous on the intervals [1, ∞) and (−∞, 5], respectively. As f is continuous only where each term is continuous, f is continuous on [1, 5], the intersection of these two intervals. A graph of f is given in Figure 1.35. 2. The functions y = x and y = sin x are each continuous everywhere, hence their product is, too. 3. Theorem 10 states that f (x) = tan x is continuous “on its domain.” Its domain includes all real numbers except odd multiples of π/2. Thus f (x) = tan x is continuous on      π 3π π π 3π π , − , , , ,..., ... − ,− 2 2 2 2 2 2 or, equivalently, on D = {x ∈ R | x 6= n · π2 , n is an odd integer}. √ 4. The domain of y = x is [0, ∞). The range of y = ln x is (−∞, ∞), but if we restrict its domain to [1, ∞) its range is [0, ∞). So restricting y = ln x to√the domain of [1, ∞) restricts its output is√[0, ∞), on which y = x is defined. Thus the domain of f (x) = ln x is [1, ∞). A common way of thinking of a continuous function is that “its graph can be sketched without lifting your pencil.” That is, its graph forms a “continuous” curve, without holes, breaks or jumps. While beyond the scope of this text, this pseudo–definition glosses over some of the finer points of continuity. Very strange functions are continuous that one would be hard pressed to actually sketch by hand. Notes: Page 56 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.6 Continuity This intuitive notion of continuity does help us understand another important concept as follows. Suppose f is defined on [1, 2] and f (1) = −10 and f (2) = 5. If f is continuous on [1, 2] (i.e., its graph can be sketched as a continuous curve from (1, −10) to (2, 5)) then we know intuitively that somewhere on [1, 2] f must be equal to −9, and −8, and −7, −6, . . . , 0, 1/2, etc. In short, f takes on all intermediate values between −10 and 5. It may take on more values; f may actually equal 6 at some time, for instance, but we are guaranteed all values between −10 and 5. While this notion seems intuitive, it is not trivial to prove and its importance is profound. Therefore the concept is stated in the form of a theorem. Theorem 11 Intermediate Value Theorem Let f be a continuous function on [a, b] and, without loss of generality, let f (a) < f (b). Then for every value y, where f (a) < y < f (b), there is a value c in [a, b] such that f (c) = y. One important application of the Intermediate Value Theorem is root finding. Given a function f , we are often interested in finding values of x where f (x) = 0. These roots may be very difficult to find exactly. Good approximations can be found through successive applications of this theorem. Suppose through direct computation we find that f (a) < 0 and f (b) > 0, where a < b. The Intermediate Value Theorem states that there is a c in [a, b] such that f (c) = 0. The theorem does not give us any clue as to where that value is in the interval [a, b], just that it exists. There is a technique that produces a good approximation of c. Let d be the midpoint of the interval [a, b] and consider f (d). There are three possibilities: 1. f (d) = 0 – we got lucky and stumbled on the actual value. We stop as we found a root. 2. f (d) < 0 Then we know there is a root of f on the interval [d, b] – we have halved the size of our interval, hence are closer to a good approximation of the root. 3. f (d) > 0 Then we know there is a root of f on the interval [a, d] – again,we have halved the size of our interval, hence are closer to a good approximation of the root. Successively applying this technique is called the Bisection Method of root finding. We continue until the interval is sufficiently small. We Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 57 Chapter 1 Limits demonstrate this in the following example. y 0.5 x 0.5 Example 33 Using the Bisection Method Approximate the root of f (x) = x − cos x, accurate to three places after the decimal. 1 Consider the graph of f (x) = x − cos x, shown in Figure Solution 1.36. It is clear that the graph crosses the x-axis somewhere near x = 0.8. To start the Bisection Method, pick an interval that contains 0.8. We choose [0.7, 0.9]. Note that all we care about are signs of f (x), not their actual value, so this is all we display. −0.5 −1 . Figure 1.36: Graphing a root of f (x) = x − cos x. Iteration 1: f (0.7) < 0, f (0.9) > 0, and f (0.8) > 0. So replace 0.9 with 0.8 and repeat. Iteration 2: f (0.7) < 0, f (0.8) > 0, and at the midpoint, 0.75, we have f (0.75) > 0. So replace 0.8 with 0.75 and repeat. Note that we don’t need to continue to check the endpoints, just the midpoint. Thus we put the rest of the iterations in Table 1.37. Iteration # 1 2 3 4 5 6 7 8 9 10 11 12 Interval [0.7, 0.9] [0.7, 0.8] [0.7, 0.75] [0.725, 0.75] [0.7375, 0.75] [0.7375, 0.7438] [0.7375, 0.7407] [0.7375, 0.7391] [0.7383, 0.7391] [0.7387, 0.7391] [0.7389, 0.7391] [0.7390, 0.7391] Notice that in the 12th iteration we have the endpoints of the interval Midpoint Sign each starting with 0.739. Thus we have narrowed the zero down to an f (0.8) > 0 accuracy of the first three places after the decimal. Using a computer, we f (0.75) > 0 have f (0.725) < 0 f (0.7390) = −0.00014, f (0.7391) = 0.000024. f (0.7375) < 0 f (0.7438) > 0 f (0.7407) > 0 f (0.7391) > 0 f (0.7383) < 0 f (0.7387) < 0 f (0.7389) < 0 f (0.7390) < 0 Figure 1.37: Iterations of the Bisection Method of Root Finding Either endpoint of the interval gives a good approximation of where f is 0. The Intermediate Value Theorem states that the actual zero is still within this interval. While we do not know its exact value, we know it starts with 0.739. This type of exercise is rarely done by hand. Rather, it is simple to program a computer to run such an algorithm and stop when the endpoints differ by a preset small amount. One of the authors did write such a program and found the zero of f , accurate to 10 places after the decimal, to be 0.7390851332. While it took a few minutes to write the program, it took less than a thousandth of a second for the program to run the necessary 35 iterations. In less than 8 hundredths of a second, the zero was calculated to 100 decimal places (with less than 200 iterations). It is a simple matter to extend the Bisection Method to solve problems similar to “Find x, where f (x) = 0.” For instance, we can find x, where f (x) = 1. It actually works very well to define a new function g where g(x) = f (x) − 1. Then use the Bisection Method to solve g(x) = 0. Similarly, given two functions f and g, we can use the Bisection Method to solve f (x) = g(x). Once again, create a new function h where h(x) = f (x) − g(x) and solve h(x) = 0. Notes: Page 58 Custom Made for Math 1174 at Langara College by P. Anhaouy 1.6 Continuity There is another equation solving method, called Newton’s Method. In many cases, Newton’s Method is much faster. It relies on more advanced mathematics, though, so we will wait before introducing it. This section formally defined what it means to be a continuous function. “Most” functions that we deal with are continuous, so often it feels odd to have to formally define this concept. Regardless, it is important, and forms the basis of the next chapter. To finish this section off, we look at how continuity plays a role in business. Example 34 An income tax schedule shows the following taxes due on these ranges of adjusted gross income for single taxpayers. Income Range $10, 000 ≤ x < $15, 000 $15, 000 ≤ x < $20, 000 $20, 000 ≤ x < $25, 000 Tax $500 + 5% of amount over $10, 000 $800 + 6% of amount over $15, 000 $1200 + 7% of amount over $20, 000 If T (x) denotes the tax due on income level x, is T a continues function for all x? Why or why not? The T (x) function is defined by Solution T (x) =   500 + 0.05(x − 10, 000), 10, 000 ≤ x < 15, 000, 800 + 0.06(x − 15, 000), 15, 000 ≤ x < 20, 000,  1200 + 0.07(x − 20, 000), 20, 000 ≤ x < 25, 000. From here we can see that T may be or may be not continuous at x = 15, 000 and 20, 000. Then, lim T (x) = lim T (x) = x→15,000− x→15,000+ lim [500 + 0.05(15, 000 − 10, 000)] = 750. lim [800 + 0.06(15, 000 − 15, 000)] = 800. x→15,000− x→15,000+ Since T (15, 000) = 800, we can see that lim x→15,000− T (x) 6= lim x→15,000+ T (x) = f (15, 000). So, T (x) is not continuous at x = 15, 000. A similar calculation shows that T (x) is not continuous at x = 20, 000 either. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 59 Chapter 1 Limits Example 35 The supply S(p) of an electronic razors that manufacturers will supply weekly at price p is given by the function  1   p, 0 ≤ p < 20,    2 2 S(p) = p − 10, 20 ≤ p < 40,  3     4 p − 110 , 40 ≤ p. 3 3 For what values of p is this function discontinuous? S(p) is continuous at p = 40, but not at p = 20. Why Solution not? See below.   4 110 50 lim+ S(p) = lim+ p− = . 3 3 3 p→40 p→40   2 50 p − 10 = . lim S(p) = lim 3 3 p→40− p→40− and S(40) = 50 10 as well. But, for p = 20, we have S(20) = and 3 3 1 lim− S(p) = lim− p = 10. p→20 p→20 2   2 10 lim S(p) = lim+ p − 10 = . 3 3 p→20+ p→20 Example 36 Suppose that we invest $5000 in a saving account for 10 years (120 months), with an annual interest rate of r, compounded monthly. The amount of r money in the account after 10 years is given by A(r) = 5000(1 + )120 . 12 Show that there is an interest rate between 0% and 8% that allows the money to grow to the amount of $7000 in this ten years period. A is a continuous function of r on [0, .08] because it Solution is polynomial function of degree 120. Since A(0) = 5000 and A(0.08) ≈ 11098.2, (and 7000 is an intermediate value between these two numbers) the Intermediate Value Theorem guarantees a value of r between 0% and 8% where A(r) = 7000. Notes: Page 60 Custom Made for Math 1174 at Langara College by P. Anhaouy Exercises 1.6 Terms and Concepts y 2 1.5 1. In your own words, describe what it means for a function to be continuous. 1 0.5 2. In your own words, describe what the Intermediate Value Theorem states. 3. What is a “root” of a function? x . 0.5 1 1.5 2 0.5 1 1.5 2 0.5 1 1.5 2 0.5 1 1.5 2 13. a = 1 y 4. Given functions f and g on an interval I, how can the Bisection Method be used to find a value c where f (c) = g(c)? 2 1.5 1 5. T/F: If f is defined on an open interval containing c, and lim f (x) exists, then f is continuous at c. 0.5 x→c x . 6. T/F: If f is continuous at c, then lim f (x) exists. x→c 14. a = 0 y 7. T/F: If f is continuous at c, then lim f (x) = f (c). 2 x→c+ 1.5 8. T/F: If f is continuous on [a, b], then lim f (x) = x→a− f (a). 1 0.5 9. T/F: If f is continuous on [0, 1) and [1, 2), then f is continuous on [0, 2). 10. T/F: The sum of continuous functions is also continuous. x . 15. a = 1 y 2 1.5 Problems 1 In Exercises 11 – 17, a graph of a function f is given along with a value a. Determine if f is continuous at a; if it is not, state why it is not. 0.5 . x 16. a = 4 11. a = 1 y y 4 2 2 1.5 x 1 −4 −3 −2 −1 . x 0.5 1 1.5 3 4 −4 2 17. 12. a = 1 2 −2 0.5 . 1 (a) a = −2 (b) a = 0 Custom Made for Math 1174 at Langara College by P. Anhaouy Page 61 (c) a = 2 In Exercises 18 – 21, determine if f is continuous at the indicated values. If not, explain why. 1 sin x x ( 18. f (x) = x=0 x>0 (a) x = 0 (b) x = π x3 − x x−2  19. f (x) = x<1 x≥1 (a) x = 0 (b) x = 1  2  x + 5x + 4 2 20. f (x) = +2  x + 3x 3 x 6= −1 x = −1 (a) x = −1 (b) x = 10 21. f (x) =   x2 − 64 x2 − 11x + 24 5  x 6= 8 x=8 (a) x = 0 (b) x = 8 In Exercises 22 – 32, give the intervals on which the given function is continuous. 22. f (x) = x2 − 3x + 9 23. g(x) = p x2 − 4 24. h(k) = √ √ 1−k+ k+1 25. f (t) = p 26. g(t) = √ 27. g(x) = 5t2 − 30 1 1 − t2 1 1 + x2 28. f (x) = ex 29. g(s) = ln s 30. h(t) = cos t y 4 31. f (k) = p 1 − ek 2 x −4 −3 −2 −1 −2 . −4 1 2 3 32. f (x) = sin(ex + x2 ) 4 33. Let f be continuous on [1, 5] where f (1) = −2 and f (5) = −10. Does a value 1 < c < 5 exist such that f (c) = −9? Why/why not? Page 62 Custom Made for Math 1174 at Langara College by P. Anhaouy 34. Let g be continuous on [−3, 7] where g(0) = 0 and g(2) = 25. Does a value −3 < c < 7 exist such that g(c) = 15? Why/why not? 35. Let f be continuous on [−1, 1] where f (−1) = −10 and f (1) = 10. Does a value −1 < c < 1 exist such that f (c) = 11? Why/why not? 36. Let h be a function on [−1, 1] where h(−1) = −10 and h(1) = 10. Does a value −1 < c < 1 exist such that h(c) = 0? Why/why not? In Exercises 37 – 40, use the Bisection Method to approximate, accurate to two decimal places, the value of the root of the given function in the given interval. 37. f (x) = x2 + 2x − 4 on [1, 1.5]. 40. f (x) = cos x − sin x on [0.7, 0.8]. Review  41. Let f (x) = (a) x2 − 5 5x lim f (x) x→5− (b) lim f (x) x→5+ x<5 . x≥5 (c) lim f (x) x→5 (d) f (5) 42. Numerically approximate the following limits: (a) x2 − 8.2x − 7.2 2 x→−4/5+ x + 5.8x + 4 (b) x2 − 8.2x − 7.2 2 x→−4/5− x + 5.8x + 4 lim lim 38. f (x) = sin x − 1/2 on [0.5, 0.55] 43. Give an example of function f (x) for which lim f (x) 39. f (x) = ex − 2 on [0.65, 0.7]. x→0 does not exist. Custom Made for Math 1174 at Langara College by P. Anhaouy Page 63 2: Derivatives The previous chapter introduced the most fundamental of calculus topics: the limit. This chapter introduces the second most fundamental of calculus topics: the derivative. Limits describe where a function is going; derivatives describe how fast the function is going. 2.1 Introduction to Derivative A common amusement park ride lifts riders to a height then allows them to freefall a certain distance before safely stopping them. Suppose such a ride drops riders from a height of 150 feet. Students of physics may recall that the height (in feet) of the riders, t seconds after freefall (and ignoring air resistance, etc.) can be accurately modeled by f (t) = −16t2 + 150. Using this formula, it is easy to verify √ that, without intervention, the riders will hit the ground at t = 2.5 1.5 ≈ 3.06 seconds. Suppose the designers of the ride decide to begin slowing the riders’ fall after 2 seconds (corresponding to a height of 86 ft.). How fast will the riders be traveling at that time? We have been given a position function, but what we want to compute is a velocity at a specific point in time, i.e., we want an instantaneous velocity. We do not currently know how to calculate this. However, we do know from common experience how to calculate an average velocity. (If we travel 60 miles in 2 hours, we know we had an average velocity of 30 mph.) We looked at this concept in Section 1.1 when we introduced the difference quotient. We have “ rise ” change in distance = = average velocity. change in time run We can approximate the instantaneous velocity at t = 2 by considering the average velocity over some time period containing t = 2. If we make the time interval small, we will get a good approximation. (This fact is commonly used. For instance, high speed cameras are used to track fast moving objects. Distances are measured over a fixed number of frames to generate an accurate approximation of the velocity.) Consider the interval from t = 2 to t = 3 (just before the riders hit the ground). On that interval, the average velocity is f (3) − f (2) f (3) − f (2) = = −80 ft/s, 3−2 1 Chapter 2 Derivatives where the minus sign indicates that the riders are moving down. By narrowing the interval we consider, we will likely get a better approximation of the instantaneous velocity. On [2, 2.5] we have f (2.5) − f (2) f (2.5) − f (2) = = −72 ft/s. 2.5 − 2 0.5 We can do this for smaller and smaller intervals of time. For instance, over a time span of 1/10th of a second, i.e., on [2, 2.1], we have f (2.1) − f (2) f (2.1) − f (2) = = −65.6 ft/s. 2.1 − 2 0.1 Over a time span of 1/100th of a second, on [2, 2.01], the average velocity is f (2.01) − f (2) f (2.01) − f (2) = = −64.16 ft/s. 2.01 − 2 0.01 What we are really computing is the average velocity on the interval [2, 2 + h] for small values of h. That is, we are computing f (2 + h) − f (2) h h Average Velocity ft/s 1 0.5 0.1 0.01 0.001 −80 −72 −65.6 −64.16 −64.016 Figure 2.1: Approximating the instantaneous velocity with average velocities over a small time period h. where h is small. What we really want is for h = 0, but this, of course, returns the familiar “0/0” indeterminate form. So we employ a limit, as we did in Section 1.1. We can approximate the value of this limit numerically with small values of h as seen in Figure 2.1. It looks as though the velocity is approaching −64 ft/s. Computing the limit directly gives f (2 + h) − f (2) −16(2 + h)2 + 150 − (−16(2)2 + 150) = lim h→0 h→0 h h −64h − 16h2 = lim h→0 h = lim −64 − 16h lim h→0 = −64. Graphically, we can view the average velocities we computed numerically as the slopes of secant lines on the graph of f going through the points (2, f (2)) and (2 + h, f (2 + h)). In Figure 2.2, the secant line corresponding to h = 1 is shown in three contexts. Figure 2.2(a) shows a “zoomed out” version of f with its secant line. In (b), we zoom in around Notes: Page 66 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.1 Introduction to Derivative the points of intersection between f and the secant line. Notice how well this secant line approximates f between those two points – it is a common practice to approximate functions with straight lines. As h → 0, these secant lines approach the tangent line, a line that goes through the point (2, f (2)) with the special slope of −64. In parts (c) and (d) of Figure 2.2, we zoom in around the point (2, 86). In (c) we see the secant line, which approximates f well, but not as well the tangent line shown in (d). y y 150 100 100 50 50 x 1 2 3 . −50 . (a) 2.5 3 (b) . y y 100 100 50 50 . 1.5 x 2 x 2 (c) 2.5 . 1.5 x 2 2.5 (d) Figure 2.2: Parts (a), (b) and (c) show the secant line to f (x) with h = 1, zoomed in different amounts. Part (d) shows the tangent line to f at x = 2. We have just introduced a number of important concepts that we will flesh out more within this section. First, we formally define two of them. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 67 Chapter 2 Derivatives Definition 7 Derivative at a Point Let f be a continuous function on an open interval I and let c be in I. The derivative of f at c, denoted f 0 (c), is lim h→0 f (c + h) − f (c) , h provided the limit exists. If the limit exists, we say that f is differentiable at c; if the limit does not exist, then f is not differentiable at c. If f is differentiable at every point in I, then f is differentiable on I. Definition 8 Tangent Line Let f be continuous on an open interval I and differentiable at c, for some c in I. The line with equation `(x) = f 0 (c)(x − c) + f (c) is the tangent line to the graph of f at c; that is, it is the line through (c, f (c)) whose slope is the derivative of f at c. Some examples will help us understand these definitions. Example 37 Finding derivatives and tangent lines Let f (x) = 3x2 + 5x − 7. Find: 1. f 0 (1) 3. f 0 (3) 2. The equation of the tangent line to the graph of f at x = 1. 4. The equation of the tangent line to the graph f at x = 3. Solution Notes: Page 68 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.1 Introduction to Derivative 1. We compute this directly using Definition 7. f (1 + h) − f (1) h 3(1 + h)2 + 5(1 + h) − 7 − (3(1)2 + 5(1) − 7) = lim h→0 h 3h2 + 11h = lim h→0 h = lim 3h + 11 = 11. f 0 (1) = lim h→0 y 60 h→0 40 2. The tangent line at x = 1 has slope f 0 (1) and goes through the point (1, f (1)) = (1, 1). Thus the tangent line has equation, in point-slope form, y = 11(x−1)+1. In slope-intercept form we have y = 11x−10. 20 x . 3. Again, using the definition, f (3 + h) − f (3) h 3(3 + h)2 + 5(3 + h) − 7 − (3(3)2 + 5(3) − 7) = lim h→0 h 3h2 + 23h = lim h→0 h = lim 3h + 23 1 2 3 4 Figure 2.3: A graph of f (x) = 3x2 + 5x − 7 and its tangent lines at x = 1 and x = 3. f 0 (3) = lim h→0 h→0 = 23. 4. The tangent line at x = 3 has slope 23 and goes through the point (3, f (3)) = (3, 35). Thus the tangent line has equation y = 23(x − 3) + 35 = 23x − 34. A graph of f is given in Figure 2.3 along with the tangent lines at x = 1 and x = 3. Another important line that can be created using information from the derivative is the normal line. It is perpendicular to the tangent line, hence its slope is the opposite–reciprocal of the tangent line’s slope. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 69 Chapter 2 Derivatives y Definition 9 3 Normal Line Let f be continuous on an open interval I and differentiable at c, for some c in I. The normal line to the graph of f at c is the line with equation −1 n(x) = 0 (x − c) + f (c), f (c) 2 1 . x 1 2 3 4 Figure 2.4: A graph of f (x) = 3x2 + 5x − 7, along with its normal line at x = 1. where f 0 (c) 6= 0. When f 0 (c) = 0, the normal line is the vertical  line through c, f (c) ; that is, x = c. Example 38 Finding equations of normal lines Let f (x) = 3x2 + 5x − 7, as in Example 37. Find the equations of the normal lines to the graph of f at x = 1 and x = 3. In Example 37, we found that f 0 (1) = 11. Hence at x = 1, the normal line will have slope −1/11. An equation for the normal line is −1 (x − 1) + 1. n(x) = 11 Solution . The normal line is plotted with y = f (x) in Figure 2.4. Note how the line looks perpendicular to f . (A key word here is “looks.” Mathematically, we say that the normal line is perpendicular to f at x = 1 as the slope of the normal line is the opposite–reciprocal of the slope of the tangent line. However, normal lines may not always look perpendicular. The aspect ratio of the picture of the graph plays a big role in this.) We also found that f 0 (3) = 23, so the normal line to the graph of f at x = 3 will have slope −1/23. An equation for the normal line is n(x) = −1 (x − 3) + 35. 23 Linear functions are easy to work with; many functions that arise in the course of solving real problems are not easy to work with. A common practice in mathematical problem solving is to approximate difficult functions with not–so–difficult functions. Lines are a common choice. It turns out that at any given point on the graph of a differentiable function f , the best linear approximation to f is its tangent line. That is one reason we’ll spend considerable time finding tangent lines to functions. One type of function that does not benefit from a tangent–line approximation is a line; it is rather simple to recognize that the tangent line to Notes: Page 70 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.1 Introduction to Derivative a line is the line itself. We look at this in the following example. Example 39 Finding the Derivative of a Line Consider f (x) = 3x + 5. Find the equation of the tangent line to f at x = 1 and x = 7. Solution We find the slope of the tangent line by using Definition 7. f (1 + h) − f (1) h 3(1 + h) + 5 − (3 + 5) = lim h→0 h 3h = lim h→0 h = lim 3 f 0 (1) = lim h→0 y h→0 1 = 3. 0.5 0 We just found that f (1) = 3. That is, we found the instantaneous rate of change of f (x) = 3x + 5 is 3. This is not surprising; lines are characterized by being the only functions with a constant rate of change. That rate of change is called the slope of the line. Since their rates of change are constant, their instantaneous rates of change are always the same; they are all the slope. So given a line f (x) = ax + b, the derivative at any point x will be a; that is, f 0 (x) = a. It is now easy to see that the tangent line to the graph of f at x = 1 is just f , with the same being true for x = 7. x −π − π2 π 2 π −0.5 . −1 Figure 2.5: f (x) = sin x graphed with an approximation to its tangent line at x = 0. We often desire to find the tangent line to the graph of a function without knowing the actual derivative of the function. In these cases, the best we may be able to do is approximate the tangent line. We demonstrate this in the next example. Example 40 Numerical Approximation of the Tangent Line Approximate the equation of the tangent line to the graph of f (x) = sin x at x = 0. In order to find the equation of the tangent line, we Solution need a slope and a point. The point is given to us: (0, sin 0) = (0, 0). To compute the slope, we need the derivative. This is where we will make an Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 71 Chapter 2 Derivatives approximation. Recall that f 0 (0) ≈ sin(0 + h) − sin 0 h for a small value of h. We choose (somewhat arbitrarily) to let h = 0.1. Thus sin(0.1) − sin 0 f 0 (0) ≈ ≈ 0.9983. 0.1 Thus our approximation of the equation of the tangent line is y = 0.9983(x− 0) + 0 = 0.9983x; it is graphed in Figure 2.5. The graph seems to imply the approximation is rather good. sin x = 1, meaning for values of x x near 0, sin x ≈ x. Since the slope of the line y = x is 1 at x = 0, it should seem reasonable that “the slope of f (x) = sin x” is near 1 at x = 0. In fact, since we approximated the value of the slope to be 0.9983, we might guess the actual value is 1. We’ll come back to this later. Recall from Section 1.3 that lim x→0 Consider again Example 37. To find the derivative of f at x = 1, we needed to evaluate a limit. To find the derivative of f at x = 3, we needed to again evaluate a limit. We have this process: input specific number c do something to f and c return number f 0 (c) This process describes a function; given one input (the value of c), we return exactly one output (the value of f 0 (c)). The “do something” box is where the tedious work (taking limits) of this function occurs. Instead of applying this function repeatedly for different values of c, let us apply it just once to the variable x. We then take a limit just once. The process now looks like: input variable x do something to f and x return function f 0 (x) The output is the “derivative function,” f 0 (x). The f 0 (x) function will take a number c as input and return the derivative of f at c. This calls for a definition. Notes: Page 72 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.1 Definition 10 Introduction to Derivative Derivative Function Let f be a differentiable function on an open interval I. The function f (x + h) − f (x) f 0 (x) = lim h→0 h is the derivative of f . Notation: Let y = f (x). The following notations all represent the derivative: f 0 (x) = y 0 = dy df d d = = (f ) = (y). dx dx dx dx dy Important: The notation is one symbol; it is not the fraction “dy/dx”. dx The notation, while somewhat confusing at first, was chosen with care. A fraction–looking symbol was chosen because the derivative has many fraction–like properties. Among other places, we see these properties at work when we talk about the units of the derivative, when we discuss the Chain Rule, and when we learn about integration (topics that appear in later sections and chapters). Examples will help us understand this definition. Example 41 Finding the derivative of a function Let f (x) = 3x2 + 5x − 7 as in Example 37. Find f 0 (x). Solution We apply Definition 10. f (x + h) − f (x) h 3(x + h)2 + 5(x + h) − 7 − (3x2 + 5x − 7) = lim h→0 h 3h2 + 6xh + 5h = lim h→0 h = lim 3h + 6x + 5 f 0 (x) = lim h→0 h→0 = 6x + 5 So f 0 (x) = 6x+5. Recall earlier we found that f 0 (1) = 11 and f 0 (3) = 23. Note our new computation of f 0 (x) affirm these facts. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 73 Chapter 2 Derivatives Example 42 Finding the derivative of a function 1 Let f (x) = . Find f 0 (x). x+1 We apply Definition 10. Solution f 0 (x) = lim h→0 = lim h→0 f (x + h) − f (x) h 1 1 x+h+1 − x+1 h Now find common denominator then subtract; pull 1/h out front to facilitate reading.   x+1 x+h+1 1 · − h→0 h (x + 1)(x + h + 1) (x + 1)(x + h + 1)   1 x + 1 − (x + h + 1) = lim · h→0 h (x + 1)(x + h + 1)   −h 1 = lim · h→0 h (x + 1)(x + h + 1) −1 = lim h→0 (x + 1)(x + h + 1) −1 = (x + 1)(x + 1) −1 = (x + 1)2 = lim So f 0 (x) = state −1 . To practice using our notation, we could also (x + 1)2   d 1 −1 = . dx x + 1 (x + 1)2 Notes: Page 74 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.1 Introduction to Derivative Example 43 Finding the derivative of a piecewise defined function Find the derivative of the absolute value function,  −x x < 0 f (x) = |x| = . x x≥0 y 1 See Figure 2.6. f (x + h) − f (x) . As f is piecewise– h→0 h defined, we need to consider separately the limits when x < 0 and when x > 0. Solution 0.5 We need to evaluate lim x −1 . When x < 0: −0.5 0.5 1 Figure 2.6: The absolute value function, f (x) = |x|. Notice how the slope of the lines (and hence the tangent lines) abruptly changes at x = 0.  −(x + h) − (−x) d − x = lim h→0 dx h −h = lim h→0 h = lim −1 h→0 = −1. When x > 0, a similar computation shows that d  x = 1. dx We need to also find the derivative at x = 0. By the definition of the derivative at a point, we have f 0 (0) = lim h→0 f (0 + h) − f (0) . h Since x = 0 is the point where our function’s definition switches from one piece to other, we need to consider left and right-hand limits. Consider the following, where we compute the left and right hand limits side by side. lim h→0− f (0 + h) − f (0) = h −h − 0 lim = h h→0− lim− −1 = −1 h→0 lim h→0+ f (0 + h) − f (0) = h h−0 lim = h h→0+ lim+ 1 = 1 h→0 The last lines of each column tell the story: the left and right hand limits are not equal. Therefore the limit does not exist at 0, and f is not y 1 x −1 . −0.5 0.5 1 −1 Figure 2.7: A graph of the derivative of f (x) = |x|. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 75 Chapter 2 Derivatives differentiable at 0. So we have f 0 (x) =  −1 1 x<0 . x>0 At x = 0, f 0 (x) does not exist; there is a jump discontinuity at 0; see Figure 2.7. So f (x) = |x| is differentiable everywhere except at 0. This section defined the derivative; in some sense, it answers the question of “What is the derivative?” The next section addresses the question “What does the derivative mean?” Notes: Page 76 Custom Made for Math 1174 at Langara College by P. Anhaouy Exercises 2.1 Terms and Concepts 17. f (x) = 3x2 − x + 4, at x = −1. 1. T/F: Let f be a position function. The average rate of change on [a, b] is the slope of the line through the points (a, f (a)) and (b, f (b)). 18. r(x) = 1 , at x = −2. x 19. r(x) = 1 , at x = 3. x−2 2. T/F: The definition of the derivative of a function at a point involves taking a limit. 3. In your own words, explain the difference between the average rate of change and instantaneous rate of change. 4. In your own words, explain the difference between Definitions 7 and 10. 5. Let y = f (x). Give three different notations equivalent to “f 0 (x).” In Exercises 20 – 23, a function f and an x–value a are given. Approximate the equation of the tangent line to the graph of f at x = a by numerically approximating f 0 (a), using h = 0.1. 20. f (x) = x2 + 2x + 1, x = 3 21. f (x) = 10 ,x=9 x+1 22. The graph of f (x) = x2 − 1 is shown. Problems In Exercises 6 – 12, use the definition of the derivative to compute the derivative of the given function. (a) Use the graph to approximate the slope of the tangent line to f at the following points: (−1, 0), (0, −1) and (2, 3). (b) Using the definition, find f 0 (x). (c) Find the slope of the tangent line at the points (−1, 0), (0, −1) and (2, 3). 6. f (x) = 6 7. f (x) = 2x y 3 8. f (t) = 4 − 3t 2 9. g(x) = x2 1 2 10. f (x) = 3x − x + 4 x −1 −2 11. r(x) = 1 x 1 −1 . 23. The graph of f (x) = 1 12. r(s) = s−2 In Exercises 13 – 19, a function and an x–value c are given. (Note: these functions are the same as those given in Exercises 6 through 12.) (a) Find the tangent line to the graph of the function at c. (b) Find the normal line to the graph of the function at c. 2 1 is shown. x+1 (a) Use the graph to approximate the slope of the tangent line to f at the following points: (0, 1) and (1, 0.5). (b) Using the definition, find f 0 (x). (c) Find the slope of the tangent line at the points (0, 1) and (1, 0.5). y 5 13. f (x) = 6, at x = −2. 4 3 14. f (x) = 2x, at x = 3. 2 15. f (x) = 4 − 3x, at x = 7. 16. g(x) = x2 , at x = 2. 1 . −1 x 1 2 3 Custom Made for Math 1174 at Langara College by P. Anhaouy Page 77 In Exercises 24 – 27, a graph of a function f (x) is given. Using the graph, sketch f 0 (x). Review 29. Approximate lim x2 + 2x − 35 x→5 x2 − 10.5x + 27.5 y . 3 30. Use the Bisection Method to approximate, accurate to two decimal places, the root of g(x) = x3 + x2 + x − 1 on [0.5, 0.6]. 2 24. 1 x −1 −2 1 2 31. Use the graph of f (x) provided to answer the following. 4 3 −1 . (a) (b) y (c) lim f (x) =? (d) Where is f continuous? x→−3+ y 2 3 25. x −6 −4 −2 2 2 1 −2 . x −4 . y 5 26. x −2 −1 1 2 −5 . y 1 0.5 27. x −2π π −π 2π . −0.5 . −1 . 28. Using the graph of g(x) below, answer the following questions. (a) Where is g(x) > 0? (c) Where is g 0 (x) < 0? (b) Where is g(x) < 0? (d) Where is g 0 (x) > 0? (c) Where is g(x) = 0? (e) Where is g 0 (x) = 0? y 5 x −2 −1 1 lim f (x) =? lim f (x) =? x→−3− 2 −5 . Page 78 Custom Made for Math 1174 at Langara College by P. Anhaouy −2 −1 x→−3 2.2 2.2 Interpretations of the Derivative Interpretations of the Derivative The previous section defined the derivative of a function and gave examples of how to compute it using its definition (i.e., using limits). The section also started with a brief motivation for this definition, that is, finding the instantaneous velocity of a falling object given its position function. The next section will give us more accessible tools for computing the derivative, tools that are easier to use than repeated use of limits. This section falls in between the “What is the definition of the derivative?” and “How do I compute the derivative?” sections. Here we are concerned with “What does the derivative mean?”, or perhaps, when read with the right emphasis, “What is the derivative?” We offer two interconnected interpretations of the derivative, hopefully explaining why we care about it and why it is worthy of study. Interpretation of the Derivative #1: Instantaneous Rate of Change The previous section started with an example of using the position of an object (in this case, a falling amusement–park rider) to find the object’s velocity. This type of example is often used when introducing the derivative because we tend to readily recognize that velocity is the instantaneous rate of change of position. In general, if f is a function of x, then f 0 (x) measures the instantaneous rate of change of f with respect to x. Put another way, the derivative answers “When x changes, at what rate does f change?” Thinking back to the amusement–park ride, we asked “When time changed, at what rate did the height change?” and found the answer to be “By −64 feet per second.” Now imagine driving a car and looking at the speedometer, which reads “60 mph.” Five minutes later, you wonder how far you have traveled. Certainly, lots of things could have happened in those 5 minutes; you could have intentionally sped up significantly, you might have come to a complete stop, you might have slowed to 20 mph as you passed through construction. But suppose that you know, as the driver, none of these things happened. You know you maintained a fairly consistent speed over those 5 minutes. What is a good approximation of the distance traveled? One could argue the only good approximation, given the information provided, would be based on “distance = rate × time.” In this case, we assume a constant rate of 60 mph with a time of 5/60 hours. Hence we would approximate the distance traveled as 5 miles. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 79 Chapter 2 Derivatives Referring back to the falling amusement–park ride, knowing that at t = 2 the velocity was −64 ft/s, we could reasonably assume that 1 second later the riders’ height would have dropped by about 64 feet. Knowing that the riders were accelerating as they fell would inform us that this is an under–approximation. If all we knew was that f (2) = 86 and f 0 (2) = −64, we’d know that we’d have to stop the riders quickly otherwise they would hit the ground! Units of the Derivative It is useful to recognize the units of the derivative function. If y is a function of x, i.e., y = f (x) for some function f , and y is measured in feet and x in seconds, then the units of y 0 = f 0 are “feet per second,” commonly written as “ft/s.” In general, if y is measured in units P and x is measured in units Q, then y 0 will be measured in units “P per Q”, or “P/Q.” Here we see the fraction–like behavior of the derivative in the notation: the units of dy dx are units of y . units of x Example 44 The meaning of the derivative: World Population Let P (t) represent the world population t minutes after 12:00 a.m., January 1, 2012. It is fairly accurate to say that P (0) = 7, 028, 734, 178 (www.prb.org). It is also fairly accurate to state that P 0 (0) = 156; that is, at midnight on January 1, 2012, the population of the world was growing by about 156 people per minute (note the units). Twenty days later (or, 28,800 minutes later) we could reasonably assume the population grew by about 28, 800 · 156 = 4, 492, 800 people. Example 45 The meaning of the derivative: Manufacturing The term widget is an economic term for a generic unit of manufacturing output. Suppose a company produces widgets and knows that the market supports a price of $10 per widget. Let P (n) give the profit, in dollars, earned by manufacturing and selling n widgets. The company likely cannot make a (positive) profit making just one widget; the start–up costs will likely exceed $10. Mathematically, we would write this as P (1) < 0. What do P (1000) = 500 and P 0 (1000) = 0.25 mean? Approximate P (1100). The equation P (1000) = 500 means that selling 1,000 Solution widgets returns a profit of $500. We interpret P 0 (1000) = 0.25 as meaning that the profit is increasing at rate of $0.25 per widget (the units are Notes: Page 80 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.2 Interpretations of the Derivative “dollars per widget.”) Since we have no other information to use, our best approximation for P (1100) is: P (1100) ≈ P (1000) + P 0 (1000) × 100 = $500 + 100 · 0.25 = $525. We approximate that selling 1,100 widgets returns a profit of $525. The previous examples made use of an important approximation tool that we first used in our previous “driving a car at 60 mph” example at the beginning of this section. Five minutes after looking at the speedometer, our best approximation for distance traveled assumed the rate of change was constant. In Examples 44 and 45 we made similar approximations. We were given rate of change information which we used to approximate total change. Notationally, we would say that f (c + h) ≈ f (c) + f 0 (c) · h. This approximation is best when h is “small.” “Small” is a relative term; when dealing with the world population, h = 22 days = 28,800 minutes is small in comparison to years. When manufacturing widgets, 100 widgets is small when one plans to manufacture thousands. The Derivative and Motion One of the most fundamental applications of the derivative is the study of motion. Let s(t) be a position function, where t is time and s(t) is distance. For instance, s could measure the height of a projectile or the distance an object has traveled. Let’s let s(t) measure the distance traveled, in feet, of an object after t seconds of travel. Then s 0 (t) has units “feet per second,” and s 0 (t) measures the instantaneous rate of distance change – it measures velocity. Now consider v(t), a velocity function. That is, at time t, v(t) gives the velocity of an object. The derivative of v, v 0 (t), gives the instantaneous rate of velocity change – acceleration. (We often think of acceleration in terms of cars: a car may “go from 0 to 60 in 4.8 seconds.” This is an average acceleration, a measurement of how quickly the velocity changed.) If velocity is measured in feet per second, and time is measured in seconds, then the units of acceleration (i.e., the units of v 0 (t)) are “feet per second per second,” or (ft/s)/s. We often shorten this to “feet per second squared,” or ft/s2 , but this tends to obscure the meaning of the units. Perhaps the most well known acceleration is that of gravity. In this text, we use g = 32ft/s2 or g = 9.8m/s2 . What do these numbers mean? Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 81 Chapter 2 Derivatives A constant acceleration of 32(ft/s)/s means that the velocity changes by 32ft/s each second. For instance, let v(t) measures the velocity of a ball thrown straight up into the air, where v has units ft/s and t is measured in seconds. The ball will have a positive velocity while traveling upwards and a negative velocity while falling down. The acceleration is thus −32ft/s2 . If v(1) = 20ft/s, then when t = 2, the velocity will have decreased by 32ft/s; that is, v(2) = −12ft/s. We can continue: v(3) = −44ft/s, and we can also figure that v(0) = 52ft/s. We now consider the second interpretation of the derivative given in this section. This interpretation is not independent from the first by any means; many of the same concepts will be stressed, just from a slightly different perspective. y Interpretation of the Derivative #2: The Slope of the Tangent Line 16 12 8 4 . x 1 2 3 4 Figure 2.8: A graph of f (x) = x2 . f (c + h) − f (c) Given a function y = f (x), the difference quotient h gives a change in y values divided by a change in x values; i.e., it is a measure of the “rise over run,” or “slope,” of the line that   goes through two points on the graph of f : c, f (c) and c + h, f (c + h) . As h shrinks to 0, these two points come close together; in the limit we find f 0 (c), the slope of a special line called the tangent line that intersects f only once near x = c. Lines have a constant rate of change, their slope. Nonlinear functions do not have a constant rate of change, but we can measure their instantaneous rate of change at a given x value c by computing f 0 (c). We can get an idea of how f is behaving by looking at the slopes of its tangent lines. We explore this idea in the following example. y Example 46 Understanding the derivative: the rate of change Consider f (x) = x2 as shown in Figure 2.8. It is clear that at x = 3 the function is growing faster than at x = 1, as it is steeper at x = 3. How much faster is it growing? 16 12 8 4 .. x 1 2 3 4 Figure 2.9: A graph of f (x) = x2 and tangent lines. We can answer this directly after the following section, Solution where we learn to quickly compute derivatives. For now, we will answer graphically, by considering the slopes of the respective tangent lines. With practice, one can fairly effectively sketch tangent lines to a curve at a particular point. In Figure 2.9, we have sketched the tangent lines to f at x = 1 and x = 3, along with a grid to help us measure the slopes of Notes: Page 82 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.2 Interpretations of the Derivative these lines. At x = 1, the slope is 2; at x = 3, the slope is 6. Thus we can say not only is f growing faster at x = 3 than at x = 1, it is growing three times as fast. y f(x) 5 ′ f (x) Example 47 Understanding the graph of the derivative Consider the graph of f (x) and its derivative, f 0 (x), in Figure 2.10(a). Use these graphs to find the slopes of the tangent lines to the graph of f at x = 1, x = 2, and x = 3. x 1 2 3 −5 To find the appropriate slopes of tangent lines to the Solution graph of f , we need to look at the corresponding values of f 0 . The slope of the tangent line to f at x = 1 is f 0 (1); this looks to be about −1. The slope of the tangent line to f at x = 2 is f 0 (2); this looks to be about 4. The slope of the tangent line to f at x = 3 is f 0 (3); this looks to be about 3. Using these slopes, the tangent lines to f are sketched in Figure 2.10(b). Included on the graph of f 0 in this figure are filled circles where x = 1, x = 2 and x = 3 to help better visualize the y value of f 0 at those points. . (a) . y f(x) 5 f ′ (x) x 1 2 3 −5 . Example 48 Approximation with the derivative Consider again the graph of f (x) and its derivative f 0 (x) in Example 47. Use the tangent line to f at x = 3 to approximate the value of f (3.1). Figure 2.11 shows the graph of f along with its tangent Solution line, zoomed in at x = 3. Notice that near x = 3, the tangent line makes an excellent approximation of f . Since lines are easy to deal with, often it works well to approximate a function with its tangent line. (This is especially true when you don’t actually know much about the function at hand, as we don’t in this example.) While the tangent line to f was drawn in Example 47, it was not explicitly computed. Recall that the tangent line to f at x = c is y = f 0 (c)(x − c) + f (c). While f is not explicitly given, by the graph it looks like f (3) = 4. Recalling that f 0 (3) = 3, we can compute the tangent line to be approximately y = 3(x − 3) + 4. It is often useful to leave the tangent line in point–slope form. To use the tangent line to approximate f (3.1), we simply evaluate y at 3.1 instead of f . f (3.1) ≈ y(3.1) = 3(3.1 − 3) + 4 = .1 ∗ 3 + 4 = 4.3. (b) . Figure 2.10: Graphs of f and f 0 in Example 47, along with tangent lines in (b). y 4 3 2 . x 2.8 3 3.2 Figure 2.11: Zooming in on f at x = 3 for the function given in Examples 47 and 48. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 83 Chapter 2 Derivatives We approximate f (3.1) ≈ 4.3. To demonstrate the accuracy of the tangent line approximation, we now state that in Example 48, f (x) = −x3 + 7x2 − 12x + 4. We can evaluate f (3.1) = 4.279. Had we known f all along, certainly we could have just made this computation. In reality, we often only know two things: 1. What f (c) is, for some value of c, and 2. what f 0 (c) is. For instance, we can easily observe the location of an object and its instantaneous velocity at a particular point in time. We do not have a “function f ” for the location, just an observation. This is enough to create an approximating function for f . This last example has a direct connection to our approximation method explained above after Example 45. We stated there that f (c + h) ≈ f (c) + f 0 (c) · h. If we know f (c) and f 0 (c) for some value x = c, then computing the tangent line at (c, f (c)) is easy: y(x) = f 0 (c)(x − c) + f (c). In Example 48, we used the tangent line to approximate a value of f . Let’s use the tangent line at x = c to approximate a value of f near x = c; i.e., compute y(c + h) to approximate f (c + h), assuming again that h is “small.” Note:  y(c + h) = f 0 (c) (c + h) − c + f (c) = f 0 (c) · h + f (c). This is the exact same approximation method used above! Not only does it make intuitive sense, as explained above, it makes analytical sense, as this approximation method is simply using a tangent line to approximate a function’s value. The importance of understanding the derivative cannot be understated. When f is a function of x, f 0 (x) measures the instantaneous rate of change of f with respect to x and gives the slope of the tangent line to f at x. Notes: Page 84 Custom Made for Math 1174 at Langara College by P. Anhaouy Exercises 2.2 Terms and Concepts 1. What is the instantaneous rate of change of position called? In Exercises 15 – 18, graphs of functions f (x) and g(x) are given. Identify which function is the derivative of the other.) y 2. Given a function y = f (x), in your own words describe how to find the units of f 0 (x). g(x) 4 f(x) 2 15. 3. What functions have a constant rate of change? x −4 −2 4 2 −2 Problems −4 . 4. Given f (5) = 10 and f 0 (5) = 2, approximate f (6). y 0 5. Given P (100) = −67 and P (100) = 5, approximate P (110). 6. Given z(25) = 187 and z 0 (25) = 17, approximate z(20). 4 g(x) 16. x 2 −2 7. Knowing f (10) = 25 and f 0 (10) = 5 and the methods described in this section, which approximation is likely to be most accurate: f (10.1), f (11), or f (20)? Explain your reasoning. −4 . y 5 8. Given f (7) = 26 and f (8) = 22, approximate f 0 (7). 0 9. Given H(0) = 17 and H(2) = 29, approximate H (2). f(x) 2 f(x) 17. . x −5 10. Let V (x) measure the volume, in decibels, measured inside a restaurant with x customers. What are the units of V 0 (x)? 5 g(x) −5 . y 11. Let v(t) measure the velocity, in ft/s, of a car moving in a straight line t seconds after starting. What are the units of v 0 (t)? f(x) 1 18. x 12. The height H, in feet, of a river is recorded t hours after midnight, April 1. What are the units of H 0 (t)? −4 13. P is the profit, in thousands of dollars, of producing and selling c cars. . −2 2 −1 (a) What are the units of P 0 (c)? 4 g(x) Review (b) What is likely true of P (0)? 14. T is the temperature in degrees Fahrenheit, h hours after midnight on July 4 in Sidney, NE. (a) What are the units of T 0 (h)? (b) Is T 0 (8) likely greater than or less than 0? Why? (c) Is T (8) likely greater than or less than 0? Why? In Exercises 19 – 20, use the definition to compute the derivatives of the following functions. 19. f (x) = 5x2 20. f (x) = (x − 2)3 Custom Made for Math 1174 at Langara College by P. Anhaouy Page 85 Chapter 2 Derivatives 2.3 Basic Differentiation Rules Let’s practice abstraction on linear functions, y = mx + b. What is y 0 ? Without limits, recognize that linear function are characterized by being functions with a constant rate of change (the slope). The derivative, y 0 , gives the instantaneous rate of change; with a linear function, this is constant, m. Thus y 0 = m. Let’s abstract once more. Let’s find the derivative of the general quadratic function, f (x) = ax2 + bx + c. Using the definition of the derivative, we have: a(x + h)2 + b(x + h) + c − (ax2 + bx + c) h→0 h ah2 + 2ahx + bh = lim h→0 h = lim ah + 2ax + b f 0 (x) = lim h→0 = 2ax + b. So if y = 6x2 + 11x − 13, we can immediately compute y 0 = 12x + 11. In this section (and in some sections to follow) we will learn some of what mathematicians have already discovered about the derivatives of certain functions and how derivatives interact with arithmetic operations. We start with a theorem. Theorem 12 Derivatives of Common Functions 1. Constant Rule: d  c = 0, where c is a constant. dx 2. Power Rule: d (xn ) = nxn−1 , where n is an integer, n > 0. dx 5. Natural Exponential Rule: d x (e ) = ex dx 6. Linear Function Rule: d (mx + b) = m dx 3. Reciprocal Rule: d 1 1 =− 2 dx x x Notes: Page 86 Custom Made for Math 1174 at Langara College by P. Anhaouy 4. Square Root Rule: d √  1 x = √ dx 2 x 2.3 Basic Differentiation Rules This theorem starts by stating an intuitive fact: constant functions have no rate of change as they are constant. Therefore their derivative is 0 (they change at the rate of 0). The theorem then states some fairly amazing things. The Power Rule states that the derivatives of Power Functions (of the form y = xn ) are very straightforward: multiply by the power, then subtract 1 from the power. We see something incredible about the function y = ex : it is its own derivative. We also see a new connection between the sine and cosine functions. One special case of the Power Rule is when n = 1, i.e., when f (x) = x. What is f 0 (x)? According to the Power Rule, f 0 (x) = d 1 d  x = x = 1 · x0 = 1. dx dx In words, we are asking “At what rate does f change with respect to x?” Since f is x, we are asking “At what rate does x change with respect to x?” The answer is: 1. They change at the same rate. Let’s practice using this theorem. Example 49 Let f (x) = x3 . Using Theorem 12 to find, and use, derivatives y 0 1. Find f (x). 4 2. Find the equation of the line tangent to the graph of f at x = −1. 2 3. Use the tangent line to approximate (−1.1)3 . 4. Sketch f , f 0 and the found tangent line on the same axis. Solution x −2 2. To find the equation of the line tangent to the graph of f at x = −1, we need a point and the slope. The point is (−1, f (−1)) = (−1, −1). The slope is f 0 (−1) = 3. Thus the tangent line has equation y = 3(x − (−1)) + (−1) = 3x + 2. 1 2 −2 . 1. The Power Rule states that if f (x) = x3 , then f 0 (x) = 3x2 . −1 −4 Figure 2.12: A graph of f (x) = x3 , 0 2 along . with its derivative f (x) = 3x and its tangent line at x = −1. 3. We can use the tangent line to approximate (−1.1)3 as −1.1 is close to −1. We have (−1.1)3 ≈ 3(−1.1) + 2 = −1.3. We can easily find the actual answer; (−1.1)3 = −1.331. 4. See Figure 2.12. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 87 Chapter 2 Derivatives Theorem 12 gives useful information, but we will need much more. For instance, using the theorem, we can easily find the derivative of y = x3 , but it does not tell how to compute the derivative of y = 2x3 , y = x3 +sin x nor y = x3 sin x. The following theorem helps with the first two of these examples (the third is answered in the next section). Theorem 13 Properties of the Derivative Let f and g be differentiable on an open interval I and let c be a real number. Then: 1. Sum/Difference Rule:    d  d  d  f (x) ± g(x) = f (x) ± g(x) = f 0 (x) ± g 0 (x) dx dx dx 2. Constant Multiple Rule:   d  d  c · f (x) = c · f (x) = c · f 0 (x). dx dx Theorem 13 allows us to find the derivatives of a wide variety of functions. It can be used in conjunction with the Power Rule to find the derivatives of any polynomial. Recall in Example 41 that we found, using the limit definition, the derivative of f (x) = 3x2 + 5x − 7. We can now find its derivative without expressly using limits: Note: Definition 11 comes with the caveat “Where the corresponding limits exist.” With f differentiable on I, it is possible that f 0 is not differentiable on all of I, and so on.  d  2 d  2 d   d   3x + 5x + 7 = 3 x +5 x + 7 dx dx dx dx = 3 · 2x + 5 · 1 + 0 = 6x + 5. We were a bit pedantic here, showing every step. Normally we would do  d  2 all the arithmetic and steps in our head and readily find 3x +5x+7 = dx 6x + 5. Higher Order Derivatives The derivative of a function f is itself a function, therefore we can take its derivative. The following definition gives a name to this concept and introduces its notation. Notes: Page 88 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.3 Definition 11 Basic Differentiation Rules Higher Order Derivatives Let y = f (x) be a differentiable function on I. 1. The second derivative of f is: f 00 (x) =   d dy d 2y d  0  f (x) = = y 00 . = dx dx dx dx2 2. The third derivative of f is: f 000 (x) =   d  00  d d 2y d 3y f (x) = = = y 000 . dx dx dx2 dx3 3. The nth derivative of f is: f (n) (x) =   d ny d  (n−1)  d d n−1 y f = = y (n) . (x) = n−1 dx dx dx dxn In general, when finding the fourth derivative and on, we resort to the f (4) (x) notation, not f 0000 (x); after a while, too many ticks is too confusing. Let’s practice using this new concept. Example 50 Finding higher order derivatives Find the first four derivatives of the following functions: 2. f (x) = 5ex 1. f (x) = 4x2 Solution 1. Using the Power and Constant Multiple Rules, we have: f 0 (x) = 8x. Continuing on, we have f 00 (x) =  d 8x = 8; dx f 000 (x) = 0; f (4) (x) = 0. Notice how all successive derivatives will also be 0. 2. Employing Theorem 12 and the Constant Multiple Rule, we can see that f 0 (x) = f 00 (x) = f 000 (x) = f (4) (x) = 5ex . Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 89 Chapter 2 Derivatives Interpreting Higher Order Derivatives What do higher order derivatives mean? What is the practical interpretation? Our first answer is a bit wordy, but is technically correct and beneficial to understand. That is, The second derivative of a function f is the rate of change of the rate of change of f . One way to grasp this concept is to let f describe a position function. Then, f 0 describes the rate of position change: velocity. We now consider f 00 , which describes the rate of velocity change. Sports car enthusiasts talk of how fast a car can go from 0 to 60 mph; they are bragging about the acceleration of the car. We started this chapter with amusement–park riders free–falling with position function f (t) = −16t2 + 150. It is easy to compute f 0 (t) = −32t ft/s and f 00 (t) = −32 (ft/s)/s. We may recognize this latter constant; it is the acceleration due to gravity. In keeping with the unit notation introduced in the previous section, we say the units are “feet per second per second.” This is usually shortened to “feet per second squared,” written as “ft/s2 .” It can be difficult to consider the meaning of the third, and higher order, derivatives. The third derivative is “the rate of change of the rate of change of the rate of change of f .” That is essentially meaningless to the uninitiated. In the context of our position/velocity/acceleration example, the third derivative is the “rate of change of acceleration,” commonly referred to as “jerk.” Make no mistake: higher order derivatives have great importance even if their practical interpretations are hard (or “impossible”) to understand. The mathematical topic of series makes extensive use of higher order derivatives. Notes: Page 90 Custom Made for Math 1174 at Langara College by P. Anhaouy Exercises 2.3 Terms and Concepts 13. p(s) = 41 s4 + 31 s3 + 12 s2 + s + 1 1. What is the name of the rule which states that d n x = nxn−1 , where n > 0 is an integer? dx 14. g(t) = (1 + 3t)2 15. g(x) = (2x − 5)3 0 2. Give an example of a function f (x) where f (x) = f (x). 16. f (x) = (1 − x)3 3. Give an example of a function f (x) where f 0 (x) = 0. 17. f (x) = (2 − 3x)2 4. Explain in your own words how to find the third derivative of a function f (x). In Exercises 18 – 23, compute the first four derivatives of the given function. 5. Give an example of a function where f 0 (x) 6= 0 and f 00 (x) = 0. 18. f (x) = x6 6. Explain in your own words what the second derivative “means.” 19. h(t) = t2 − et 20. p(θ) = θ4 − θ3 0 7. If f (x) describes a position function, then f (x) describes what kind of function? What kind of function is f 00 (x)? 8. Let f (x) be a function measured in pounds, where x is measured in feet. What are the units of f 00 (x)? In Exercises 21 – 26, find the equations of the tangent and normal lines to the graph of the function at the given point. 21. f (x) = x3 − x at x = 1 22. f (t) = et + 3 at t = 0 Problems In Exercises 9 – 23, compute the derivative of the given function. 23. f (x) = 2x + 3 at x = 5 9. f (x) = 7x2 − 5x + 7 Review 10. g(x) = 14x3 + 7x2 + 11x − 29 24. Given that e0 = 1, approximate the value of e0.1 using the tangent line to f (x) = ex at x = 0. 11. m(t) = 9t5 − 81 t3 + 3t − 8 r 12. f (r) = 6e 25. Approximate the value of (3.01)4 using the tangent line to f (x) = x4 at x = 3. Custom Made for Math 1174 at Langara College by P. Anhaouy Page 91 Chapter 2 Derivatives 2.4 The Product and Quotient Rules The previous section showed that, in some ways, derivatives behave nicely. The Constant Multiple and Sum/Difference Rules established that the derivative of f (x) = 5x2 + sin x was not complicated. We neglected com5x2 puting the derivative of things like g(x) = 5x2 sin x and h(x) = sin x on purpose; their derivatives are not as straightforward. (If you had to guess what their respective derivatives are, you would probably guess wrong.) For these, we need the Product and Quotient Rules, respectively, which are defined in this section. We begin with the Product Rule. Theorem 14 Product Rule Let f and g be differentiable functions on an open interval I. Then f g is a differentiable function on I, and  d  f (x)g(x) = f 0 (x)g(x) + f (x)g 0 (x). dx  d  Important: f (x)g(x) 6= f 0 (x)g 0 (x)! While this answer is simdx pler than the Product Rule, it is wrong. We practice using this new rule in an example, followed by an example that demonstrates why this theorem is true. Example 51 Using the Product Rule Use the Product Rule to compute the derivative of y = 5x2 ex . Evaluate the derivative at x = 1. To make our use of the Product Rule explicit, let’s set Solution f (x) = 5x2 and g(x) = ex . We easily compute/recall that f 0 (x) = 10x and g 0 (x) = ex . Employing the rule, we have d  2 x 5x e = 10xex + 5x2 ex = 5xex (2 + x). dx At x = 1, we have y 0 (1) = 5(1)e1 (2 + 1) = 15e. Notes: Page 92 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.4 The Product and Quotient Rules We now investigate why the Product Rule is true. Example 52 A proof of the Product Rule Use the definition of the derivative to prove Theorem 14. Solution By the limit definition, we have  d  f (x + h)g(x + h) − f (x)g(x) f (x)g(x) = lim . h→0 dx h We now do something a bit unexpected; add 0 to the numerator (so that nothing is changed) in the form of −f (x + h)g(x) + f (x + h)g(x), then do some regrouping as shown.  f (x + h)g(x + h) − f (x)g(x) d  f (x)g(x) = lim (now add 0 to the numerator) h→0 dx h f (x + h)g(x + h) − f (x + h)g(x) + f (x + h)g(x) − f (x)g(x) = lim (regroup) h→0 h     f (x + h)g(x + h) − f (x + h)g(x) + f (x + h)g(x) − f (x)g(x) = lim h→0 h f (x + h)g(x + h) − f (x + h)g(x) f (x + h)g(x) − f (x)g(x) = lim + lim (factor) h→0 h→0 h h g(x + h) − g(x) f (x + h) − f (x) = lim f (x + h) + lim g(x) (apply limits) h→0 h→0 h h 0 0 = f (x)g (x) + f (x)g(x) It is often true that we can recognize that a theorem is true through its proof yet somehow doubt its applicability to real problems. In the following example, we compute the derivative of a product of functions in two ways to verify that the Product Rule is indeed “right.” Example 53 Exploring alternate derivative methods Let y = (x2 + 3x + 1)(2x2 − 3x + 1). Find y 0 two ways: first, by expanding the given product and then taking the derivative, and second, by applying the Product Rule. Verify that both methods give the same answer. We first expand the expression for y; a little algebra Solution shows that y = 2x4 + 3x3 − 6x2 + 1. It is easy to compute y 0 ; y 0 = 8x3 + 9x2 − 12x. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 93 Chapter 2 Derivatives Now apply the Product Rule. y 0 = (x2 + 3x + 1)(4x − 3) + (2x + 3)(2x2 − 3x + 1)   = 4x3 + 9x2 − 5x − 3 + 4x3 − 7x + 3 = 8x3 + 9x2 − 12x. The uninformed usually assume that “the derivative of the product is the product of the derivatives.” Thus we are tempted to say that y 0 = (2x + 3)(4x − 3) = 8x2 + 6x − 9. Obviously this is not correct. Example 54 functions  Let y = x3 ex Using the Product Rule with a product of three  1 √ + x . Find y 0 . No need to simplify the result. x We have a product of three functions while the Product Solution Rule only specifies how to handle a product of two functions. Our method of handling this problem is to simply group the latter two functions together, and consider y = x3 ln x cos x . Following the Product Rule, we have h 1 √ i h  1 √  i0 y 0 = (3x2 ) ex + x3 ex + x + x x x  To evaluate ex  √ 0 1 x ] , we apply the Product Rule again: x +  i h   1 √  1 1 + x + ex − 2 + √ = (3x2 ) ex ex x x 2 x Recognize the pattern in our answer above: when applying the Product Rule to a product of three functions, there are three terms added together in the final derivative. Each terms contains only one derivative of one of the original functions, and each function’s derivative shows up in only one term. It is straightforward to extend this pattern to finding the derivative of a product of 4 or more functions. Notes: Page 94 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.4 The Product and Quotient Rules We have learned how to compute the derivatives of sums, differences, and products of functions. We now learn how to find the derivative of a quotient of functions. Theorem 15 Quotient Rule Let f and g be functions defined on an open interval I, where g(x) 6= 0 on I. Then f /g is differentiable on I, and   d f (x) f 0 (x)g(x) − f (x)g 0 (x) = dx g(x) g(x)2 The Quotient Rule is not hard to use, although it might be a bit tricky to remember. First, we write the quotient rule in the following way.   g(x)f 0 (x) − f (x)g 0 (x) d f (x) = . dx g(x) g(x)2 A useful mnemonic works as follows. Consider a fraction’s numerator and denominator as “HI” and “LO”, respectively. Then   d HI LO· dHI – HI· dLO , = dx LO LOLO read “low dee high minus high dee low, over low low.” Said fast, that phrase can roll off the tongue, making it easy to memorize. The “dee high” and “dee low” parts refer to the derivatives of the numerator and denominator, respectively. Let’s practice using the Quotient Rule. Example 55 Using the Quotient Rule 5x2 . Find f 0 (x). No need to simplify the result. Let f (x) = x e + x3 Directly applying the Quotient Rule gives:     ex + x3 · 10x − 5x2 · ex + 3x2 5x2 d = dx ex + x3 (ex + x3 )2 Solution Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 95 Chapter 2 Derivatives The Quotient Rule allows us to fill in holes in our understanding of derivatives of the common trigonometric functions. We start with finding the derivative of the tangent function. Example 56 Using the Rule √ Quotient x + ex Find the derivative of y = at x = 1 where g(1) = g 0 (1) = 2. g(x) We can apply the Quotient Rule.   √ 1 √  √x + ex  + ex g(x) − ( x + ex ) g 0 (x) d 2 x 0 = y = dx g(x) (g 0 (x))2 Solution At x = 1, we get  y 0 (1) = 1 √ + e1 2 1  g(1) − √ (g 0 (1))2  1 + e1 g 0 (1) = 1 + 2e − 2 − 2e 1 =− 2 2 4 . The Quotient Rule gives other useful results, as show in the next example. Example 57 Using the Quotient Rule to expand the Power Rule Find the derivatives of the following functions. 1. f (x) = 1 x 2. f (x) = 1 , where n > 0 is an integer. xn Solution We employ the Quotient Rule. 1. f 0 (x) = 1 x·0−1·1 = − 2. 2 x x 2. f 0 (x) = xn · 0 − 1 · nxn−1 nxn−1 n = − = − n+1 . n 2 2n (x ) x x Notes: Page 96 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.4 The Product and Quotient Rules 1 The derivative of y = n turned out to be rather nice. It gets better. x Consider:   d 1 d  −n  (apply result from Example 57) = x dx xn dx n = − n+1 (rewrite algebraically) x = −nx−(n+1) = −nx−n−1 . This is reminiscent of the Power Rule: multiply by the power, then subtract 1 from the power. We now add to our previous Power Rule, which had the restriction of n > 0. Theorem 16 Power Rule with Integer Exponents Let f (x) = xn , where n 6= 0 is an integer. Then f 0 (x) = n · xn−1 . Taking the derivative of many functions is relatively straightforward. It is clear (with practice) what rules apply and in what order they should be applied. Other functions present multiple paths; different rules may be applied depending on how the function is treated. One of the beautiful things about calculus is that there is not “the” right way; each path, when applied correctly, leads to the same result, the derivative. We demonstrate this concept in an example. Example 58 Exploring alternate derivative methods x2 − 3x + 1 . Find f 0 (x) in each of the following ways: Let f (x) = x 1. By applying the Quotient Rule,  2. by viewing f as f (x) = x2 − 3x + 1 · x−1 and applying the Product and Power Rules, and 3. by “simplifying” first through division. Verify that all three methods give the same result. Solution Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 97 Chapter 2 Derivatives 1. Applying the Quotient Rule gives:   x · 2x − 3 − x2 − 3x + 1 · 1 x2 − 1 1 0 f (x) = = = 1 − 2. 2 2 x x x 2. By rewriting f , we can apply the Product and Power Rules as follows:   f 0 (x) = x2 − 3x + 1 · (−1)x−2 + 2x − 3 · x−1 x2 − 3x + 1 2x − 3 + x2 x x2 − 3x + 1 2x2 − 3x =− + x2 x2 2 x −1 1 = = 1 − 2, x2 x =− the same result as above. 1 3. As x 6= 0, we can divide through by x first, giving f (x) = x − 3 + . x Now apply the Power Rule. f 0 (x) = 1 − 1 , x2 the same result as before. Example 58 demonstrates three methods of finding f 0 . One is hard pressed to argue for a “best method” as all three gave the same result without too much difficulty, although it is clear that using the Product Rule required more steps. Ultimately, the important principle to take away from this is: reduce the answer to a form that seems “simple” and easy to interpret. In that example, we saw different expressions for f 0 , including:     x · 2x − 3 − x2 − 3x + 1 · 1 1 = x2 −3x+1 ·(−1)x−2 + 2x−3 ·x−1 . 1− 2 = x x2 They are equal; they are all correct; only the first is “clear.” Work to make answers clear. Notes: Page 98 Custom Made for Math 1174 at Langara College by P. Anhaouy Exercises 2.4 Terms and Concepts In Exercises 13 – 27, compute the derivative of the given function. 1. T/F: The derivatives of the trigonometric functions that start with “c” have minus signs in them. 13. g(x) = x+7 x−5 2. What derivative rule is used to extend the Power Rule to include negative integer exponents? 14. h(t) = 7t2 + 6t − 2 3. T/F: Regardless of the function, there is always exactly one right way of computing its derivative. 15. f (x) = 4. In your own words, explain what it means to make your answers “clear.” 16. f (x) = (16x3 + 24x2 + 3x) Problems In Exercises 17 – 20, find the equations of the tangent and normal lines to the graph of g at the indicated point. In Exercises 5 – 8: (a) Use the Product Rule to differentiate the function. (b) Manipulate the function algebraically and differentiate without the Product Rule. (c) Show that the answers from (a) and (b) are equivalent. 5. f (x) = x(x2 + 3x) 7x − 1 16x3 + 24x2 + 3x 17. g(s) = es (s2 + 2) at (0, 2). 18. g(x) = x2 at (2, 4) x−1 In Exercises 19 – 22, find the x–values where the graph of the function has a horizontal tangent line. 19. f (x) = 6x2 − 18x − 24 6. g(x) = 2x2 (5x3 ) 20. f (x) = x x+1 21. f (x) = x2 x+1 7. h(s) = (2s − 1)(s + 4) 2 x4 + 2x3 x+2 3 8. f (x) = (x + 5)(3 − x ) In Exercises 9 – 12: (a) Use the Quotient Rule to differentiate the function. (b) Manipulate the function algebraically and differentiate without the Quotient Rule. (c) Show that the answers from (a) and (b) are equivalent. 9. f (x) = x2 + 3 x 10. g(x) = x3 − 2x2 2x2 11. h(s) = 3 4s3 12. f (t) = t2 − 1 t+1 Custom Made for Math 1174 at Langara College by P. Anhaouy Page 99 Chapter 2 Derivatives 2.5 Derivative as Rates of Change 2.5.1 Linear Motion Suppose that an object moves along a straight line with its position at any time t is given by s(t) and it is oriented from a reference point. Definition 12 Average Velocity The average velocity of the object over a very small time interval [t, t + ∆t], we have v̄ = s(t + ∆t) − s(t) . ∆t Letting ∆t → 0, it becomes the instantaneous velocity v(t) = lim ∆t→0 s(t + ∆t) − s(t) . ∆t This is in fact the limit definition of derivative of the function s(t). So, we have the following definition of velocity. Definition 13 Velocity The velocity v(t) of an object moving in a straight line (also called its instantaneous velocity) is defined as v(t) = s0 (t) = ds , dt which is the rate at which its position changes with respect to time. The speed is the magnitude of the velocity: |v(t)|. Notes: Page 100 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.5 Definition 14 Derivative as Rates of Change Acceleration The rate at which the velocity changes with respect to time is called the acceleration, and it often denoted by a(t). That is, a(t) = v 0 (t) = dv . dt The velocity of a moving object is increasing if a(t) > 0 and decreasing if a(t) < 0. Notice that if we take derivative of s(t) twice, we get the acceleration. So, a(t) is the second derivative of the position s(t). We write a(t) = s00 (t). Example 59 Suppose that a particle moves along the x-axis so that its coordinate at time t is given by the formula x(t) = t3 − t2 − 5t + 10 for t ≥ 0. 1. What is its instantaneous velocity when t = 2? when t = 4? 2. What is its average velocity between t = 2 and t = 4? 3. In which direction (left or right) is the particle moving when t = 1? when t = 3? 4. At what time does the particle change direction? Solution 1. v(t) = x0 (t) = 3t2 − 2t − 5 ⇒ v(2) = 3(22 ) − 2(2) − 5 = 3 and v(4) = 35. 2. The average velocity between t = 2 and t = 4 is v̄ = x(4) − x(2) 38 − 4 = = 17. 4−2 2 3. When t = 1, v(1) = −4 < 0, it moves to the left. When t = 3, v(3) = 16 > 0, it moves to the right. 5 5 4. v(t) = 0 ⇒ 3t2 − 2t − 5 = 0 ⇒ t = , −1. But, t ≥ 0, we take t = . 3 3 5 5 Observe that v < 0 if 0 ≤ t ≤ and v > 0 if t > . Therefore, the 3 3 5 particle changes direction at t = . 3 Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 101 Chapter 2 Derivatives Example 60 Suppose an object is thrown vertically upward and its height at any time t, in seconds, is given by h(t) = −6t2 + 64t + 192, in meters. 1. When does the object reach the highest point? Find the maximum height. 2. Find the velocity of the object when it hits the ground. Solution 16 . The object reach the highest point 1. v(t) = −12t + 64 = 0 ⇒ t = 3   16 when t = seconds. The maximum height is h 16 = 362.67 3 3 meters. 2. The object hits the ground when h(t) = 0 ⇒ t = 13.1 seconds. So, the velocity is v(13.1) = −93.3. Thus, the velocity when it hits the ground is 93.3 meters per second. 2.5.2 Marginal Analysis In economics, the rates at which the cost, revenue, and profit are changing are called marginal cost, marginal revenue, and marginal profit. So, the term derivative is applied. We then have the following: Notes: Page 102 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.5 Derivative as Rates of Change Definition 15 Suppose that C(x), R(x), and P (x) are the cost, revenue, and profit functions, respectively. • The marginal cost is the derivative of the cost function: M C(x) = C 0 (x). • The marginal revenue is the derivative of the revenue function: M R(x) = R0 (x). • The marginal profit is the derivative of the profit function: M P (x) = P 0 (x). • Since P (x) = R(x) − C(x), we have M P (x) = M R(x) − M C(x). Revenue and Demand: When the number of items that a particular merchant or company sell in a unit is small relative to the total consumer demand, the selling price per item is often a fixed price p regardless of the number x of items sold. In this case the total revenue function is simply R(x) = px and M R(x) = p. However, when a company’s position in the market approaches a monopoly, the selling price p and the sales level x are often related to each other via a demand equation. The simplest equation is the linear one, i.e., p = a−bx. Example 61 A small company that produces toys has fixed costs of $1000 per week. In addition, each toy costs $4.50 to make. The weekly revenue is given by the formula R(x) = 10x − 0.01x2 for 0 ≤ x ≤ 800, where x is the number of toys produced and sold weekly. Find the marginal cost, revenue, and profit. The cost function is C(x) = 1000 + 4.50x ⇒ M C(x) = Solution C 0 (x) = 4.50. Also, M R(x) = R0 (x) = 10 − 0.02x. Then, M P (x) = M R(x) − M C(x) = 5.5 − 0.02x, 0 < x < 800. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 103 Chapter 2 Derivatives Example 62 The marginal cost and revenue arising in the production and sale of a certain item are given by M C(x) = 350 and M R(x) = 800 − 2x, where x is the number item produced and sold per week. 1. What is the marginal profit? 2. If the company is currently producing 160 items per week, should it increase or decrease production in order to raise its profit? Explain your answer. 3. What if the company is currently producing 200 items per week? 240 items per week? Solution 1. M P (x) = M R(x) − M C(x) = (800 − 2x) − 350 = 450 − 2x. 2. The company should increase production in order to raise the profit because M P (160) = 130 > 0, which means the rate at which the profit is changing is positive and a small increase in x will result in higher P . 3. Since M P (200) = 50 > 0, the company should increase production at level x = 200 for the same reason above. But, P M (240) = −30 < 0 implies that the profit is changing negatively and small increase in x will result in lower profit. So, the company should decrease the production. Example 63 Suppose that the selling price p of an item and the weekly sales levels x items are related by the demand equation p(x) = 60 − x. Suppose also the cost of producing x items per week is given by C(x) = 40 + 5x + 41 x2 . Find the weekly marginal profit, if the production level is x = 20. R(x) = p · x = (60 − x)x = 60x − x2 . Then, P (x) = 1 5 R(x) − C(x) = 60x − x2 − 40 − 5x − x2 = −40 + 55x − x2 ⇒ M P (x) = 4 4 5 5 P 0 (x) = 55 x ⇒ M P (20) = 55 − (20) = 5, meaning that at production 2 2 level x = 20 profit is increasing at rate of $5 per additional item. Solution Notes: Page 104 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.5 Derivative as Rates of Change Example 64 The cost to produce x units of cook book per year is give by C(x) = 1800x + 357x2 , where x > 0. C(x) of producing each book if 1. What is the average cost, C̄(x) = x the annual production level is 50 books? 2. Find the marginal cost of the production if the annual production level is 50. 3. If the production level increases from 50 to 51 books, what would happen to the average cost per book? Solution C̄(x) = C(x) = 1800 + 357x. x 1. C̄(50) = 1800 + 357(50) per book 2. M C(x) = 1800 + 714x ⇒ M C(50) = $37, 500 per book. 3. The cost will increase by $37, 500 per book. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 105 Exercises 2.5 Problems In Exercises 7 – 12, do the following marginal analysis problems. In Exercises 1 – 7, do the following linear motion problems. 1. Suppose that an object moves in a straight line in such a way its position is given by s(t) = 2t3 − 9t2 + 12t, where s is in cm and t is in s. Analyze the motion of the object. 2 7. Suppose that a store selling out cameras under-priced at p = −3x2 +600x in dollars each, where x is the number of the cameras sold, during the down turn of economic crisis. The more you sell, the more you get less money for each unit. (a) Find the marginal revenue if the store sells 100 cameras. What does this answer tell you. 2. Use Gallileo’s formula for the distance fallen, s(t) = ct , where c = 4.9 m/s2 , to compute the instantaneous velocity of a falling object a minute later. (b) Find the marginal revenue if the store sells 300 cameras. What does this answer tell you. 3. An object is shot vertically into the air and its height s is given by s(t) = 100t − 5t2 , s is in meter and t is in second. 3 8. Suppose that the price is at p = 100 − x in dollars for 4 each unit of a dumb phone, where x is the number of the dumb sold. (a) What is the average velocity between t = 1 and t = 4? (a) Find the marginal revenue at x = 10. What does this answer tell you. (b) What is the maximum height reached? (c) What is the velocity at t = 2 and t = 15? (d) What is the total distance travelled by the object between t = 5 and t = 12? 4. For each of the following motions where s is measured in meters and t is measured in seconds, find the velocity at time t = 2 and the average velocity over the given interval. (a) s(t) = 3t2 + 5 and t changes from 2 to 3 seconds. (b) s(t) = t3 − 3t2 from t = 3 to t = 5 seconds. (c) s(t) = 2t2 + 5t − 3 on [1, 2]. 5. A ball is dropped from a building, 490 meters tall above the ground. Its height,h, at time t, in seconds, is known to follow the relationship h(t) = 490 − 12gt2 , where g = 9.8 m/s2. (a) Find the average velocity of the falling ball between t = 1 and t = 2 seconds. (b) Find the average velocity between t seconds and t +  seconds where 0 ≤  ≤ 1 is some small time increment (assume that the ball is in the air during this time interval). (c) Determine the time at which the ball hits the ground. 6. Suppose an rock is thrown upward from the roof of 10 m tall building. It goes up and then back down, Its hight above the ground t seconds later is given by h(t) = −4.9t2 + 8t + 10. (a) What is the greatest height above the ground that the rock reaches? (b) With what speed does it hit the ground? (b) Find the marginal revenue at x = 30. What does this answer tell you. 9. The cost to produce x units of bicycles per month is give by C(x) = 4, 200 + 5.40x − 0.001x2 + 0.000002x3 , where x > 0. C(x) (a) What is the average cost, C̄(x) = of prox ducing each bicycle if the daily production level is 1000 units? (b) Find the marginal cost of the production if the daily production level is 1000. (c) If the production level increases from 1000 to 1001 units, what would happen to the average cost per unit? 10. The cost to produce x units of cook book per year is give by C(x) = 1800x + 357x2 , where x > 0. C(x) (a) What is the average cost, C̄(x) = of prox ducing each book if the annual production level is 50 books? (b) Find the marginal cost of the production if the annual production level is 50. (c) If the production level increases from 50 to 51 books, what would happen to the average cost per book? 11. Suppose that a company has a price function p(x) = −3x2 + 600x and a cost function C(x) = 1800 + 357x2 (a) Find the marginal profit at x = 10. What does this answer tell you. (b) Find the marginal profit at x = 100. What does this answer tell you. Page 106 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.6 2.6 Derivatives of Trigonometric Functions Derivatives of Trigonometric Functions In this section, we study the derivatives of the trigonometric functions. We first consider the following example. Example 65 Finding the derivative of a function Find the derivative of f (x) = sin x. By the limit definition of the derivative 10, we have Solution sin(x + h) − sin x h→0 h sin x cos h + cos x sin h − sin x = lim h→0 h sin x(cos h − 1) + cos x sin h = lim h→0 h   sin x(cos h − 1) cos x sin h + = lim h→0 h h f 0 (x) = lim  Use trig identity sin(x + h) = sin x cos h + cos x sin h  (regroup) (split into two fractions) cos h − 1 h→0 h  sin h lim =1 h→0 h  use lim = 0  and  = sin x · 0 + cos x · 1 = cos x ! We have found that when f (x) = sin x, f 0 (x) = cos x. This should be somewhat surprising; the result of a tedious limit process and the sine function is a nice function. Then again, perhaps this is not entirely surprising. The sine function is periodic – it repeats itself on regular intervals. Therefore its rate of change also repeats itself on the same regular intervals. We should have known the derivative would be periodic; we now know exactly which periodic function it is. y Example 66 tion Finding the derivative of a piecewise defined func sin x x ≤ π/2 Find the derivative of f (x), where f (x) = . See Figure 1 x > π/2 2.13. 1 0.5 Using Example 65, we know that when x < π/2, f 0 (x) = cos x. It is easy to verify that when x > π/2, f 0 (x) = 0; consider: Solution f (x + h) − f (x) 1−1 = lim = lim 0 = 0. h→0 h→0 h→0 h h lim So far we have f 0 (x) =  x π 2 . Figure 2.13: A graph of f (x) as defined in Example 66. cos x x < π/2 . 0 x > π/2 Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 107 Chapter 2 Derivatives We still need to find f 0 (π/2). Notice at x = π/2 that both pieces of f 0 are 0, meaning we can state that f 0 (π/2) = 0. Being more rigorous, we can again evaluate the difference quotient limit at x = π/2, utilizing again left and right–hand limits: y f (π/2 + h) − f (π/2) = h sin(π/2 + h) − sin(π/2) lim = − h h→0 sin( π2 ) cos(h) + sin(h) cos( π2 ) − sin( π2 ) lim = − h h→0 1 · cos(h) + sin(h) · 0 − 1 lim = h h→0− 0 lim 1 h→0− 0.5 x π 2 . Figure 2.14: A graph of f 0 (x) in Example 66. lim h→0+ f (π/2 + h) − f (π/2) = h 1−1 = lim h h→0+ 0 lim = h→0+ h 0 Since both the left and right hand limits are 0 at x = π/2, the limit exists and f 0 (π/2) exists (and is 0). Therefore we can fully write f 0 as  cos x x ≤ π/2 f 0 (x) = . 0 x > π/2 See Figure 2.14 for a graph of this function. Example 67 Using the Product Rule Use the Product Rule to compute the derivative of y = 5x2 sin x. Evaluate the derivative at x = π/2. To make our use of the Product Rule explicit, let’s set Solution f (x) = 5x2 and g(x) = sin x. We easily compute/recall that f 0 (x) = 10x and g 0 (x) = cos x. Employing the rule, we have y 20 15  d  2 5x sin x = 5x2 cos x + 10x sin x. dx 10 At x = π/2, we have 5 . y 0 (π/2) = 5 x π 2 π Figure 2.15: A graph of y = 5x2 sin x and its tangent line at x = π/2.  π 2 2 cos π 2 + 10 π π sin = 5π. 2 2 We graph y and its tangent line at x = π/2, which has a slope of 5π, in Figure 2.15. While this does not prove that the Product Rule is the correct way to handle derivatives of products, it helps validate its truth. . Notes: Page 108 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.6 Derivatives of Trigonometric Functions Example 68 Using the Quotient Rule 5x2 . Find f 0 (x). Let f (x) = sin x Solution Directly applying the Quotient Rule gives:  2 d 5x sin x · 10x − 5x2 · cos x = dx sin x sin2 x 10x sin x − 5x2 cos x = . sin2 x The Quotient Rule allows us to fill in holes in our understanding of derivatives of the common trigonometric functions. We start with finding the derivative of the tangent function. y 10 5 x − π2 − π2 π 4 π 2 −5 −10 . . Figure 2.16: A graph of y = tan x along with its tangent line at x = π/4.  d Example 69 Using the Quotient Rule to find dx tan x . Find the derivative of y = tan x. At first, one might feel unequipped to answer this quesSolution tion. But recall that tan x = sin x/ cos x, so we can apply the Quotient Rule.    d  d sin x tan x = dx dx cos x cos x cos x − sin x(− sin x) = cos2 x 2 cos x + sin2 x = cos2 x 1 = cos2 x = sec2 x. This is a beautiful result. To confirm its truth, we can find the equation of the tangent line to y = tan x at x = π/4. The slope is sec2 (π/4) = 2; y = tan x, along with its tangent line, is graphed in Figure 2.16. We include this result in the following theorem about the derivatives of the trigonometric functions. Recall we found the derivative of y = sin x in Example 65 and using the same limit definition of derivative, we d can find the derivative of the cosine function to be (cos x) = − sin x. dx The derivatives of the cotangent, cosecant and secant functions can all be computed directly using the derivatives of sine and cosine functions and the Quotient Rule. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 109 Chapter 2 Derivatives Theorem 17 Derivatives of Trigonometric Functions  d sin x = cos x dx  d 3. tan x = sec2 x dx  d 5. sec x = sec x tan x dx 1.  d cos x = − sin x dx  d 4. cot x = − csc2 x dx  d 6. csc x = − csc x cot x dx 2. To remember the above, it may be helpful to keep in mind that the derivatives of the trigonometric functions that start with “c” have a minus sign in them. Example 70 Exploring alternate derivative methods 5x2 In Example 68 the derivative of f (x) = was found using the Quotient sin x 2 0 Rule. Rewriting f as f (x) = 5x csc x, find f using Theorem 17 and verify the two answers are the same. 10x sin x − 5x2 cos x . sin2 x 0 2 We now find f using the Product Rule, considering f as f (x) = 5x csc x. Solution We found in Example 68 that the f 0 (x) =  d  2 5x csc x dx = 5x2 (− csc x cot x) + 10x csc x −1 cos x 10x = 5x2 · · + sin x sin x sin x −5x2 cos x 10x = + sin x sin2 x 10x sin x − 5x2 cos x = sin2 x f 0 (x) = (now rewrite trig functions) (get common denominator) Finding f 0 using either method returned the same result. At first, the answers looked different, but some algebra verified they are the same. In general, there is not one final form that we seek; the immediate result from the Product Rule is fine. Work to “simplify” your results into a form that is most readable and useful to you. Notes: Page 110 Custom Made for Math 1174 at Langara College by P. Anhaouy Exercises 2.6 Terms and Concepts 12. g(x) = e2 sin(π/4) − 1  d 2 1. T/F: The Product Rule states that x sin x = dx 2x cos x. 13. g(t) = 4t3 et − sin t cos t 14. h(t) = d 2. T/F: The Quotient Rule states that dx cos x . 2x   2 x sin x t2 sin t + 3 t2 cos t + 2 = Problems In Exercises 3 – 16, compute the derivative of the given function. 3. f (θ) = 9 sin θ + 10 cos θ  15. f (x) = x2 ex tan x 16. g(x) = 2x sin x sec x In Exercises 17 – 20, find the equations of the tangent and normal lines to the graph of function at the indicated point. 17. f (x) = 4 sin x at x = π/2 18. f (x) = −2 cos x at x = π/4 4 4. g(t) = 10t − cos t + 7 sin t , − 3π ) 19. g(t) = t sin t at ( 3π 2 2 5. h(t) = et − sin t − cos t 20. g(θ) = 6. f (x) = x sin x 7. f (t) = 1 (csc t − 4) t2 8. g(t) = t5 cos t − 2t2 cos θ − 8θ at (0, −5) θ+1 In Exercises 21 – 21, find the x–values where the graph of the function has a horizontal tangent line. 21. f (x) = x sin x on [−1, 1] In Exercises 22 – 25, find the requested derivative. x 9. h(x) = cot x − e 10. f (t) = t5 (sec t + et ) 11. f (x) = sin x cos x + 3 22. f (x) = x sin x; find f 00 (x). 23. f (x) = x sin x; find f (4) (x). 24. f (x) = csc x; find f 00 (x). Custom Made for Math 1174 at Langara College by P. Anhaouy Page 111 Chapter 2 Derivatives 2.7 The Chain Rule We have covered almost all of the derivative rules that deal with combinations of two (or more) functions. The operations of addition, subtraction, multiplication (including by a constant) and division led to the Sum and Difference rules, the Constant Multiple Rule, the Power Rule, the Product Rule and the Quotient Rule. To complete the list of differentiation rules, we look at the last way two (or more) functions can be combined: the process of composition (i.e. one function “inside” another). One example of a composition of functions is f (x) = cos(x2 ). We currently do not know how to compute this derivative. If forced to guess, one would likely guess f 0 (x) = − sin(2x), where we recognize − sin x as the derivative of cos x and 2x as the derivative of x2 . However, this is not the case; f 0 (x) 6= − sin(2x). In Example 74 we’ll see the correct answer, which employs the new rule this section introduces, the Chain Rule. Before we define this new rule, recall the notation for composition of functions. We write (f ◦ g)(x) or f (g(x)), read as “f of g of x,” to denote composing f with g. In shorthand, we simply write f ◦ g or f (g) and read it as “f of g.” Before giving the corresponding differentiation rule, we note that the rule extends to multiple compositions like f (g(h(x))) or f (g(h(j(x)))), etc. To motivate the rule, let’s look at three derivatives we can already compute. Example 71 Exploring similar derivatives Find the derivatives of F1 (x) = (1 − x)2 , F2 (x) = (1 − x)3 , and F3 (x) = (1−x)4 . (We’ll see later why we are using subscripts for different functions and an uppercase F .) In order to use the rules we already have, we must first Solution expand each function as F1 (x) = 1 − 2x + x2 , F2 (x) = 1 − 3x + 3x2 − x3 and F3 (x) = 1 − 4x + 6x2 − 4x3 + x4 . It is not hard to see that: F 01 (x) = −2 + 2x, F 02 (x) = −3 + 6x − 3x2 and F 03 (x) = −4 + 12x − 12x2 + 4x3 . An interesting fact is that these can be rewritten as F 01 (x) = −2(1 − x), F 02 (x) = −3(1 − x)2 and F 03 (x) = −4(1 − x)3 . Notes: Page 112 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.7 The Chain Rule A pattern might jump out at you. Recognize that each of these functions is a composition, letting g(x) = 1 − x: F1 (x) = f1 (g(x)), where f1 (x) = x2 , F2 (x) = f2 (g(x)), where f2 (x) = x3 , F3 (x) = f3 (g(x)), where f3 (x) = x4 . We’ll come back to this example after giving the formal statements of the Chain Rule; for now, we are just illustrating a pattern. Theorem 18 The Chain Rule Let y = f (u) be a differentiable function of u and let u = g(x) be a differentiable function of x. Then y = f (g(x)) is a differentiable function of x, and y 0 = f 0 (g(x)) · g 0 (x). To help understand the Chain Rule, we return to Example 71. Example 72 Using the Chain Rule Use the Chain Rule to find the derivatives of the following functions, as given in Example 71. Example 71 ended with the recognition that each of the Solution given functions was actually a composition of functions. To avoid confusion, we ignore most of the subscripts here. F1 (x) = (1 − x)2 : We found that y = (1 − x)2 = f (g(x)), where f (x) = x2 and g(x) = 1 − x. To find y 0 , we apply the Chain Rule. We need f 0 (x) = 2x and g 0 (x) = −1. Part of the Chain Rule uses f 0 (g(x)). This means substitute g(x) for x in the equation for f 0 (x). That is, f 0 (x) = 2(1 − x). Finishing out the Chain Rule we have y 0 = f 0 (g(x)) · g 0 (x) = 2(1 − x) · (−1) = −2(1 − x) = 2x − 2. F2 (x) = (1 − x)3 : Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 113 Chapter 2 Derivatives Let y = (1 − x)3 = f (g(x)), where f (x) = x3 and g(x) = (1 − x). We have f 0 (x) = 3x2 , so f 0 (g(x)) = 3(1 − x)2 . The Chain Rule then states y 0 = f 0 (g(x)) · g 0 (x) = 3(1 − x)2 · (−1) = −3(1 − x)2 . F3 (x) = (1 − x)4 : Finally, when y = (1 − x)4 , we have f (x) = x4 and g(x) = (1 − x). Thus f 0 (x) = 4x3 and f 0 (g(x)) = 4(1 − x)3 . Thus y 0 = f 0 (g(x)) · g 0 (x) = 4(1 − x)3 · (−1) = −4(1 − x)3 . Example 72 demonstrated a particular pattern: when f (x) = xn , then y 0 = n · (g(x))n−1 · g 0 (x). This is called the Generalized Power Rule. Theorem 19 Generalized Power Rule Let g(x) be a differentiable function and let n 6= 0 be an integer. Then  n−1 0 d  g(x)n = n · g(x) · g (x). dx This allows us to quickly find the derivative of functions like y = (3x2 − 5x + 7 + sin x)20 . While it may look intimidating, the Generalized Power Rule states that y 0 = 20(3x2 − 5x + 7 + sin x)19 · (6x − 5 + cos x). Treat the derivative–taking process step–by–step. In the example just given, first multiply by 20, then rewrite the inside of the parentheses, raising it all to the 19th power. Then think about the derivative of the expression inside the parentheses, and multiply by that. We now consider more examples that employ the Chain Rule. Example 73 Using the Chain Rule Find the derivatives of the following functions: 1. y = sin 2x 2. y = cos 4x3 3. y = e−x 2 Solution 1. Consider y = sin 2x. Recognize that this is a composition of functions, where f (x) = sin x and g(x) = 2x. Thus y 0 = f 0 (g(x)) · g 0 (x) = cos(2x) · 2 = 2 cos 2x. Notes: Page 114 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.7 The Chain Rule 2. Recognize that y = cos(4x3 ) is the composition of f (x) = cos x and g(x) = 4x3 . Also, recall that  d  cos x = − sin x. dx This leads us to: y 0 = − sin 4x3 · (12x2 ) = −12x2 sin 4x3 . 2 3. Recognize that y = e−x is the composition of f (x) = ex and g(x) = −x2 . Remembering that f 0 (x) = ex , we have 2 2 y 0 = e−x · (−2x) = (−2x)e−x . y Example 74 Using the Chain Rule to find a tangent line Let f (x) = cos x2 . Find the equation of the line tangent to the graph of f at x = 1. 1 0.5 The tangent line goes through the point (1, f (1)) ≈ (1, 0.54) with slope f 0 (1). To find f 0 , we need the Chain Rule. f 0 (x) = − sin(x2 ) · (2x) = −2x sin x2 . Evaluated at x = 1, we have 0 f (1) = −2 sin 1 ≈ −1.68. Thus the equation of the tangent line is Solution x −2 −0.5 y = −1.68(x − 1) + 0.54. The tangent line is sketched along with f in Figure 2.17. The Chain Rule is used often in taking derivatives. Because of this, one can become familiar with the basic process and learn patterns that facilitate finding derivatives quickly. For instance, 2 . −1 Figure 2.17: f (x) = cos x2 sketched along with its tangent line at x = 1.  d  (anything)0 1 · (anything)0 = . ln(anything) = dx anything anything A concrete example of this is 15 x d  3x15 −cos x+ex  e = e3x −cos x+e · (45x14 + sin x + ex ). dx While the derivative may look intimidating at first, look for the pattern. The denominator is the same as what was inside the natural log function; the numerator is simply its derivative. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 115 Chapter 2 Derivatives This pattern recognition process can be applied to lots of functions. In general, instead of writing “anything”, we use u as a generic function of x. We then say d  u e = eu · u0 . dx The following is a short list of how the Chain Rule can be quickly applied to familiar functions. d  n u = n · un−1 · u 0 . dx  d  2. sin u = cos u · u 0 . dx 1 d 1 = − 2 · u 0. 3. dx u u 1.  d  cos u = − sin u · u 0 . dx  d  5. tan u = sec2 u · u 0 . dx d √  1 6. u = √ · u 0. dx 2 u 4. Of course, the Chain Rule can be applied in conjunction with any of the other rules we have already learned. We practice this next. Example 75 Using the Product, Quotient and Chain Rules Find the derivatives of the following functions. 1. f (x) = x5 sin 2x3 2. f (x) = 5x3 . e−x2 Solution 1. We must use the Product and Chain Rules. Do not think that you must be able to “see” the whole answer immediately; rather, just proceed step–by–step.   f 0 (x) = x5 6x2 cos 2x3 + 5x4 sin 2x3 = 6x7 cos 2x3 + 5x4 sin 2x3 . 2. We must employ the Quotient Rule along with the Chain Rule. Again, proceed step–by–step.   2 2 2 e−x 10x4 + 15x2 e−x 15x2 − 5x3 (−2x)e−x 0 = f (x) = 2 e−2x2 e−x2  2 = ex 10x4 + 15x2 . A key to correctly working these problems is to break the problem down into smaller, more manageable pieces. For instance, when using Notes: Page 116 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.7 The Chain Rule the Product and Chain Rules together, just consider the first part of the Product Rule at first: f (x)g 0 (x). Just rewrite f (x), then find g 0 (x). Then move on to the f 0 (x)g(x) part. Don’t attempt to figure out both parts at once. Likewise, using the Quotient Rule, approach the numerator in two steps and handle the denominator after completing that. Only simplify afterward. We can also employ the Chain Rule itself several times, as shown in the next example. Example 76 Using the Chain Rule multiple times Find the derivative of y = tan5 (6x3 − 7x). Recognize that we have the g(x) = tan(6x3 − 7x) func5 tion “inside” the f (x) = x5 function; that is, we have y = tan(6x3 −7x) . We begin using the Generalized Power Rule; in this first step, we do not fully compute the derivative. Rather, we are approaching this step–by– step. 4 y 0 = 5 tan(6x3 − 7x) · g 0 (x). Solution We now find g 0 (x). We again need the Chain Rule; g 0 (x) = sec2 (6x3 − 7x) · (18x2 − 7). Combine this with what we found above to give y 0 = 5 tan(6x3 − 7x) 4 · sec2 (6x3 − 7x) · (18x2 − 7) = (90x2 − 35) sec2 (6x3 − 7x) tan4 (6x3 − 7x). This function is frankly a ridiculous function, possessing no real practical value. It is very difficult to graph, as the tangent function has many vertical asymptotes and 6x3 − 7x grows so very fast. The important thing to learn from this is that the derivative can be found. In fact, it is not “hard;” one must take several simple steps and be careful to keep track of how to apply each of these steps. It is a traditional mathematical exercise to find the derivatives of arbitrarily complicated functions just to demonstrate that it can be done. Just break everything down into smaller pieces. Now, let’s see how chain rule is used in business applications problems. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 117 Chapter 2 Derivatives Example 77 Chain Rule in Business Suppose that the demand function for a certain product is given by q = 5000e−0.01p . Find the rate of change of the revenue at any price p. The revenue function is given by Solution R(p) = p · q = p · 5000e−0.01p = 5000pe−0.01p ⇒   dR = 5000e−0.01p + 5000pe−0.01p (−0.01) = 5000e−0.01p 1 − 0.01p . dp Example 78 Chain Rule in our Food A chain of food stores sells a delicacy prepared from a rare fish species. Suppose that the amount of the delicacy available at any time t during the 2 16-week season is w = 1, 000te−0.02t , 0 ≤ t ≤ 16, where w is the number of pounds t is in weeks. 1. Find dw . dt 2. Suppose that the price per pound is p = 500 − 0.08w. How fast (in dollars per week) is the revenue from the delicacy changing at the end of 8 weeks? Solution 1. 2 2 2 dw = 1000e−0.02t +1000te−0.02t (−0.04t) = 1000e−0.02t (1−0.04t2 ). dt 2. R(t) = p·w = (500−0.08w)w = 500w−0.08w2 ⇒ R0 (t) = (500−0.16w) dw = dt 2 1000e−0.02(8) (1 − 0.04(8)2 ) = −433.74. Therefore, R(8) = (500 − 0.1 · 2224.30) · (−433.74) = −120, 392.71 dollars. 2 Now, when t = 8, w(8) = 1000(8)e−0.02(8) = 2224.30 and Notes: Page 118 Custom Made for Math 1174 at Langara College by P. Anhaouy dw dt . 2.7 The Chain Rule Alternate Chain Rule Notation It is instructive to understand what the Chain Rule “looks like” using dy ” notation instead of y 0 notation. Suppose that y = f (u) is a function “ dx of u, where u = g(x) is a function of x, as stated in Theorem 18. Then, through the composition f ◦ g, we can think of y as a function of x, as y = f (g(x)). Thus the derivative of y with respect to x makes sense; we dy can talk about dx . This leads to an interesting progression of notation: x 0 0 0 y = f (g(x)) · g (x) dy = y 0 (u) · u 0 (x) (since y = f (u) and u = g(x)) dx dy dy du = (using “fractional” notation for the derivative) · dx du dx Here the “fractional” aspect of the derivative notation stands out. On the right hand side, it seems as though the “du” terms cancel out, leaving dy dy = . dx dx It is important to realize that we are not canceling these terms; the derivady tive notation of dx is one symbol. It is equally important to realize that this notation was chosen precisely because of this behavior. It makes applying the Chain Rule easy with multiple variables. For instance, dy =6 dx du =2 dx u y dy =3 du Figure 2.18: A series of gears to demonstrate the Chain Rule. Note dy dy how dx = du · du dx dy dy d d4 = · · . dt d d4 dt where and 4 are any variables you’d like to use. One of the most common ways of “visualizing” the Chain Rule is to consider a set of gears, as shown in Figure 2.18. The gears have 36, 18, and 6 teeth, respectively. That means for every revolution of the x gear, the u gear revolves twice. That is, the rate at which the u gear makes a revolution is twice as fast as the rate at which the x gear makes a revolution. Using the terminology of calculus, the rate of u-change, with respect to x, is du dx = 2. dy Likewise, every revolution of u causes 3 revolutions of y: du = 3. How does y change with respect to x? For each revolution of x, y revolves 6 times; that is, dy dy du = · = 2 · 3 = 6. dx du dx Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 119 Chapter 2 Derivatives We can then extend the Chain Rule with more variables by adding more gears to the picture. It is difficult to overstate the importance of the Chain Rule. So often the functions that we deal with are compositions of two or more functions, requiring us to use this rule to compute derivatives. It is often used in practice when actual functions are unknown. Rather, through measuredy ment, we can calculate du and du dx . With our knowledge of the Chain Rule, dy finding dx is straightforward. In the next section, we use the Chain Rule to justify another differentiation technique. There are many curves that we can draw in the plane that fail the “vertical line test.” For instance, consider x2 + y 2 = 1, which describes the unit circle. We may still be interested in finding slopes of tangent lines to the circle at various points. The next section shows how dy without first “solving for y.” While we can in this instance, we can find dx in many other instances solving for y is impossible. In these situations, implicit differentiation is indispensable. Notes: Page 120 Custom Made for Math 1174 at Langara College by P. Anhaouy Exercises 2.7 Terms and Concepts 12. h(t) = sin4 (2t) 1. T/F: The Chain Rule describes how to evaluate the derivative of a composition of functions. 13. p(t) = cos3 (t2 + 3t + 1) 14. f (x) = x2 sin(5x) 2. T/F: The Generalized Power  n−1 d  g(x)n = n g(x) . dx Rule states that 15. g(t) = cos(t2 + 3t) sin(5t − 7) 2 16. g(t) = cos( 1t )e5t dx dx dt 3. T/F: = · dy dt dy 4. T/F: Taking the derivative of f (x) = x2 sin(5x) requires the use of both the Product and Chain Rules. In Exercises 17 – 20, find the equations of tangent and normal lines to the graph of the function at the given point. Note: the functions here are the same as in Exercises 5 through 8. Problems 17. f (x) = (4x3 − x)10 at x = 0 In Exercises 5 – 26, compute the derivative of the given function. 18. f (t) = (3t − 2)5 at t = 1 19. g(θ) = (sin θ + cos θ)3 at θ = π/2 5. f (x) = (4x3 − x)10 2 6. f (t) = (3t − 2) 5 7. g(θ) = (sin θ + cos θ)3 20. h(t) = e3t +t−1 at t = −1 Review 2 8. h(t) = e3t +t−1  1 4 9. f (x) = x + x 10. f (x) = cos(3x) 11. g(x) = tan(5x) 21. The “wind chill factor” is a measurement of how cold it “feels” during cold, windy weather. Let W (w) be the wind chill factor, in degrees Fahrenheit, when it is 25◦ F outside with a wind of w mph. (a) What are the units of W 0 (w)? (b) What would you expect the sign of W 0 (10) to be? Custom Made for Math 1174 at Langara College by P. Anhaouy Page 121 Chapter 2 Derivatives 2.8 y 2 x −2 2 −2 . Figure 2.19: A graph of the implicit function sin(y) + y 3 = 6 − x3 . Implicit Differentiation dy In the previous sections we learned to find the derivative, dx , or y 0 , when y is given explicitly as a function of x. That is, if we know y = f (x) for some function f , we can find y 0 . For example, given y = 3x2 − 7, we can easily find y 0 = 6x. (Here we explicitly state how x and y are related. Knowing x, we can directly find y.) Sometimes the relationship between y and x is not explicit; rather, it is implicit. For instance, we might know that x2 − y = 4. This equality defines a relationship between x and y; if we know x, we could figure out y. Can we still find y 0 ? In this case, sure; we solve for y to get y = x2 − 4 (hence we now know y explicitly) and then differentiate to get y 0 = 2x. Sometimes the implicit relationship between x and y is complicated. Suppose we are given sin(y)+y 3 = 6−x3 . A graph of this implicit function is given in Figure 2.19. In this case there is absolutely no way to solve for y in terms of elementary functions. The surprising thing is, however, that we can still find y 0 via a process known as implicit differentiation. Implicit differentiation is a technique based on the Chain Rule that is used to find a derivative when the relationship between the variables is given implicitly rather than explicitly (solved for one variable in terms of the other). We begin by reviewing the Chain Rule. Let f and g be functions of x. Then  d  f (g(x)) = f 0 (g(x)) · g 0 (x). dx Suppose now that y = g(x). We can rewrite the above as   d  dy d  f (y)) = f 0 (y)) · y 0 , or f (y)) = f 0 (y) · . (2.1) dx dx dx These equations look strange; the key concept to learn here is that we can find y 0 even if we don’t exactly know how y and x relate. We demonstrate this process in the following example. Example 79 Using Implicit Differentiation Find y 0 given that sin(y) + y 3 = 6 − x3 . We start by taking the derivative of both sides (thus Solution maintaining the equality.) We have :   d  d  sin(y) + y 3 = 6 − x3 . dx dx Notes: Page 122 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.8 Implicit Differentiation The right hand side is easy; it returns −3x2 . The left hand side requires more consideration. We take the derivative term–by–term. Using the technique derived from Equation 2.1 above, we can see that  d  sin y = cos y · y 0 . dx We apply the same process to the y 3 term. d  3 d  3 y = (y) = 3(y)2 · y 0 . dx dx Putting this together with the right hand side, we have Now solve for y 0 . cos(y)y 0 + 3y 2 y 0 = −3x2 . cos(y)y 0 + 3y 2 y 0 = −3x2 .  cos y + 3y 2 y 0 = −3x2 y0 = −3x2 cos y + 3y 2 This equation for y 0 probably seems unusual for it contains both x and y terms. How is it to be used? We’ll address that next. Implicit functions are generally harder to deal with than explicit functions. With an explicit function, given an x value, we have an explicit formula for computing the corresponding y value. With an implicit function, one often has to find x and y values at the same time that satisfy the equation. It is much easier to demonstrate that a given point satisfies the equation than to actually find such a point. √ For instance, we can affirm easily that the point ( 3 6, 0) lies on the graph of the implicit function sin y + y 3 √ = 6 − x3 . Plugging in 0 for y, we see the left hand side is 0. Setting x = 3 6, we see the right hand side is also 0; the equation is satisfied. The following example finds the equation of the tangent line to this function at this point. Example 80 Using Implicit Differentiation to find a tangent line Find the equation of the line tangent to the√curve of the implicitly defined function sin y + y 3 = 6 − x3 at the point ( 3 6, 0). Solution In Example 79 we found that y0 = −3x2 . cos y + 3y 2 Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 123 Chapter 2 Derivatives y 2 x −2 2 −2 . √ We find the slope of the tangent line at the point ( 3 6, 0) by substituting √ √ 3 6 for x and 0 for y. Thus at the point ( 3 6, 0), we have the slope as √ √ −3 3 36 −3( 3 6)2 0 = y = ≈ −9.91. cos 0 + 3 · 02 1 Therefore the equation of the tangent√line to the implicitly defined function sin y + y 3 = 6 − x3 at the point ( 3 6, 0) is √ √ 3 3 y = −3 36(x − 6) + 0 ≈ −9.91x + 18. The curve and this tangent line are shown in Figure 2.20. 3 Figure 2.20: The function sin y +y = 3 6−x and its tangent line at the point √ 3 ( 6, 0). This suggests a general method for implicit differentiation. For the steps below assume y is a function of x. 1. Take the derivative of each term in the equation. Treat the x terms like normal. When taking the derivatives of y terms, the usual rules apply except that, because of the Chain Rule, we need to multiply each term by y 0 . 2. Get all the y 0 terms on one side of the equal sign and put the remaining terms on the other side. 3. Factor out y 0 ; solve for y 0 by dividing. Practical Note: When working by hand, it may be beneficial to use the dy symbol dx instead of y 0 , as the latter can be easily confused for y or y 1 . Example 81 Using Implicit Differentiation Given the implicitly defined function y 3 + x2 y 4 = 1 + 2x, find y 0 . We will take the implicit derivatives term by term. The Solution derivative of y 3 is 3y 2 y 0 . The second term, x2 y 4 , is a little tricky. It requires the Product Rule as it is the product of two functions of x: x2 and y 4 . Its derivative is x2 (4y 3 y 0 ) + 2xy 4 . The first part of this expression requires a y 0 because we are taking the derivative of a y term. The second part does not require it because we are taking the derivative of x2 . The derivative of the right hand side is easily found to be 2. In all, we get: 3y 2 y 0 + 4x2 y 3 y 0 + 2xy 4 = 2. Move terms around so that the left side consists only of the y 0 terms and the right side consists of all the other terms: 3y 2 y 0 + 4x2 y 3 y 0 = 2 − 2xy 4 . Notes: Page 124 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.8 y Factor out y 0 from the left side and solve to get y0 = Implicit Differentiation 2 − 2xy 4 . 3y 2 + 4x2 y 3 x 5 To confirm the validity of our work, let’s find the equation of a tangent line to this function at a point. It is easy to confirm that the point (0, 1) lies on the graph of this function. At this point, y 0 = 2/3. So the equation of the tangent line is y = 2/3(x − 0) + 1. The function and its tangent line are graphed in Figure 2.21. Notice how our function looks much different than other functions we have seen. For one, it fails the vertical line test. Such functions are important in many areas of mathematics, so developing tools to deal with them is also important. 10 −5 −10 . Figure 2.21: A graph of the implicitly defined function y 3 + x2 y 4 = 1 + 2x along with its tangent line at the point (0, 1). Example 82 Using Implicit Differentiation Given the implicitly defined function sin(x2 y 2 ) + y 3 = x + y, find y 0 . Differentiating term by term, we find the most difficulty Solution in the first term. It requires both the Chain and Product Rules.  d  2 2 d  sin(x2 y 2 ) = cos(x2 y 2 ) · x y dx dx  = cos(x2 y 2 ) · x2 (2yy 0 ) + 2xy 2 y 1 = 2(x2 yy 0 + xy 2 ) cos(x2 y 2 ). x We leave the derivatives of the other terms to the reader. After taking the derivatives of both sides, we have −1 −1 2(x2 yy 0 + xy 2 ) cos(x2 y 2 ) + 3y 2 y 0 = 1 + y 0 . We now have to be careful to properly solve for y 0 , particularly because of the product on the left. It is best to multiply out the product. Doing this, we get 1 . Figure 2.22: A graph of the implicitly defined function sin(x2 y 2 ) + y 3 = x + y. 2x2 y cos(x2 y 2 )y 0 + 2xy 2 cos(x2 y 2 ) + 3y 2 y 0 = 1 + y 0 . From here we can safely move around terms to get the following: 2x2 y cos(x2 y 2 )y 0 + 3y 2 y 0 − y 0 = 1 − 2xy 2 cos(x2 y 2 ). Then we can solve for y 0 to get y0 = 1 − 2xy 2 cos(x2 y 2 ) 2x2 y cos(x2 y 2 ) + 3y 2 − 1 . Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 125 Chapter 2 Derivatives y 1 x −1 1 −1 A graph of this implicit function is given in Figure 2.22. It is easy to verify that the points (0, 0), (0, 1) and (0, −1) all lie on the graph. We can find the slopes of the tangent lines at each of these points using our formula for y 0 . At (0, 0), the slope is −1. At (0, 1), the slope is 1/2. At (0, −1), the slope is also 1/2. The tangent lines have been added to the graph of the function in Figure 2.23. . Figure 2.23: A graph of the implicitly defined function sin(x2 y 2 )+y 3 = x+y and certain tangent lines. Quite a few “famous” curves have equations that are given implicitly. We can use implicit differentiation to find the slope at various points on those curves. We investigate two such curves in the next examples. Example 83 Finding slopes of tangent lines to a circle Find √ the slope of the tangent line to the circle x2 + y 2 = 1 at the point (1/2, 3/2). Solution gives: y 1 (1/2, √ y0 = 3/2) x −1 . Taking derivatives, we get 2x + 2yy 0 = 0. Solving for y 0 1 This is a clever formula. Recall that the slope of the line through the origin and the point (x, y) on the circle will be y/x. We have found that the slope of the tangent line to the circle at that point is the opposite reciprocal of y/x, namely, −x/y. Hence these two lines are always perpendicular. √ At the point (1/2, 3/2), we have the tangent line’s slope as −1/2 −1 y0 = √ = √ ≈ −0.577. 3/2 3 −1 Figure 2.24: The unit √ circle with its tangent line at (1/2, 3/2). −x . y √ A graph of the circle and its tangent line at (1/2, 3/2) is given in Figure 2.24, along with a thin dashed line from the origin that is perpendicular to the tangent line. (It turns out that all normal lines to a circle pass through the center of the circle.) This section has shown how to find the derivatives of implicitly defined functions, whose graphs include a wide variety of interesting and unusual shapes. Implicit differentiation can also be used to further our understanding of “regular” differentiation. One hole in our current understanding of derivatives is this: what is Notes: Page 126 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.8 Implicit Differentiation the derivative of the square root function? That is, d √  d 1/2  x = x =? dx dx We allude to a possible solution, as we can write the square root function as a power function with a rational (or, fractional) power. We are then tempted to apply the Power Rule and obtain 1 d 1/2  1 −1/2 x = x = √ . dx 2 2 x The trouble with this is that the Power Rule was initially defined only for positive integer powers, n > 0. While we did not justify this at the time, generally the Power Rule is proved using something called the Binomial Theorem, which deals only with positive integers. The Quotient Rule allowed us to extend the Power Rule to negative integer powers. Implicit Differentiation allows us to extend the Power Rule to rational powers, as shown below. Let y = xm/n , where m and n are integers with no common factors (so m = 2 and n = 5 is fine, but m = 2 and n = 4 is not). We can rewrite this explicit function implicitly as y n = xm . Now apply implicit differentiation. y = xm/n n y = xm d m d n y = x dx dx n−1 0 n·y · y = m · xm−1 m xm−1 (now substitute xm/n for y) n y n−1 m xm−1 = (apply lots of algebra) n (xm/n )n−1 y0 = = m (m−n)/n x n = m m/n−1 x . n The above derivation is the key to the proof extending the Power Rule to rational powers. Using limits, we can extend this once more to include all powers, including irrational (even transcendental!) powers, giving the following theorem. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 127 Chapter 2 Derivatives y Theorem 20 20 Power Rule for Differentiation n Let f (x) = x , where n 6= 0 is a real number. Then f is a differentiable function, and f 0 (x) = n · xn−1 . x −20 20 This theorem allows us to say the derivative of xπ is πxπ−1 . We now apply this final version of the Power Rule in the next example, the second investigation of a “famous” curve. −20 . Figure 2.25: An astroid, traced out by a point on the smaller circle as it rolls inside the larger circle. y 20 Example 84 Using the Power Rule Find the slope of x2/3 + y 2/3 = 8 at the point (8, 8). This is a particularly interesting curve called an astroid. Solution It is the shape traced out by a point on the edge of a circle that is rolling around inside of a larger circle, as shown in Figure 2.25. To find the slope of the astroid at the point (8, 8), we take the derivative implicitly. 2 −1/3 2 −1/3 0 x + y y =0 3 3 2 2 −1/3 0 y y = − x−1/3 3 3 x−1/3 y 0 = − −1/3 y r y y 1/3 0 y = − 1/3 = − 3 . x x (8, 8) x −20 . 20 −20 Figure 2.26: An astroid with a tangent line. Plugging in x = 8 and y = 8, we get a slope of −1. The astroid, with its tangent line at (8, 8), is shown in Figure 2.26. Implicit Differentiation and the Second Derivative We can use implicit differentiation to find higher order derivatives. In dy theory, this is simple: first find dx , then take its derivative with respect to x. In practice, it is not hard, but it often requires a bit of algebra. We demonstrate this in an example. Example 85 Finding the second derivative d2 y Given x + y = 1, find = y 00 . dx2 2 Solution 2 dy We found that y 0 = dx = −x/y in Example 83. To find Notes: Page 128 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.8 Implicit Differentiation y 00 , we apply implicit differentiation to y 0 . d 0 y dx   d x = − (Now use the Quotient Rule.) dx y y(1) − x(y 0 ) =− y2 y 00 = replace y 0 with −x/y: y − x(−x/y) y2 y + x2 /y . =− y2 =− While this is not a particularly simple expression, it is usable. We can see that y 00 > 0 when y < 0 and y 00 < 0 when y > 0. In Section 3.4, we will see how this relates to the shape of the graph. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 129 Exercises 2.8 Terms and Concepts 20. (3x2 + 2y 3 )4 = 2 1. In your own words, explain the difference between implicit functions and explicit functions. 21. (y 2 + 2y − x)2 = 200 2. Implicit differentiation is based on what other differentiation rule? 22. x2 + y = 17 x + y2 3. T/F: Implicit differentiation can be used to find the √ derivative of y = x. 23. sin(x) + y =1 cos(y) + x 4. T/F: Implicit differentiation can be used to find the derivative of y = x3/4 . dy 24. Show that is the same for each of the following dx implicitly defined functions. (a) xy = 1 Problems (b) x2 y 2 = 1 (c) sin(xy) = 1 In Exercises 5 – 12, compute the derivative of the given function. √ 5. f (x) = 1 x+ √ x √ 6. f (x) = 3 x + x2/3 7. f (t) = 8. g(t) = √ 1 − t2 √ In Exercises 25 – 29, find the equation of the tangent line to the graph of the implicitly defined function at the indicated points. As a visual aid, each function is graphed. 25. x2/5 + y 2/5 = 1 (a) At (1, 0). t sin t (b) At (0.1, 0.281) (which does not exactly lie on the curve, but is very close). 9. h(x) = x1.5 y 10. f (x) = xπ + x1.9 + π 1.9 1 x+7 11. g(x) = √ x √ 12. f (t) = 5 t(sec t + et ) dy In Exercises 13 – 25, find using implicit differentiadx tion. 0.5 (0.1, 0.281) x −1 −0.5 0.5 1 −0.5 −1 . 26. x4 + y 4 = 1 4 2 13. x + y + y = 7 2/5 14. x +y 2/5 (a) At (1, 0). √ √ (b) At ( 0.6, 0.8). =1 (c) At (0, 1). 15. cos(x) + sin(y) = 1 16. 17. x = 10 y y 1 y = 10 x 18. x2 e2 + 2y = 5 19. x2 tan y = 50 √ √ ( 0.6, 0.8) 0.5 x −1 −0.5 0.5 −0.5 . Page 130 Custom Made for Math 1174 at Langara College by P. Anhaouy −1 1 27. (x2 + y 2 − 4)3 = 108y 2 In Exercises 30 – 33, an implicitly defined function is d2 y . Note: these are the same problems given. Find dx2 used in Exercises 13 through 16. (a) At (0, 4). √ (b) At (2, − 4 108). y 4 30. x4 + y 2 + y = 7 2 31. x2/5 + y 2/5 = 1 x −4 −2 4 2 32. cos x + sin y = 1 −2 √ (2, − 4 108) −4 . 33. x = 10 y 28. (x2 + y 2 + x)2 = x2 + y 2 (a) At (0, 1). √   3 3 3 (b) At − , . 4 4 y ( √ − 34 , 3 4 3 ) 1 x −1 −2 −1 . 29. (x − 2)2 + (y − 3)2 = 9 √   7 6+3 3 (a) At , . 2 2 √   4+3 3 3 , . (b) At 2 2 y 6 ( √ 3 3.5, 6+32 4 2 ( 4+3√3 2 . ) 2 ) , 1.5 4 x 6 Custom Made for Math 1174 at Langara College by P. Anhaouy Page 131 Chapter 2 Derivatives 2.9 Derivatives of Exponential and Logarithmic Functions Derivative of Inverse Function y (1, 1.5) 1 (1.5, 1) (−0.5, 0.375) x −1 1 (0.375, −0.5) −1 . Figure 2.27: A function f along with its inverse f −1 . (Note how it does not matter which function we refer to as f ; the other is f −1 .) y 1 (a, b) Recall that a function y = f (x) is said to be one to one if it passes the horizontal line test; that is, for two different x values x1 and x2 , we do not have f (x1 ) = f (x2 ). In some cases the domain of f must be restricted so that it is one to one. For instance, consider f (x) = x2 . Clearly, f (−1) = f (1), so f is not one to one on its regular domain, but by restricting f to (0, ∞), f is one to one. Now recall that one to one functions have inverses. That is, if f is one to one, it has an inverse function, denoted by f −1 , such that if f (a) = b, then f −1 (b) = a. The domain of f −1 is the range of f , and vice-versa. For ease of notation, we set g = f −1 and treat g as a function of x. Since f (a) = b implies g(b) = a, when we compose f and g we get a nice result:  f g(b) = f (a) = b.   In general, f g(x) = x and g f (x) = x. This gives us a convenient way to check if two functions are inverses of each other: compose them and if the result is x, then they are inverses (on the appropriate domains.) When the point (a, b) lies on the graph of f , the point (b, a) lies on the graph of g. This leads us to discover that the graph of g is the reflection of f across the line y = x. In Figure 2.27 we see a function graphed along with its inverse. See how the point (1, 1.5) lies on one graph, whereas (1.5, 1) lies on the other. Because of this relationship, whatever we know about f can quickly be transferred into knowledge about g. For example, consider Figure 2.28 where the tangent line to f at the point (a, b) is drawn. That line has slope f 0 (a). Through reflection across y = x, we can see that the tangent line to g at the point (b, a) should have 1 1 slope 0 . This then tells us that g 0 (b) = 0 . f (a) f (a) Consider: x −1 1 (b, a) . −1 Figure 2.28: Corresponding tangent lines drawn to f and f −1 . Information about f Information about g = f −1 (−0.5, 0.375) lies on f Slope of tangent line to f at x = −0.5 is 3/4 (0.375, −0.5) lies on g Slope of tangent line to g at x = 0.375 is 4/3 f 0 (−0.5) = 3/4 g 0 (0.375) = 4/3 Notes: Page 132 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.9 Derivatives of Exponential and Logarithmic Functions We have discovered a relationship between f 0 and g 0 in a mostly graphical way. We can realize this relationship analytically as well. Let y = g(x), where again g = f −1 . We want to find y 0 . Since y = g(x), we know that f (y) = x. Using the Chain Rule and Implicit Differentiation, take the derivative of both sides of this last equality.  d   d  f (y) = x dx dx f 0 (y) · y 0 = 1 1 y0 = 0 f (y) 1 y0 = 0 f (g(x)) This leads us to the following theorem. Theorem 21 Derivatives of Inverse Functions Let f be differentiable and one to one on an open interval I, where f 0 (x) 6= 0 for all x in I, let J be the range of f on I, let g be the inverse function of f , and let f (a) = b for some a in I. Then g is a differentiable function on J, and in particular, 0 • f −1 (b) = g 0 (b) = 0 • f −1 (x) = g 0 (x) = 1 f 0 (a) 1 f 0 (g(x)) The results of Theorem 21 are not trivial; the notation may seem confusing at first. Careful consideration, along with examples, should earn understanding. Derivatives of Exponential Functions Recall that the derivative of the natural exponential function is give by d x (e ) = ex . It deserves it own ”box.” dx Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 133 Chapter 2 Derivatives Theorem 22 tions Derivative of Natural Exponential Func- Let f (x) = ex . Then f is differentiable for all real numbers and f 0 (x) = ex . The Chain Rule also has theoretic value. That is, it can be used to find the derivatives of functions that we have not yet learned as we do in the following example. Example 86 The Chain Rule and exponential functions Use the Chain Rule to find the derivative of y = ax where a > 0, a 6= 1 is constant. We only know how to find the derivative of one expoSolution nential function: y = ex ; this problem is asking us to find the derivative of functions such as y = 2x . This can be accomplished by rewriting ax in terms of e. Recalling that x e and ln x are inverse functions, we can write a = eln a and so x y = ax = eln(a ) . By the exponent property of logarithms, we can “bring down” the power to get y = ax = ex(ln a) . The function is now the composition y = f (g(x)), with f (x) = ex and g(x) = x(ln a). Since f 0 (x) = ex and g 0 (x) = ln a, the Chain Rule gives y 0 = ex(ln a) · ln a. Recall that the ex(ln a) term on the right hand side is just ax , our original function. Thus, the derivative contains the original function itself. We have y 0 = y · ln a = ax · ln a. The Chain Rule, coupled with the derivative rule of ex , allows us to find the derivatives of all exponential functions. The previous example produced a result worthy of its own “box.” Notes: Page 134 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.9 Theorem 23 Derivatives of Exponential and Logarithmic Functions Derivatives of Exponential Functions x Let f (x) = a , for a > 0, a 6= 1. Then f is differentiable for all real numbers and f 0 (x) = ln a · ax . Note that Chain rule can be used with derivatives of the exponential functions. d u d u (e ) = eu · u0 and (a ) = ln a · au · u0 dx dx Example 87 Finding the derivative of y = ax Use Theorem 23 and Chain rule to compute derivative of each of the following functions 2 2. y = 2cos 4x 1. y = esin 2x 3. y = 5−x Solution 1. We have 2. We have 3. We have y 0 = esin 2x · cos(2x) · 2. y 0 = ln 2 · 2cos 4x · (− sin 4x) · (4). 2 y 0 = ln 5 · 5−x · (−2x). Derivatives of Logarithmic Functions Here, we find derivatives of logarithmic functions, which are the inverses of the exponential functions. We start with the natural logarithmic function, g(x) = ln x, which is the inverse of the exponential function f (x) = ex . In fact, we have g(x) = f −1 (x) = ln x. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 135 Chapter 2 Derivatives Finding the derivative of y = ln x  d ln x . Use Theorem 21 to compute dx Example 88 View y = ln x as the inverse of y = ex . Therefore, using our standard notation, let f (x) = ex and g(x) = ln x. We wish to find g 0 (x). Theorem 21 gives: Solution g 0 (x) = 1 1 1 = ln x = . f 0 (g(x)) e x This example produced a result worthy of its own “box.” Theorem 24 tion Derivatives of Natural Logarithmic Func- Let f (x) = ln x. Then f is differentiable for all x > 0 and f 0 (x) = 1 . x d 1 It also can be shown that (loga x) = , where this is a logarithdx x ln a mic function with based a. Theorem 25 Derivatives of Logarithmic Functions Let f (x) = loga x, for a > 0, a 6= 1. Then f is differentiable for all x > 0 and 1 f 0 (x) = . x ln a Note that Chain rule can be used with derivatives of the logarithmic functions. d 1 d 1 (ln u) = · u0 and (loga u) = · u0 . dx u dx u ln a Notes: Page 136 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.9 Derivatives of Exponential and Logarithmic Functions Example 89  d ln(4x3 − 2x2 ) . Compute dx Solution We have  1 d ln(4x3 − 2x2 ) = 3 · (12x2 − 4x). dx 4x − 2x2 Example 90 Using the Product Rule with a product of three functions Let y = x3 ln x cos x. Find y 0 . We have a product of three functions while the Product Solution Rule only specifies how to handle a product of two functions. Our method of handling this problem is to simply group the latter two functions together, and consider y = x3 ln x cos x . Following the Product Rule, we have 0  y 0 = (x3 ) ln x cos x + 3x2 ln x cos x 0 To evaluate ln x cos x , we apply the Product Rule again:   1 cos x + 3x2 ln x cos x x 3 31 = x ln x(− sin x) + x cos x + 3x2 ln x cos x x = (x3 ) ln x(− sin x) + Example 91 Using the Product, Quotient and Chain Rules x cos(x−2 ) − sin2 (e4x ) . Find the derivative of f (x) = ln(x2 + 5x4 ) This function likely has no practical use outside of demonSolution strating derivative skills. The answer is given below without simplification. It employs the Quotient Rule, the Product Rule, and the Chain Rule three times. f 0 (x) =   h i   ln(x2 + 5x4 ) · x · (− sin(x−2 )) · (−2x−3 ) + 1 · cos(x−2 ) − 2 sin(e4x ) · cos(e4x ) · (4e4x )    3 − x cos(x−2 ) − sin2 (e4x ) · 2x+20x x2 +5x4 .  ln(x2 + 5x4 ) 2 Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 137 Chapter 2 Derivatives Example 92 Using the Product Rule Find the derivatives of the following functions. 1. f (x) = x ln x 2. g(x) = x ln x − x. Recalling that the derivative of ln x is 1/x, we use the Product Rule to find our answers.  d  x ln x = x · 1/x + 1 · ln x = 1 + ln x. 1. dx Solution 2. Using the result from above, we compute  d  x ln x − x = 1 + ln x − 1 = ln x. dx This seems significant; if the natural log function ln x is an important function (it is), it seems worthwhile to know a function whose derivative is ln x. We have found one. (We leave it to the reader to find another; a correct answer will be very similar to this one.) y 4 Logarithmic Differentiation 3 Consider the function y = xx ; it is graphed in Figure 2.29. It is well– defined for x > 0 and we might be interested in finding equations of lines tangent and normal to its graph. How do we take its derivative? The function is not a power function: it has a “power” of x, not a constant. It is not an exponential function: it has a “base” of x, not a constant. A differentiation technique known as logarithmic differentiation becomes useful here. The basic principle is this: take the natural log of both sides of an equation y = f (x), then use implicit differentiation to find y 0 . We demonstrate this in the following example. 2 1 . x 1 2 Figure 2.29: A plot of y = xx . Example 93 Using Logarithmic Differentiation Given y = xx , use logarithmic differentiation to find y 0 . Solution As suggested above, we start by taking the natural log Notes: Page 138 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.9 Derivatives of Exponential and Logarithmic Functions of both sides then applying implicit differentiation. y = xx ln(y) = ln(xx ) y 4 (apply logarithm rule) ln(y) = x ln x (now use implicit differentiation)   d  d  ln(y) = x ln x dx dx y0 1 = ln x + x · y x y0 = ln x + 1 y  y 0 = y ln x + 1 (substitute y = xx )  y 0 = xx ln x + 1 . 3 2 1 . x 1 2 Figure 2.30: A graph of y = xx and its tangent line at x = 1.5. To “test” our answer, let’s use it to find the equation of the tangent line at x = 1.5. The point on the graph our tangent line must pass through is (1.5, 1.51.5 ) ≈ (1.5, 1.837). Using the equation for y 0 , we find the slope as  y 0 = 1.51.5 ln 1.5 + 1 ≈ 1.837(1.405) ≈ 2.582. Thus the equation of the tangent line is y = 1.6833(x−1.5)+1.837. Figure 2.25 graphs y = xx along with this tangent line. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 139 Exercises 2.9 Terms and Concepts 1. T/F: Every function has an inverse. 14. f (x) = 6e3x Point= (0, 6) 0 Evaluate f −1 (6) 2. In your own words explain what it means for a function to be “one to one.” In Exercises 15 – 25, compute the derivative of the given function. 3. If (1, 10) lies on the graph of y = f (x), what can be said about the graph of y = f −1 (x)? 15. f (x) = ln(cos x) 16. f (x) = ln(x2 ) 0 4. If (1, 10) lies on the graph of y = f (x) and f (1) = 5, what can be said about y = f −1 (x)? 17. f (x) = 2 ln(x) 18. g(r) = 4r Problems 19. g(t) = 5cos t In Exercises 5 – 8, verify that the given functions are inverses. 20. f (t) = ln tet 5. f (x) = 2x + 6 and g(x) = 21 x − 3 21. g(t) = 152 6. f (x) = x2 + 6x + 11, x ≥ 3 and √ g(x) = x − 2 − 3, x ≥ 2 22. m(w) = 3w 2w 3 7. f (x) = , x 6= 5 and x−5 3 + 5x , x 6= 0 g(x) = x 23. m(w) = 3w + 1 2w 8. f (x) = x+1 , x 6= 1 and g(x) = f (x) x−1 In Exercises 9 – 14, an invertible function f (x) is given along with a point that 0 lies on its graph. Using Theorem 21, evaluate f −1 (x) at the indicated value. 9. f (x) = 5x + 10 Point= (2, 20) 0 Evaluate f −1 (20) 2 24. f (x) = 3x + x 2x2 25. h(t) = 2t + 3 3t + 2  d ln(kx) two ways: dx (a) Using the Chain Rule, and 26. Compute (b) by first using the logarithm rule ln(ab) = ln a + ln b, then taking the derivative.  d ln(xk ) two ways: dx (a) Using the Chain Rule, and 27. Compute 10. f (x) = x2 − 2x + 4, x ≥ 1 Point= (3, 7) 0 Evaluate f −1 (7) (b) by first using the logarithm rule ln(ap ) = p ln a, then taking the derivative. 11. f (x) = sin 2x, −π/4 ≤ x ≤ π/4 √ Point= (π/6, 3/2) 0 √ Evaluate f −1 ( 3/2) In Exercises 28 – 33, use logarithmic differentiation to dy find , then find the equation of the tangent line at dx the indicated x–value. 12. f (x) = x3 − 6x2 + 15x − 2 Point= (1, 8) 0 Evaluate f −1 (8) 28. y = (1 + x)1/x , 1 ,x≥0 1 + x2 Point= (1, 1/2) 0 Evaluate f −1 (1/2) 13. f (x) = 2 x=1 29. y = (2x)x , x=1 xx , x+1 x=1 30. y = 31. y = xsin(x)+2 , Page 140 Custom Made for Math 1174 at Langara College by P. Anhaouy x = π/2 32. y = x+1 , x+2 33. y = (x + 1)(x + 2) , (x + 3)(x + 4) Review dy , where x2 y − y 2 x = 1. 34. Find dx x=1 x=0 35. Find the equation of the line tangent to the graph of x2 + y 2 + xy = 7 at the point (1, 2). 36. Let f (x) = x3 + x. f (x + s) − f (x) Evaluate lim . s→0 s Custom Made for Math 1174 at Langara College by P. Anhaouy Page 141 Chapter 2 Derivatives 2.10 1 In the next example we apply Theorem 21 to the arcsine function. x Example 94 Finding the derivative of an inverse trigonometric function Let y = arcsin x = sin−1 x. Find y 0 using Theorem 21. y . √ Derivatives of Inverse Trigonometric Functions 1 − x2 Figure 2.31: A right triangle defined by y = sin−1 (x/1) with the length of the third leg found using the Pythagorean Theorem. Adopting our previously defined notation, let g(x) = Solution arcsin x and f (x) = sin x. Thus f 0 (x) = cos x. Applying the theorem, we have g 0 (x) = y 1 ( π3 , √ 3 2 ) x − π2 − π4 y = sin x π 4 π 2 −1 . π 2 √ ( 23 , π3 ) −1 . This last expression is not immediately illuminating. Drawing a figure will help, as shown in Figure 2.31. Recall that the sine function can be viewed as taking in an angle and returning a ratio of sides of a right triangle, specifically, the ratio “opposite over hypotenuse.” This means that the arcsine function takes as input a ratio of sides and returns an angle. The equation y = arcsin x can be rewritten as y = arcsin(x/1); that is, consider a right triangle where the hypotenuse has length 1 and the side opposite of the angle with measure y has length x. This means √ the final side has length 1 − x2 , using the Pythagorean Theorem. Therefore cos(sin−1 x) = cos y = π 4 y = sin−1 x 1 1 = . f 0 (g(x)) cos(arcsin x) − π2 Figure 2.32: Graphs of sin x and sin−1 x along with corresponding tangent lines. 1 − x2 /1 = √ 1 − x2 , resulting in  d 1 arcsin x = g 0 (x) = √ . dx 1 − x2 1 − π4 √ Remember that the input x of the arcsine function is a ratio of a side of a right triangle to its hypotenuse; the absolute value of this ratio will never be greater than 1. Therefore the inside of the square root will never be negative. In order to make y = sin x one to one, we restrict its domain to [−π/2, π/2]; on this domain, the range is [−1, 1]. Therefore the domain of y = arcsin x is [−1, 1] and the range is [−π/2, π/2]. When x = ±1, note how the derivative of the arcsine function is undefined; this corresponds to the fact that as x → ±1, the tangent lines to arcsine approach vertical lines with undefined slopes. Notes: Page 142 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.10 Derivatives of Inverse Trigonometric Functions In Figure 2.32 we see f (x) = sin x and f −1 = sin−1 x graphed √ on their respective domains. The line tangent to sin x at the point (π/3, 3/2) has slope cos π/3 = 1/2. The slope of the corresponding point on sin−1 x, the √ 1 1 1 1 = p = p = point ( 3/2, π/3), is q = 2, √ 1/2 1 − 3/4 1/4 1 − ( 3/2)2 verifying yet again that at corresponding points, a function and its inverse have reciprocal slopes. Using similar techniques, we can find the derivatives of all the inverse trigonometric functions. In Figure 2.33 we show the restrictions of the domains of the standard trigonometric functions that allow them to be invertible. Function Domain Range Inverse Function −1 Domain Range sin x [−π/2, π/2] [−1, 1] sin x [−1, 1] [−π/2, π/2] cos x [0, π] [−1, 1] cos−1 (x) [−1, 1] [0, π] (−∞, ∞) −1 (−∞, ∞) (−π/2, π/2) tan x csc x sec x cot x (−π/2, π/2) [−π/2, 0) ∪ (0, π/2] [0, π/2) ∪ (π/2, π] (−∞, −1] ∪ [1, ∞) (−∞, −1] ∪ [1, ∞) (0, π) (−∞, ∞) tan (x) −1 csc sec x (−∞, −1] ∪ [1, ∞) [−π/2, 0) ∪ (0, π/2] −1 (x) (−∞, −1] ∪ [1, ∞) [0, π/2) ∪ (π/2, π] −1 (x) (−∞, ∞) (0, π) cot Figure 2.33: Domains and ranges of the trigonometric and inverse trigonometric functions. Theorem 26 tions Derivatives of Inverse Trigonometric Func- The inverse trigonometric functions are differentiable on all open sets contained in their domains (as listed in Figure 2.33) and their derivatives are as follows: 1 d sin−1 (x) = √ dx 1 − x2  1 d √ sec−1 (x) = 2. dx |x| x2 − 1  d 1 3. tan−1 (x) = dx 1 + x2 1.   d 1 cos−1 (x) = − √ dx 1 − x2  d 5. csc−1 (x) = dx 1 − √ |x| x2 − 1  d 1 cot−1 (x) = − 6. dx 1 + x2 4. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 143 Chapter 2 Derivatives Example 95 Finding the derivative of an inverse tangent function  d Compute sin−1 (x3 + x2 ) . dx Solution Using Theorem 26 and Chain Rule gives:   1 d sin−1 (x3 + x2 ) = p · 3x2 + 2x . 3 2 2 dx 1 − (x + x ) Example 96 Finding the derivative of an inverse tangent function  d Compute tan−1 (ln x) . dx Solution Using Theorem 26 and Chain Rule gives:    1 d 1 −1 tan (ln x) = · . dx 1 + (ln x)2 x Example 97 Finding the derivative of an inverse tangent function  d Compute cos−1 (ex ) . dx Solution Using Theorem 26 and Chain Rule gives:  −1 d cos−1 (ex ) = √ · ex . dx 1 − e2x Example 98 Finding the derivative of an inverse tangent function dy π Compute if x and y are related in the equation tan−1 x + tan−1 y = . dx 2 Solution Implicitly differentiating both sides of the equation above gives: 1 1 dy dy 1 + y2 + · = 0 ⇒ = − . 1 + x2 1 + y 2 dx dx 1 + x2 We restate the most important of these in the following theorem, intended to be a reference for further work. Notes: Page 144 Custom Made for Math 1174 at Langara College by P. Anhaouy 2.10 Theorem 27 Functions Derivatives of Inverse Trigonometric Functions Glossary of Derivatives of Elementary Let u and v be differentiable functions, and let a, c and n be real numbers, a > 0, n 6= 0.  d cu = cu0 dx  d 3. u · v = uv 0 + u0 v dx  d u(v) = u0 (v)v 0 5. dx d  7. x =1 dx d 1 1 9. =− 2 dx x x d x 11. e = ex dx  1 d ln x = 13. dx x  d 15. sin x = cos x dx  d 17. csc x = − csc x cot x dx  d 19. tan x = sec2 x dx  1 d 21. sin−1 x = √ dx 1 − x2 1. 23. 25.  1 d csc−1 x = − √ dx |x| x2 − 1  d 1 tan−1 x = dx 1 + x2 2.  d u ± v = u0 ± v 0 dx d u  u0 v − uv 0 = dx v v2 d  6. c =0 dx d n 8. x = nxn−1 dx 1 d √  x = √ 10. dx 2 x 4. d x a = ln a · ax dx  1 1 d loga x = · 14. dx ln a x  d 16. cos x = − sin x dx  d sec x = sec x tan x 18. dx  d 20. cot x = − csc2 x dx  d 1 22. cos−1 x = − √ dx 1 − x2 12. 24. 26.  1 d √ sec−1 x = dx |x| x2 − 1  d 1 cot−1 x = − dx 1 + x2 Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 145 Exercises 2.10 Terms and Concepts In Exercises 10 – 12, compute the derivative of the given function in two ways: Problems (a) By simplifying first, then taking the derivative, and In Exercises 1 – 10, compute the derivative of the given function. (b) by using the Chain Rule first then simplifying. −1 1. h(t) = sin (2t) −1 2. f (t) = sec 10. f (x) = sin(sin−1 x) (2t) −1 3. g(x) = tan Verify that the two answers are the same. 11. f (x) = tan−1 (tan x) (2x) 4. f (x) = x sin−1 x 5. g(t) = sin t cos−1 t sin−1 x cos−1 x √ 7. g(x) = tan−1 ( x) 6. h(x) = 12. f (x) = sin(cos−1 x) In Exercises 13 – 14, find the equation of the line tangent to the graph of f at the indicated x value. 13. f (x) = sin−1 x 14. f (x) = cos−1 (2x) 8. f (x) = sec−1 (1/x) 9. f (x) = sin(sin−1 x) Review Page 146 Custom Made for Math 1174 at Langara College by P. Anhaouy √ at x= 2 2 √ at x= 3 4 3: The Graphical Behaviour of Functions 3.1 y Extreme Values Given any quantity described by a function, we are often interested in the largest and/or smallest values that quantity attains. For instance, if a function describes the speed of an object, it seems reasonable to want to know the fastest/slowest the object traveled. If a function describes the value of a stock, we might want to know how the highest/lowest values the stock attained over the past year. We call such values extreme values. 4 2 ( .−2 Definition 16 −1 Extreme Values 4 2. f (c) is the maximum (also, absolute maximum) of f on I if f (c) ≥ f (x) for all x in I. 2 The maximum and minimum values are the extreme values, or extrema, of f on I. [ −2 . −1 x ] x 2 1 2 (b) Consider Figure 3.1. The function displayed in (a) has a maximum, but no minimum, as the interval over which the function is defined is open. In (b), the function has a minimum, but no maximum; there is a discontinuity in the “natural” place for the maximum to occur. Finally, the function shown in (c) has both a maximum and a minimum; note that the function is continuous and the interval on which it is defined is closed. It is possible for discontinuous functions defined on an open interval to have both a maximum and minimum value, but we have just seen examples where they did not. On the other hand, continuous functions on a closed interval always have a maximum and minimum value. y 4 2 [ −2 . Let f be a continuous function defined on a closed interval I. Then f has both a maximum and minimum value on I. ] 1 y 1. f (c) is the minimum (also, absolute minimum) of f on I if f (c) ≤ f (x) for all x in I. The Extreme Value Theorem x 2 (a) Let f be defined on an interval I containing c. Theorem 28 ) 1 −1 (c) Figure 3.1: Graphs of functions with and without extreme values. Note: The extreme values of a function are “y” values, values the function attains, not the input values. Chapter 3 The Graphical Behaviour of Functions This theorem states that f has extreme values, but it does not offer any advice about how/where to find these values. The process can seem to be fairly easy, as the next example illustrates. After the example, we will draw on lessons learned to form a more general and powerful method for finding extreme values. Example 99 Approximating extreme values Consider f (x) = 2x3 − 9x2 on I = [−1, 5], as graphed in Figure 3.2. Approximate the extreme values of f . y (5, 25) 20 (0, 0) x −1 5 (−1, −11) −20 (3, −27) . Figure 3.2: A graph of f (x) = 2x3 − 9x2 as in Example 99. The graph is drawn in such a way to draw attention to Solution certain points. It certainly seems that the smallest y value is −27, found when x = 3. It also seems that the largest y value is 25, found at the endpoint of I, x = 5. We use the word seems, for by the graph alone we cannot be sure the smallest value is not less than −27. Since the problem asks for an approximation, we approximate the extreme values to be 25 and −27. Notice how the minimum value came at “the bottom of a hill,” and the maximum value came at an endpoint. Also note that while 0 is not an extreme value, it would be if we narrowed our interval to [−1, 4]. The idea that the point (0, 0) is the location of an extreme value for some interval is important, leading us to a definition. Definition 17 mum Relative Minimum and Relative Maxi- Let f be defined on an interval I containing c. 1. If there is an open interval containing c such that f (c) is the minimum value, then f (c) is a relative minimum of f . We also say that f has a relative minimum at (c, f (c)). Note: The terms local minimum and local maximum are often used as synonyms for relative minimum and relative maximum. 2. If there is an open interval containing c such that f (c) is the maximum value, then f (c) is a relative maximum of f . We also say that f has a relative maximum at (c, f (c)). The relative maximum and minimum values comprise the relative extrema of f . We briefly practice using these definitions. Notes: Page 148 Custom Made for Math 1174 at Langara College by P. Anhaouy 3.1 Extreme Values Example 100 Approximating relative extrema Consider f (x) = (3x4 − 4x3 − 12x2 + 5)/5, as shown in Figure 3.3. Approximate the relative extrema of f . At each of these points, evaluate f 0. y We still do not have the tools to exactly find the relative extrema, but the graph does allow us to make reasonable approximations. It seems f has relative minima at x = −1 and x = 2, with values of f (−1) = 0 and f (2) = −5.4. It also seems that f has a relative maximum at the point (0, 1). 6 Solution 4 2 x −1 −2 We approximate the relative minima to be 0 and −5.4; we approximate the relative maximum to be 1. It is straightforward to evaluate f 0 (x) = 51 (12x3 − 12x2 − 24x) at x = 0, 1 and 2. In each case, f 0 (x) = 0. 3 −2 −6 . Figure 3.3: A graph of f (x) = (3x4 − 4x3 − 12x2 + 5)/5 as in Example 100. y The figure implies that f does not have any relative maxima, but has a relative minimum at (1, 2). In fact, the graph suggests that not only is this point a relative minimum, y = f (1) = 2 the minimum value of the function. 3 We compute f 0 (x) = 23 (x − 1)−1/3 . When x = 1, f 0 is undefined. 1 Solution Definition 18 2 −4 Example 101 Approximating relative extrema Approximate the relative extrema of f (x) = (x−1)2/3 +2, shown in Figure 3.4. At each of these points, evaluate f 0 . What can we learn from the previous two examples? We were able to visually approximate relative extrema, and at each such point, the derivative was either 0 or it was not defined. This observation holds for all functions, leading to a definition and a theorem. 1 2 x . 1 2 Figure 3.4: A graph of f (x) = (x − 1)2/3 + 2 as in Example 101. Critical Numbers and Critical Points Let f be defined at c. The value c is a critical number (or critical value) of f if f 0 (c) = 0 or f 0 (c) is not defined. If c is a critical number of f , then the point (c, f (c)) is a critical point of f . Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 149 Chapter 3 The Graphical Behaviour of Functions y Theorem 29 1 Relative Extrema and Critical Points Let a function f have a relative extrema at the point (c, f (c)). Then c is a critical number of f . x −1 1 −1 . Figure 3.5: A graph of f (x) = x3 which has a critical value of x = 0, but no relative extrema. Be careful to understand that this theorem states “All relative extrema occur at critical points.” It does not say “All critical numbers produce relative extrema.” For instance, consider f (x) = x3 . Since f 0 (x) = 3x2 , it is straightforward to determine that x = 0 is a critical number of f . However, f has no relative extrema, as illustrated in Figure 3.5. Theorem 28 states that a continuous function on a closed interval will have absolute extrema, that is, both an absolute maximum and an absolute minimum. These extrema occur either at the endpoints or at critical values in the interval. We combine these concepts to offer a strategy for finding extrema. Key Idea 1 Finding Extrema on a Closed Interval Let f be a continuous function defined on a closed interval [a, b]. To find the maximum and minimum values of f on [a, b]: 1. Evaluate f at the endpoints a and b of the interval. 2. Find the critical numbers of f in [a, b]. 3. Evaluate f at each critical number. 4. The absolute maximum of f is the largest of these values, and the absolute minimum of f is the least of these values. y 40 We practice these ideas in the next examples. Example 102 Finding extreme values Find the extreme values of f (x) = 2x3 + 3x2 − 12x on [0, 3], graphed in Figure 3.6. 20 x . 1 2 3 We follow the steps outlined in Key Idea 1. We first Solution evaluate f at the endpoints: Figure 3.6: A graph of f (x) = 2x3 + 3x2 − 12x on [0, 3] as in Example 102. f (0) = 0 Notes: Page 150 Custom Made for Math 1174 at Langara College by P. Anhaouy and f (3) = 45. Next, we find the critical values of f on [0, 3]. f 0 (x) = 6x2 + 6x − 12 = 6(x + 2)(x − 1); therefore the critical values of f are x = −2 and x = 1. Since x = −2 does not lie in the interval [0, 3], we ignore it. Evaluating f at the only critical number in our interval gives: f (1) = −7. The table in Figure 3.7 gives f evaluated at the “important” x values in [0, 3]. We can easily see the maximum and minimum values of f : the maximum value is 45 and the minimum value is −7. 3.1 Extreme Values x f (x) 0 1 3 0 −7 45 Figure 3.7: Finding the extreme values of f in Example 102. Note that all this was done without the aid of a graph; this work followed an analytic algorithm and did not depend on any visualization. Figure 3.6 shows f and we can confirm our answer, but it is important to understand that these answers can be found without graphical assistance. We practice again. Example 103 Finding extreme values Find the maximum and minimum values of f on [−4, 2], where  (x − 1)2 x ≤ 0 f (x) = . x+1 x>0 Here f is piecewise–defined, but we can still apply Key Solution Idea 1. Evaluating f at the endpoints gives: f (−4) = 25 and f (2) = 3. x f (x) −4 0 2 25 1 3 Figure 3.8: Finding the extreme values of f in Example 103. We now find the critical numbers of f . We have to define f 0 in a piecewise manner; it is  2(x − 1) x < 0 0 f (x) = . 1 x>0 Note that while f is defined for all of [−4, 2], f 0 is not, as the derivative of f does not exist when x = 0. (From the left, the derivative approaches −2; from the right the derivative is 1.) Thus one critical number of f is x = 0. We now set f 0 (x) = 0. When x > 0, f 0 (x) is never 0. When x < 0, 0 f (x) is also never 0. (We may be tempted to say that f 0 (x) = 0 when x = 1. However, this is nonsensical, for we only consider f 0 (x) = 2(x − 1) when x < 0, so we will ignore a solution that says x = 1.) So we have three important x values to consider: x = −4, 2 and 0. Evaluating f at each gives, respectively, 25, 3 and 1, shown in Figure 3.8. Thus the absolute minimum of f is 1; the absolute maximum of f is 25. Our answer is confirmed by the graph of f in Figure 3.9. y 20 10 . x −4 −2 2 Figure 3.9: A graph of f (x) on [−4, 2] as in Example 103. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 151 Chapter 3 The Graphical Behaviour of Functions x f (x) −2 √ − π 0 √ π 2 −0.65 −1 1 −1 −0.65 Example 104 Finding extreme values Find the extrema of f (x) = cos(x2 ) on [−2, 2]. Figure 3.10: Finding the extrema of f (x) = cos(x2 ) in Example 104. y 1 0.5 x −2 −1 1 2 −0.5 . We again use Key Idea 1. Evaluating f at the endpoints Solution of the interval gives: f (−2) = f (2) = cos(4) ≈ −0.6536. We now find the critical values of f . Applying the Chain Rule, we find f 0 (x) = −2x sin(x2 ). Set f 0 (x) = 0 and solve for x to find the critical values of f . We have f 0 (x) = 0 when x = 0 and when sin(x2 ) = 0. In general, sin t = 0 when t = . . . − 2π, −π, 0, π, . . . Thus sin(x2 ) = 0 when x2 = 0, π, 2π, . . . (x2√is always positive so we ignore −π, etc.) So sin(x2 ) = 0 √ when x = 0, ± √π, ± 2π, √. . .. The only values to fall in the given interval of [−2, 2] are − π and π, approximately ±1.77. We again construct a table of important values in √ Figure 3.10. In this example we have 5 values to consider: x = 0, ±2, ± π. From the table it is clear that the maximum value of f on [−2, 2] is 1; the minimum value is −1. The graph in Figure 3.11 confirms our results. We consider one more example. −1 Figure 3.11: A graph of f (x) = cos(x2 ) on [−2, 2] as in Example 104. x f (x) −1 0 1 0 1 0 Figure 3.13: Finding the extrema of the half–circle in Example 105. Example 105 Finding extreme √ values Find the extreme values of f (x) = 1 − x2 . A closed interval is not given, so we find the extreme Solution values of f on its domain. f is defined whenever 1 − x2 ≥ 0; thus the domain of f is [−1, 1]. Evaluating f at either endpoint returns 0. −x . The critical points of Using the Chain Rule, we find f 0 (x) = √ 1 − x2 0 0 f are found when f (x) = 0 or when f is undefined. It is straightforward to find that f 0 (x) = 0 when x = 0, and f 0 is undefined when x = ±1, the endpoints of the interval. The table of important values is given in Figure 3.13. The maximum value is 1, and the minimum value is 0. y 1 x . −1 1 Figure 3.12: A graph of f (x) = √ 1 − x2 on [−1, 1] as in Example 105. Note: We implicitly found the derivative of x2 + y 2 = 1, the unit dy circle, in Example 83 as dx = −x/y. In Example 105, half of the √ unit circle is given as y = f (x) = 1 − x2 . We found f 0 (x) = √−x 2 . Recog- We have seen that continuous functions on closed intervals always have a maximum and minimum value, and we have also developed a technique to find these values. In the next section, we further our study of the information we can glean from “nice” functions with the Mean Value Theorem. On a closed interval, we can find the average rate of change of a function (as we did at the beginning of Chapter 2). We will see that differentiable functions always have a point at which their instantaneous rate of change is same as the average rate of change. This is surprisingly useful, as we’ll see. Notes: 1−x nize that the denominator of this fraction is y; that is, we again found dy f 0 (x) = dx = −x/y. Page 152 Custom Made for Math 1174 at Langara College by P. Anhaouy Exercises 3.1 Terms and Concepts 9. f (x) = x2 p 6 − x2 y 1. Describe what an “extreme value” of a function is in your own words. √ (2, 4 2) 6 4 2. Sketch the graph of a function f on (−1, 1) that has both a maximum and minimum value. 3. Describe the difference between absolute and relative maxima in your own words. 4. Sketch the graph of a function f where f has a relative maximum at x = 1 and f 0 (1) is undefined. 2 x (0, 0) −2 . 2 10. f (x) = sin x y (π/2, 1) 5. T/F: If c is a critical value of a function f , then f has either a relative maximum or relative minimum at x = c. 1 x 4 2 Problems −1 In Exercises 6 – 7, identify each of the marked points as being an absolute maximum or minimum, a relative maximum or minimum, or none of the above. (3π/2, −1) . √ 11. f (x) = x2 4 − x y y ( 16 512 √ 5 , 25 5 10 B G D 2 6. ) 5 E C 6 x 4 2 6 (4, 0) F −2 A . (0, 0) . −2  12. f (x) = y C 2 x2 x5 2 x 4 x≤0 x>0 y B 1 D 7. x 4 2 0.5 E −2 A x . −1 In Exercises 8 – 14, evaluate f 0 (x) at the points indicated in the graph. (0, 0) 0.5 −0.5 . −0.5  13. f (x) = 2 8. f (x) = 2 x +1 1 x2 x x≤0 x>0 y y 1 (0, 2) 2 0.5 1 x −1 (0, 0) 0.5 −0.5 1 x −5 . 5 . −0.5 Custom Made for Math 1174 at Langara College by P. Anhaouy Page 153 14. f (x) = (x − 2)2/3 x 20. f (x) = x2 on [−3, 5]. 21. f (x) = ex cos x on [0, π]. 22. f (x) = ex sin x on [0, π]. x2 + 5 y 6 4 ( 2 6, 1 + ) √ 3 2 3 23. f (x) = (2, 1) ln x x [1, 4]. on x . 5 10 24. f (x) = x2/3 − x on [0, 2]. In Exercises 15 – 24, find the extreme values of the function on the given interval. 15. f (x) = x2 + x + 4 16. f (x) = x3 − on p 4 − x2 19. f (x) = x + 3 x on on [π/4, 2π/3]. on Review [−1, 2]. 9 2 x − 30x + 3 2 17. f (x) = 3 sin x 18. f (x) = x2 on [−2, 2]. [1, 5]. [0, 6]. dy 25. Find dx , where x2 y − y 2 x = 1. 26. Find the equation of the line tangent to the graph of x2 + y 2 + xy = 7 at the point (1, 2). 27. Let f (x) = x3 + x. Evaluate lim s→0 Page 154 Custom Made for Math 1174 at Langara College by P. Anhaouy f (x + s) − f (x) . s 3.2 3.2 The Mean Value Theorem The Mean Value Theorem We motivate this section with the following question: Suppose you leave your house and drive to your friend’s house in a city 100 miles away, completing the trip in two hours. At any point during the trip do you necessarily have to be going 50 miles per hour? In answering this question, it is clear that the average speed for the entire trip is 50 mph (i.e. 100 miles in 2 hours), but the question is whether or not your instantaneous speed is ever exactly 50 mph. More simply, does your speedometer ever read exactly 50 mph?. The answer, under some very reasonable assumptions, is “yes.” Let’s now see why this situation is in a calculus text by translating it into mathematical symbols. First assume that the function y = f (t) gives the distance (in miles) traveled from your home at time t (in hours) where 0 ≤ t ≤ 2. In particular, this gives f (0) = 0 and f (2) = 100. The slope of the secant line connecting the starting and ending points (0, f (0)) and (2, f (2)) is therefore f (2) − f (0) 100 − 0 ∆f = = = 50 mph. ∆t 2−0 2 The slope at any point on the graph itself is given by the derivative f 0 (t). So, since the answer to the question above is “yes,” this means that at some time during the trip, the derivative takes on the value of 50 mph. Symbolically, f (2) − f (0) f 0 (c) = = 50 2−0 for some time 0 ≤ c ≤ 2. How about more generally? Given any function y = f (x) and a range a ≤ x ≤ b does the value of the derivative at some point between a and b have to match the slope of the secant line connecting the points (a, f (a)) (a) have and (b, f (b))? Or equivalently, does the equation f 0 (c) = f (b)−f b−a to hold for some a < c < b? Let’s look at two functions in an example. Example 106 Comparing average and instantaneous rates of change Consider functions f1 (x) = 1 x2 and f2 (x) = |x| Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 155 Chapter 3 The Graphical Behaviour of Functions with a = −1 and b = 1 as shown in Figure 3.14(a) and (b), respectively. Both functions have a value of 1 at a and b. Therefore the slope of the secant line connecting the end points is 0 in each case. But if you look at the plots of each, you can see that there are no points on either graph where the tangent lines have slope zero. Therefore we have found that there is no c in [−1, 1] such that y 2 f 0 (c) = x −1 .. f (1) − f (−1) = 0. 1 − (−1) 1 So what went “wrong”? It may not be surprising to find that the discontinuity of f1 and the corner of f2 play a role. If our functions had been continuous and differentiable, would we have been able to find that special value c? This is our motivation for the following theorem. (a) y 1 Theorem 30 tion 0.5 . x −1 The Mean Value Theorem of Differentia- Let y = f (x) be continuous function on the closed interval [a, b] and differentiable on the open interval (a, b). There exists a value c, a < c < b, such that 1 f 0 (c) = (b) Figure 3.14: A graph of f1 (x) = 1/x2 and f2 (x) = |x| in Example 106. f (b) − f (a) . b−a That is, there is a value c in (a, b) where the instantaneous rate of change of f at c is equal to the average rate of change of f on [a, b]. Note that the reasons that the functions in Example 106 fail are indeed that f1 has a discontinuity on the interval [−1, 1] and f2 is not differentiable at the origin. We will give a proof of the Mean Value Theorem below. To do so, we use a fact, called Rolle’s Theorem, stated here. Theorem 31 Rolle’s Theorem Let f be continuous on [a, b] and differentiable on (a, b), where f (a) = f (b). There is some c in (a, b) such that f 0 (c) = 0. Notes: Page 156 Custom Made for Math 1174 at Langara College by P. Anhaouy 3.2 The Mean Value Theorem Consider Figure 3.15 where the graph of a function f is given, where f (a) = f (b). It should make intuitive sense that if f is differentiable (and hence, continuous) that there would be a value c in (a, b) where f 0 (c) = 0; that is, there would be a relative maximum or minimum of f in (a, b). Rolle’s Theorem guarantees at least one; there may be more. Rolle’s Theorem is really just a special case of the Mean Value Theorem. If f (a) = f (b), then the average rate of change on (a, b) is 0, and the theorem guarantees some c where f 0 (c) = 0. We will prove Rolle’s Theorem, then use it to prove the Mean Value Theorem. Proof of Rolle’s Theorem Let f be differentiable on (a, b) where f (a) = f (b). We consider two cases. Case 1: Consider the case when f is constant on [a, b]; that is, f (x) = f (a) = f (b) for all x in [a, b]. Then f 0 (x) = 0 for all x in [a, b], showing there is at least one value c in (a, b) where f 0 (c) = 0. Case 2: Now assume that f is not constant on [a, b]. The Extreme Value Theorem guarantees that f has a maximal and minimal value on [a, b], found either at the endpoints or at a critical value in (a, b). Since f (a) = f (b) and f is not constant, it is clear that the maximum and minimum cannot both be found at the endpoints. Assume, without loss of generality, that the maximum of f is not found at the endpoints. Therefore there is a c in (a, b) such that f (c) is the maximum value of f . By Theorem 29, c must be a critical number of f ; since f is differentiable, we have that f 0 (c) = 0, completing the proof of the theorem.  We can now prove the Mean Value Theorem. Proof of the Mean Value Theorem Define the function g(x) = f (x) − y f (b) − f (a) x. b−a 5 We know g is differentiable on (a, b) and continuous on [a, b] since f is. We can show g(a) = g(b) (it is actually easier to show g(b) − g(a) = 0, which suffices). We can then apply Rolle’s theorem to guarantee the existence of c ∈ (a, b) such that g 0 (c) = 0. But note that 0 = g 0 (c) = f 0 (c) − hence f 0 (c) = Notes: f (b) − f (a) ; b−a f (b) − f (a) , b−a x −1 . a c 1 b 2 −5 Figure 3.15: A graph of f (x) = x3 − 5x2 +3x+5, where f (a) = f (b). Note the existence of c, where a < c < b, where f 0 (c) = 0. . Custom Made for Math 1174 at Langara College by P. Anhaouy Page 157 Chapter 3 The Graphical Behaviour of Functions which is what we sought to prove.  Going back to the very beginning of the section, we see that the only assumption we would need about our distance function f (t) is that it be continuous and differentiable for t from 0 to 2 hours (both reasonable assumptions). By the Mean Value Theorem, we are guaranteed a time during the trip where our instantaneous speed is 50 mph. This fact is used in practice. Some law enforcement agencies monitor traffic speeds while in aircraft. They do not measure speed with radar, but rather by timing individual cars as they pass over lines painted on the highway whose distances apart are known. The officer is able to measure the average speed of a car between the painted lines; if that average speed is greater than the posted speed limit, the officer is assured that the driver exceeded the speed limit at some time. Note that the Mean Value Theorem is an existence theorem. It states that a special value c exists, but it does not give any indication about how to find it. It turns out that when we need the Mean Value Theorem, existence is all we need. y 40 Example 107 Using the Mean Value Theorem Consider f (x) = x3 + 5x + 5 on [−3, 3]. Find c in [−3, 3] that satisfies the Mean Value Theorem. 20 x −3 −2 −1 1 2 3 Solution −20 . f (3) − f (−3) 84 = = 14. 3 − (−3) 6 −40 Figure 3.16: Demonstrating the Mean Value Theorem in Example 107. The average rate of change of f on [−3, 3] is: We want to find c such that f 0 (c) = 14. We find f 0 (x) = 3x2 + 5. We set this equal to 14 and solve for x. f 0 (x) = 14 3x2 + 5 = 14 x2 = 3 √ x = ± 3 ≈ ±1.732 We have found 2 values c in [−3, 3] where the instantaneous rate of change is equal to the average rate of change; the Mean Value Theorem guaranteed at least one. In Figure 3.16 f is graphed with a dashed line √ representing the average rate of change; the lines tangent to f at x = ± 3 are also given. Note how these lines are parallel (i.e., have the same slope) as the dashed line. Notes: Page 158 Custom Made for Math 1174 at Langara College by P. Anhaouy Exercises 3.2 Terms and Concepts 1. Explain in your own words what the Mean Value Theorem states. In Exercises 11 – 20, a function f (x) and interval [a, b] are given. Check if the Mean Value Theorem can be applied to f on [a, b]; if so, find a value c in [a, b] guaranteed by the Mean Value Theorem. 11. f (x) = x2 + 3x − 1 on [−2, 2]. 2. Explain in your own words what Rolle’s Theorem states. Problems In Exercises 3 – 10, a function f (x) and interval [a, b] are given. Check if Rolle’s Theorem can be applied to f on [a, b]; if so, find c in [a, b] such that f 0 (c) = 0. 12. f (x) = 5x2 − 6x + 8 on [0, 5]. 13. f (x) = p 9 − x2 on [0, 3]. 14. f (x) = √ 25 − x on [0, 9]. 15. f (x) = x2 − 9 on [0, 2]. x2 − 1 16. f (x) = ln x on [1, 5]. 3. f (x) = 6 on [−1, 1]. 17. f (x) = tan x on [−π/4, π/4]. 4. f (x) = 6x on [−1, 1]. 18. f (x) = x3 − 2x2 + x + 1 on [−2, 2]. 5. f (x) = x2 + x − 6 on [−3, 2]. 19. f (x) = 2x3 − 5x2 + 6x + 1 on [−5, 2]. 6. f (x) = x2 + x − 2 on [−3, 2]. 20. f (x) = sin−1 x on [−1, 1]. 7. f (x) = x2 + x on [−2, 2]. 8. f (x) = sin x on [π/6, 5π/6]. Review 21. Find the extreme values of f (x) = x2 − 3x + 9 on [−2, 5]. 9. f (x) = cos x on [0, π]. 22. Describe the critical points of f (x) = cos x. 10. f (x) = 1 on [0, 2]. x2 − 2x + 1 23. Describe the critical points of f (x) = tan x. Custom Made for Math 1174 at Langara College by P. Anhaouy Page 159 Chapter 3 The Graphical Behaviour of Functions 3.3 y 4 2 x . 1 2 Figure 3.17: A graph of a function f used to illustrate the concepts of increasing and decreasing. Increasing and Decreasing Functions Our study of “nice” functions f in this chapter has so far focused on individual points: points where f is maximal/minimal, points where f 0 (x) = 0 or f 0 does not exist, and points c where f 0 (c) is the average rate of change of f on some interval. In this section we begin to study how functions behave between special points; we begin studying in more detail the shape of their graphs. We start with an intuitive concept. Given the graph in Figure 3.17, where would you say the function is increasing? Decreasing? Even though we have not defined these terms mathematically, one likely answered that f is increasing when x > 1 and decreasing when x < 1. We formally define these terms here. Definition 19 Increasing and Decreasing Functions Let f be a function defined on an interval I. 1. f is increasing on I if for every a < b in I, f (a) ≤ f (b). 2. f is decreasing on I if for every a < b in I, f (a) ≥ f (b). A function is strictly increasing when a < b in I implies f (a) < f (b), with a similar definition holding for strictly decreasing. y 2 (b, f(b)) 1 (a, f(a)) x a . 1 b 2 Figure 3.18: Examining the secant line of an increasing function. Informally, a function is increasing if as x gets larger (i.e., looking left to right) f (x) gets larger. Our interest lies in finding intervals in the domain of f on which f is either increasing or decreasing. Such information should seem useful. For instance, if f describes the speed of an object, we might want to know when the speed was increasing or decreasing (i.e., when the object was accelerating vs. decelerating). If f describes the population of a city, we should be interested in when the population is growing or declining. To find such intervals, we again consider secant lines. Let f be an increasing, differentiable function on an open interval I, such as the one shown in Figure 3.18, and let a < b be given in I. The secant line on the graph of f from x = a to x = b is drawn; it has a slope of (f (b) − f (a))/(b − a). But note: Average rate of f (b) − f (a) numerator > 0 slope of the ⇒ change of f on ⇒ ⇒ secant line > 0 b−a denominator > 0 [a, b] is > 0. Notes: Page 160 Custom Made for Math 1174 at Langara College by P. Anhaouy 3.3 Increasing and Decreasing Functions We have shown mathematically what may have already been obvious: when f is increasing, its secant lines will have a positive slope. Now recall the Mean Value Theorem guarantees that there is a number c, where a < c < b, such that f 0 (c) = f (b) − f (a) > 0. b−a Note: Theorem 32 also holds if f 0 (c) = 0 for a finite number of values of c in I. By considering all such secant lines in I, we strongly imply that f 0 (x) ≥ 0 on I. A similar statement can be made for decreasing functions. Our above logic can be summarized as “If f is increasing, then f 0 is probably positive.” Theorem 32 below turns this around by stating “If f 0 is postive, then f is increasing.” This leads us to a method for finding when functions are increasing and decreasing. Theorem 32 Test For Increasing/Decreasing Functions Let f be a continuous function on [a, b] and differentiable on (a, b). 1. If f 0 (c) > 0 for all c in (a, b), then f is increasing on [a, b]. 2. If f 0 (c) < 0 for all c in (a, b), then f is decreasing on [a, b]. 3. If f 0 (c) = 0 for all c in (a, b), then f is constant on [a, b]. Let a and b be in I where f 0 (a) > 0 and f 0 (b) < 0. It follows from the Intermediate Value Theorem that there must be some value c between a and b where f 0 (c) = 0. This leads us to the following method for finding intervals on which a function is increasing or decreasing. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 161 Chapter 3 The Graphical Behaviour of Functions Key Idea 2 Finding Intervals on Which f is Increasing or Decreasing Let f be a differentiable function on an interval I. To find intervals on which f is increasing and decreasing: 1. Find the critical values of f . That is, find all c in I where f 0 (c) = 0 or f 0 is not defined. 2. Use the critical values to divide I into subintervals. 3. Pick any point p in each subinterval, and find the sign of f 0 (p). (a) If f 0 (p) > 0, then f is increasing on that subinterval. (b) If f 0 (p) < 0, then f is decreasing on that subinterval. We demonstrate using this process in the following example. Example 108 Finding intervals of increasing/decreasing Let f (x) = x3 + x2 − x + 1. Find intervals on which f is increasing or decreasing. Using Key Idea 2, we first find the critical values of f . Solution We have f 0 (x) = 3x2 + 2x − 1 = (3x − 1)(x + 1), so f 0 (x) = 0 when x = −1 and when x = 1/3. f 0 is never undefined. Since an interval was not specified for us to consider, we consider the entire domain of f which is (−∞, ∞). We thus break the whole real line into three subintervals based on the two critical values we just found: (−∞, −1), (−1, 1/3) and (1/3, ∞). This is shown in Figure 3.19. f ′ > 0 incr f ′ < 0 decr f ′ > 0 incr . −1 1/3 Figure 3.19: Number line for f in Example 108. We now pick a value p in each subinterval and find the sign of f 0 (p). All we care about is the sign, so we do not actually have to fully compute f 0 (p); pick “nice” values that make this simple. Subinterval 1, (−∞, −1): We (arbitrarily) pick p = −2. We can compute f 0 (−2) directly: f 0 (−2) = 3(−2)2 + 2(−2) − 1 = 7 > 0. We conclude Notes: Page 162 Custom Made for Math 1174 at Langara College by P. Anhaouy 3.3 Increasing and Decreasing Functions that f is increasing on (−∞, −1). Note we can arrive at the same conclusion without computation. For instance, we could choose p = −100. The first term in f 0 (−100), i.e., 3(−100)2 is clearly positive and very large. The other terms are small in comparison, so we know f 0 (−100) > 0. All we need is the sign. Subinterval 2, (−1, 1/3): We pick p = 0 since that value seems easy to deal with. f 0 (0) = −1 < 0. We conclude f is decreasing on (−1, 1/3). Subinterval 3, (1/3, ∞): Pick an arbitrarily large value for p > 1/3 and note that f 0 (p) = 3p2 + 2p − 1 > 0. We conclude that f is increasing on (1/3, ∞). We can verify our calculations by considering Figure 3.20, where f is graphed. The graph also presents f 0 ; note how f 0 > 0 when f is increasing and f 0 < 0 when f is decreasing. One is justified in wondering why so much work is done when the graph seems to make the intervals very clear. We give three reasons why the above work is worthwhile. First, the points at which f switches from increasing to decreasing are not precisely known given a graph. The graph shows us something significant happens near x = −1 and x = 0.3, but we cannot determine exactly where from the graph. One could argue that just finding critical values is important; once we know the significant points are x = −1 and x = 1/3, the graph shows the increasing/decreasing traits just fine. That is true. However, the technique prescribed here helps reinforce the relationship between increasing/decreasing and the sign of f 0 . Once mastery of this concept (and several others) is obtained, one finds that either (a) just the critical points are computed and the graph shows all else that is desired, or (b) a graph is never produced, because determining increasing/decreasing using f 0 is straightforward and the graph is unnecessary. So our second reason why the above work is worthwhile is this: once mastery of a subject is gained, one has options for finding needed information. We are working to develop mastery. Finally, our third reason: many problems we face “in the real world” are very complex. Solutions are tractable only through the use of computers to do many calculations for us. Computers do not solve problems “on their own,” however; they need to be taught (i.e., programmed ) to do the right things. It would be beneficial to give a function to a computer and have it return maximum and minimum values, intervals on which the function is increasing and decreasing, the locations of relative maxima, etc. The work that we are doing here is easily programmable. It is hard to teach a computer to “look at the graph and see if it is going up or down.” y 10 f ′ (x) 5 f(x) x −2 . −1 1/3 1 2 Figure 3.20: A graph of f (x) in Example 108, showing where f is increasing and decreasing. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 163 Chapter 3 The Graphical Behaviour of Functions It is easy to teach a computer to “determine if a number is greater than or less than 0.” In Section 3.1 we learned the definition of relative maxima and minima and found that they occur at critical points. We are now learning that functions can switch from increasing to decreasing (and vice–versa) at critical points. This new understanding of increasing and decreasing creates a great method of determining whether a critical point corresponds to a maximum, minimum, or neither. Imagine a function increasing until a critical point at x = c, after which it decreases. A quick sketch helps confirm that f (c) must be a relative maximum. A similar statement can be made for relative minimums. We formalize this concept in a theorem. Theorem 33 First Derivative Test Let f be differentiable on I and let c be a critical number in I. 1. If the sign of f 0 switches from positive to negative at c, then f (c) is a relative maximum of f . 2. If the sign of f 0 switches from negative to positive at c, then f (c) is a relative minimum of f . 3. If the sign of f 0 does not change at c, then f (c) is not a relative extrema of f . Example 109 Using the First Derivative Test Find the intervals on which f is increasing and decreasing, and use the First Derivative Test to determine the relative extrema of f , where f (x) = x2 + 3 . x−1 We start by noting the domain of f : (−∞, 1) ∪ (1, ∞). Solution Key Idea 2 describes how to find intervals where f is increasing and decreasing when the domain of f is an interval. Since the domain of f in this example is the union of two intervals, we apply the techniques of Key Idea 2 to both intervals of the domain of f . Since f is not defined at x = 1, the increasing/decreasing nature of f could switch at this value. We do not formally consider x = 1 to be a critical value of f , but we will include it in our list of critical values that we find next. Using the Quotient Rule, we find f 0 (x) = Notes: Page 164 Custom Made for Math 1174 at Langara College by P. Anhaouy x2 − 2x − 3 . (x − 1)2 3.3 We need to find the critical values of f ; we want to know when f 0 (x) = 0 and when f 0 is not defined. That latter is straightforward: when the denominator of f 0 (x) is 0, f 0 is undefined. That occurs when x = 1, which we’ve already recognized as an important value. f 0 (x) = 0 when the numerator of f 0 (x) is 0. That occurs when x2 − 2x − 3 = (x − 3)(x + 1) = 0; i.e., when x = −1, 3. We have found that f has two critical numbers, x = −1, 3, and at x = 1 something important might also happen. These three numbers divide the real number line into 4 subintervals: (−∞, −1), (−1, 1), (1, 3) and (3, ∞). Pick a number p from each subinterval and test the sign of f 0 at p to determine whether f is increasing or decreasing on that interval. Again, we do well to avoid complicated computations; notice that the denominator of f 0 is always positive so we can ignore it during our work. Interval 1, (−∞, −1): Choosing a very small number (i.e., a negative number with a large magnitude) p returns p2 − 2p − 3 in the numerator of f 0 ; that will be positive. Hence f is increasing on (−∞, −1). Interval 2, (−1, 1): Choosing 0 seems simple: f 0 (0) = −3 < 0. We conclude f is decreasing on (−1, 1). Interval 3, (1, 3): Choosing 2 seems simple: f 0 (2) = −3 < 0. Again, f is decreasing. Interval 4, (3, ∞): Choosing an very large number p from this subinterval will give a positive numerator and (of course) a positive denominator. So f is increasing on (3, ∞). In summary, f is increasing on the set (−∞, −1) ∪ (3, ∞) and is decreasing on the set (−1, 1)∪(1, 3). Since at x = −1, the sign of f 0 switched from positive to negative, Theorem 33 states that f (−1) is a relative maximum of f . At x = 3, the sign of f 0 switched from negative to positive, meaning f (3) is a relative minimum. At x = 1, f is not defined, so there is no relative extrema at x = 1. f ′ > 0 incr rel. max f ′ < 0 decr f ′ < 0 decr rel. min Increasing and Decreasing Functions y 20 f(x) 10 f ′ (x) x −4 −2 2 4 −10 . −20 Figure 3.22: A graph of f (x) in Example 109, showing where f is increasing and decreasing. f ′ > 0 incr . −1 1 3 Figure 3.21: Number line for f in Example 109. This is summarized in the number line shown in Figure 3.21. Also, Figure 3.22 shows a graph of f , confirming our calculations. This figure also shows f 0 , again demonstrating that f is increasing when f 0 > 0 and decreasing when f 0 < 0. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 165 Chapter 3 The Graphical Behaviour of Functions One is often tempted to think that functions always alternate “increasing, decreasing, increasing, decreasing,. . .” around critical values. Our previous example demonstrated that this is not always the case. While x = 1 was not technically a critical value, it was an important value we needed to consider. We found that f was decreasing on “both sides of x = 1.” We examine one more example. Example 110 Using the First Derivative Test Find the intervals on which f (x) = x8/3 − 4x2/3 is increasing and decreasing and identify the relative extrema. We again start with taking derivatives. Since we know Solution we want to solve f 0 (x) = 0, we will do some algebra after taking derivatives. 2 8 f (x) = x 3 − 4x 3 8 5 8 1 f 0 (x) = x 3 − x− 3 3 3  8 −1  6 = x 3 x3 − 1 3 8 1 = x− 3 (x2 − 1) 3 8 1 = x− 3 (x − 1)(x + 1). 3 This derivation of f 0 shows that f 0 (x) = 0 when x = ±1 and f 0 is not defined when x = 0. Thus we have 3 critical values, breaking the number line into 4 subintervals as shown in Figure 3.23. Interval 1, (∞, −1): We choose p = −2; we can easily verify that f 0 (−2) < 0. So f is decreasing on (−∞, −1). Interval 2, (−1, 0): Choose p = −1/2. Once more we practice finding the sign of f 0 (p) without computing an actual value. We have f 0 (p) = (8/3)p−1/3 (p − 1)(p + 1); find the sign of each of the three terms. f 0 (p) = 8 −1 · p 3 · (p − 1) (p + 1) . 3 |{z} | {z } | {z } <0 <0 >0 We have a “negative × negative × positive” giving a positive number; f is increasing on (−1, 0). Interval 3, (0, 1): We do a similar sign analysis as before, using p in Notes: Page 166 Custom Made for Math 1174 at Langara College by P. Anhaouy 3.3 (0, 1). f 0 (p) = Increasing and Decreasing Functions 8 −1 · p 3 · (p − 1) (p + 1) . 3 |{z} | {z } | {z } >0 <0 >0 We have 2 positive factors and one negative factor; f 0 (p) < 0 and so f is decreasing on (0, 1). Interval 4, (1, ∞): Similar work to that done for the other three intervals shows that f 0 (x) > 0 on (1, ∞), so f is increasing on this interval. f ′ < 0 decr rel. min f ′ > 0 incr rel. max f ′ < 0 decr rel. min f ′ > 0 incr . −1 0 1 Figure 3.23: Number line for f in Example 110. We conclude by stating that f is increasing on (−1, 0) ∪ (1, ∞) and decreasing on (−∞, −1) ∪ (0, 1). The sign of f 0 changes from negative to positive around x = −1 and x = 1, meaning by Theorem 33 that f (−1) and f (1) are relative minima of f . As the sign of f 0 changes from positive to negative at x = 0, we have a relative maximum at f (0). Figure 3.24 shows a graph of f , confirming our result. We also graph f 0 , highlighting once more that f is increasing when f 0 > 0 and is decreasing when f 0 < 0. We have seen how the first derivative of a function helps determine when the function is going “up” or “down.” In the next section, we will see how the second derivative helps determine how the graph of a function curves. y 10 f(x) 5 f ′ (x) x −3 −2 −1 1 2 3 . Figure 3.24: A graph of f (x) in Example 110, showing where f is increasing and decreasing. . Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 167 Exercises 3.3 Terms and Concepts In Exercises 14 – 23, a function f (x) is given. (a) Give the domain of f . 1. In your own words describe what it means for a function to be increasing. (b) Find the critical numbers of f . 2. What does a decreasing function “look like”? (c) Create a number line to determine the intervals on which f is increasing and decreasing. 3. Sketch a graph of a function on [0, 2] that is increasing but not strictly increasing. (d) Use the First Derivative Test to determine whether each critical point is a relative maximum, minimum, or neither. 4. Give an example of a function describing a situation where it is “bad” to be increasing and “good” to be decreasing. 14. f (x) = x2 + 2x − 3 15. f (x) = x3 + 3x2 + 3 2 5. A function f has derivative f 0 (x) = (sin x + 2)ex +1 , where f 0 (x) > 1 for all x. Is f increasing, decreasing, or can we not tell from the given information? 16. f (x) = 2x3 + x2 − x + 3 17. f (x) = x3 − 3x2 + 3x − 1 18. f (x) = 1 x2 − 2x + 2 (a) Compute f (x). 19. f (x) = x2 − 4 x2 − 1 (b) Graph f and f 0 on the same axes (using technology is permitted) and verify Theorem 32. 20. f (x) = x x2 − 2x − 8 21. f (x) = (x − 2)2/3 x Problems In Exercises 6 – 13, a function f (x) is given. 0 6. f (x) = 2x + 3 7. f (x) = x2 − 3x + 5 8. f (x) = cos x 9. f (x) = tan x 10. f (x) = x3 − 5x2 + 7x − 1 11. f (x) = 2x3 − x2 + x − 1 12. f (x) = x4 − 5x2 + 4 13. f (x) = 1 x2 + 1 22. f (x) = sin x cos x on (−π, π). 23. f (x) = x5 − 5x Review 24. Consider f (x) = x2 − 3x + 5 on [−1, 2]; find c guaranteed by the Mean Value Theorem. 25. Consider f (x) = sin x on [−π/2, π/2]; find c guaranteed by the Mean Value Theorem. Page 168 Custom Made for Math 1174 at Langara College by P. Anhaouy 3.4 3.4 Concavity and the Second Derivative y Concavity and the Second Derivative 30 Our study of “nice” functions continues. The previous section showed how the first derivative of a function, f 0 , can relay important information about f . We now apply the same technique to f 0 itself, and learn what this tells us about f . The key to studying f 0 is to consider its derivative, namely f 00 , which is the second derivative of f . When f 00 > 0, f 0 is increasing. When f 00 < 0, f 0 is decreasing. f 0 has relative maxima and minima where f 00 = 0 or is undefined. This section explores how knowing information about f 00 gives information about f . Concavity 20 10 . x −2 2 Figure 3.25: A function f with a concave up graph. Notice how the slopes of the tangent lines, when looking from left to right, are increasing. We begin with a definition, then explore its meaning. Definition 20 Concave Up and Concave Down Let f be differentiable on an interval I. The graph of f is concave up on I if f 0 is increasing. The graph of f is concave down on I if f 0 is decreasing. If f 0 is constant then the graph of f is said to have no concavity. Note: We often state that “f is concave up” instead of “the graph of f is concave up” for simplicity. The graph of a function f is concave up when f 0 is increasing. That means as one looks at a concave up graph from left to right, the slopes of the tangent lines will be increasing. Consider Figure 3.25, where a concave up graph is shown along with some tangent lines. Notice how the tangent line on the left is steep, downward, corresponding to a small value of f 0 . On the right, the tangent line is steep, upward, corresponding to a large value of f 0 . If a function is decreasing and concave up, then its rate of decrease is slowing; it is “leveling off.” If the function is increasing and concave up, then the rate of increase is increasing. The function is increasing at a faster and faster rate. Now consider a function which is concave down. We essentially repeat the above paragraphs with slight variation. The graph of a function f is concave down when f 0 is decreasing. That means as one looks at a concave down graph from left to right, the slopes of the tangent lines will be decreasing. Consider Figure 3.26, where a concave down graph is shown along with some tangent lines. Notice how Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 169 Chapter 3 The Graphical Behaviour of Functions f ′ > 0, increasing f ′′ < 0, c. down the tangent line on the left is steep, upward, corresponding to a large value of f 0 . On the right, the tangent line is steep, downward, corresponding to a small value of f 0 . If a function is increasing and concave down, then its rate of increase is slowing; it is “leveling off.” If the function is decreasing and concave down, then the rate of decrease is decreasing. The function is decreasing at a faster and faster rate. Our definition of concave up and concave down is given in terms of when the first derivative is increasing or decreasing. We can apply the results of the previous section and to find intervals on which a graph is concave up or down. That is, we recognize that f 0 is increasing when f 00 > 0, etc. f ′ < 0, decreasing f ′′ < 0, c. down . f ′ < 0, decreasing f ′′ > 0, c. up f ′ > 0, increasing f ′′ > 0, c. up Figure 3.27: Demonstrating the 4 ways that concavity interacts with increasing/decreasing, along with the relationships with the first and second derivatives. Note: Geometrically speaking, a function is concave up if its graph lies above its tangent lines. A function is concave down if its graph lies below its tangent lines. Note: A mnemonic for remembering what concave up/down means is: “Concave up is like a cup; concave down is like a frown.” It is admittedlyy terrible, but it works. Theorem 34 Test for Concavity Let f be twice differentiable on an interval I. The graph of f is concave up if f 00 > 0 on I, and is concave down if f 00 < 0 on I. If knowing where a graph is concave up/down is important, it makes sense that the places where the graph changes from one to the other is also important. This leads us to a definition. Definition 21 Point of Inflection A point of inflection is a point on the graph of f at which the concavity of f changes. 15 Figure 3.28 shows a graph of a function with inflection points labeled. If the concavity of f changes at a point (c, f (c)), then f 0 is changing from increasing to decreasing (or, decreasing to increasing) at x = c. That means that the sign of f 00 is changing from positive to negative (or, negative to positive) at x = c. This leads to the following theorem. 10 f ′′ > 0y f ′′ > 0 c. up c. up 30 5 f ′′ < 0 c. down . 20 1 x 2 3 Theorem 35 Points of Inflection 4 10 graph of a function Figure 3.28: A with its inflection points marked. The intervals where concave up/down are x . indicated. also −2 2 Figure 3.26: A function f with a concave down graph. Notice how the slopes of the tangent lines, when looking from left to right, are decreasing. If (c, f (c)) is a point of inflection on the graph of f , then either f 00 = 0 or f 00 is not defined at c. We have identified the concepts of concavity and points of inflection. Notes: Page 170 Custom Made for Math 1174 at Langara College by P. Anhaouy 3.4 Concavity and the Second Derivative It is now time to practice using these concepts; given a function, we should be able to find its points of inflection and identify intervals on which it is concave up or down. We do so in the following examples. Example 111 Finding intervals of concave up/down, inflection points Let f (x) = x3 − 3x + 1. Find the inflection points of f and the intervals on which it is concave up/down. We start by finding f 0 (x) = 3x2 − 3 and f 00 (x) = 6x. To find the inflection points, we use Theorem 35 and find where f 00 (x) = 0 or where f 00 is undefined. We find f 00 is always defined, and is 0 only when x = 0. So the point (0, 1) is the only possible point of inflection. This possible inflection point divides the real line into two intervals, (−∞, 0) and (0, ∞). We use a process similar to the one used in the previous section to determine increasing/decreasing. Pick any c < 0; f 00 (c) < 0 so f is concave down on (−∞, 0). Pick any c > 0; f 00 (c) > 0 so f is concave up on (0, ∞). Since the concavity changes at x = 0, the point (0, 1) is an inflection point. The number line in Figure 3.29 illustrates the process of determining concavity; Figure 3.30 shows a graph of f and f 00 , confirming our results. Notice how f is concave down precisely when f 00 (x) < 0 and concave up when f 00 (x) > 0. f ′′ < 0 c. down f ′′ > 0 c. up . 0 Figure 3.29: A number line determining the concavity of f in Example 111. Solution y f(x) 2 f ′′ (x) x −2 −1 1 2 −2 . Figure 3.30: A graph of f (x) used in Example 111. Example 112 Finding intervals of concave up/down, inflection points x Let f (x) = 2 . Find the inflection points of f and the intervals on x −1 which it is concave up/down. We need to find f 0 and f 00 . Using the Quotient Rule Solution and simplifying, we find f 0 (x) = −(1 + x2 ) (x2 − 1)2 and f 00 (x) = 2x(x2 + 3) . (x2 − 1)3 To find the possible points of inflection, we seek to find where f 00 (x) = 0 and where f 00 is not defined. Solving f 00 (x) = 0 reduces to solving 2x(x2 + 3) = 0; we find x = 0. We find that f 00 is not defined when x = ±1, for then the denominator of f 00 is 0. We also note that f itself is not defined at x = ±1, having a domain of (−∞, −1) ∪ (−1, 1) ∪ (1, ∞). Since the domain of f is the union of three intervals, it makes sense that the concavity of f could switch across intervals. We technically cannot Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 171 Chapter 3 The Graphical Behaviour of Functions say that f has a point of inflection at x = ±1 as they are not part of the domain, but we must still consider these x-values to be important and will include them in our number line. The important x-values at which concavity might switch are x = −1, x = 0 and x = 1, which split the number line into four intervals as shown in Figure 3.31. We determine the concavity on each. Keep in mind that all we are concerned with is the sign of f 00 on the interval. y 10 5 f ′′ (x) f(x) x −2 2 −5 −10 . Figure 3.32: A graph of f (x) and f 00 (x) in Example 112. Interval 1, (−∞, −1): Select a number c in this interval with a large magnitude (for instance, c = −100). The denominator of f 00 (x) will be positive. In the numerator, the (c2 + 3) will be positive and the 2c term will be negative. Thus the numerator is negative and f 00 (c) is negative. We conclude f is concave down on (−∞, −1). Interval 2, (−1, 0): For any number c in this interval, the term 2c in the numerator will be negative, the term (c2 + 3) in the numerator will be positive, and the term (c2 − 1)3 in the denominator will be negative. Thus f 00 (c) > 0 and f is concave up on this interval. Interval 3, (0, 1): Any number c in this interval will be positive and “small.” Thus the numerator is positive while the denominator is negative. Thus f 00 (c) < 0 and f is concave down on this interval. Interval 4, (1, ∞): Choose a large value for c. It is evident that f 00 (c) > 0, so we conclude that f is concave up on (1, ∞). f ′′ < 0 c. down f ′′ > 0 c. up f ′′ < 0 c. down f ′′ > 0 c. up . −1 0 1 Figure 3.31: Number line for f in Example 112. We conclude that f is concave up on (−1, 0)∪(1, ∞) and concave down on (−∞, −1) ∪ (0, 1). There is only one point of inflection, (0, 0), as f is not defined at x = ±1. Our work is confirmed by the graph of f in Figure 3.32. Notice how f is concave up whenever f 00 is positive, and concave down when f 00 is negative. y 20 15 S(t) 10 5 . t 1 2 3 Figure 3.33: A graph of S(t) in Example 113, modeling the sale of a product over time. Recall that relative maxima and minima of f are found at critical points of f ; that is, they are found when f 0 (x) = 0 or when f 0 is undefined. Likewise, the relative maxima and minima of f 0 are found when f 00 (x) = 0 or when f 00 is undefined; note that these are the inflection points of f . What does a “relative maximum of f 0 ” mean? The derivative measures the rate of change of f ; maximizing f 0 means finding the where f is increasing the most – where f has the steepest tangent line. A similar statement can be made for minimizing f 0 ; it corresponds to where f has Notes: Page 172 Custom Made for Math 1174 at Langara College by P. Anhaouy 3.4 the steepest negatively–sloped tangent line. We utilize this concept in the next example. Concavity and the Second Derivative y 20 S(t) Example 113 Understanding inflection points The sales of a certain product over a three-year span are modeled by S(t) = t4 − 8t2 + 20, where t is the time in years, shown in Figure 3.33. Over the first two years, sales are decreasing. Find the point at which sales are decreasing at their greatest rate. We want to maximize the rate of decrease, which is to Solution say, we want to find where S 0 has a minimum. To do this, we find where 3 00 2 00 S 00 is 0. We find S 0 (t) = 4t p −16t and S (t) = 12t −16. Setting S (t) = 0 and solving, we get t = 4/3 ≈ 1.16 (we ignore the negative value of t since it does not lie in the domain of our function S). This is both the inflection point and the point of maximum decrease. This is the point at which things first start looking up for the company. After the inflection point, it will still take some time before sales start to increase, but at least sales are not decreasing quite as quickly as they had been. A graph of S(t) and S 0 (t) is given in Figure 3.34. When S 0 (t) < 0, sales are decreasing; note how at t ≈ 1.16, S 0 (t) is minimized. That is, sales are decreasing at the fastest rate at t ≈ 1.16. On the interval of (1.16, 2), S is decreasing but concave up, so the decline in sales is “leveling off.” Not every critical point corresponds to a relative extrema; f (x) = x3 has a critical point at (0, 0) but no relative maximum or minimum. Likewise, just because f 00 (x) = 0 we cannot conclude concavity changes at that point. We were careful before to use terminology “possible point of inflection” since we needed to check to see if the concavity changed. The canonical example of f 00 (x) = 0 without concavity changing is f (x) = x4 . At x = 0, f 00 (x) = 0 but f is always concave up, as shown in Figure 3.35. 10 t 1 2 3 S ′ (t) −10 . Figure 3.34: A graph of S(t) in Example 113 along with S 0 (t). y 1 0.5 .−1 x −0.5 0.5 1 Figure 3.35: A graph of f (x) = x4 . Clearly f is always concave up, despite the fact that f 00 (x) = 0 when x = 0. It this example, the possible point of inflection (0, 0) is not a point of inflection. y 10 The Second Derivative Test c. down ⇒ rel. max The first derivative of a function gave us a test to find if a critical value corresponded to a relative maximum, minimum, or neither. The second derivative gives us another way to test if a critical point is a local maximum or minimum. The following theorem officially states something that is intuitive: if a critical value occurs in a region where a function f is concave up, then that critical value must correspond to a relative minimum of f , etc. See Figure 3.36 for a visualization of this. Notes: 5 x −2 −1 1 −5 . 2 c. up ⇒ rel. min −10 Figure 3.36: Demonstrating the fact that relative maxima occur when the graph is concave down and relative minima occur when the graph is concave up. Custom Made for Math 1174 at Langara College by P. Anhaouy Page 173 Chapter 3 The Graphical Behaviour of Functions y Theorem 36 40 f ′′ (10) > 0 Let c be a critical value of f where f 00 (c) is defined. 20 1. If f 00 (c) > 0, then f has a local minimum at (c, f (c)). x −20 −10 10 The Second Derivative Test 20 2. If f 00 (c) < 0, then f has a local maximum at (c, f (c)). −20 f ′′ (−10) < 0 . −40 Figure 3.37: A graph of f (x) in Example 114. The second derivative is evaluated at each critical point. When the graph is concave up, the critical point represents a local minimum; when the graph is concave down, the critical point represents a local maximum. The Second Derivative Test relates to the First Derivative Test in the following way. If f 00 (c) > 0, then the graph is concave up at a critical point c and f 0 itself is growing. Since f 0 (c) = 0 and f 0 is growing at c, then it must go from negative to positive at c. This means the function goes from decreasing to increasing, indicating a local minimum at c. Example 114 Using the Second Derivative Test 100 Let f (x) = + x. Find the critical points of f and use the Second x Derivative Test to label them as relative maxima or minima. 100 200 + 1 and f 00 (x) = 3 . We set 2 x x f 0 (x) = 0 and solve for x to find the critical values (note that f 0 is not defined at x = 0, but neither is f so this is not a critical value.) We find the critical values are x = ±10. Evaluating f 00 at x = 10 gives 0.1 > 0, so there is a local minimum at x = 10. Evaluating f 00 (−10) = −0.1 < 0, determining a relative maximum at x = −10. These results are confirmed in Figure 3.37. Solution . We find f 0 (x) = − We have been learning how the first and second derivatives of a function relate information about the graph of that function. We have found intervals of increasing and decreasing, intervals where the graph is concave up and down, along with the locations of relative extrema and inflection points. In Chapter 1 we saw how limits explained asymptotic behavior. In the next section we combine all of this information to produce accurate sketches of functions. Notes: Page 174 Custom Made for Math 1174 at Langara College by P. Anhaouy Exercises 3.4 Terms and Concepts In Exercises 16 – 28, a function f (x) is given. (a) Find the possible points of inflection of f . 1. Sketch a graph of a function f (x) that is concave up on (0, 1) and is concave down on (1, 2). (b) Create a number line to determine the intervals on which f is concave up or concave down. 2. Sketch a graph of a function f (x) that is: 16. f (x) = x2 − 2x + 1 (a) Increasing, concave up on (0, 1), 17. f (x) = −x2 − 5x + 7 (b) increasing, concave down on (1, 2), 18. f (x) = x3 − x + 1 (c) decreasing, concave down on (2, 3) and 19. f (x) = 2x3 − 3x2 + 9x + 5 (d) increasing, concave down on (3, 4). x4 x3 + − 2x + 3 4 3 3. Is is possible for a function to be increasing and concave down on (0, ∞) with a horizontal asymptote of y = 1? If so, give a sketch of such a function. 20. f (x) = 4. Is is possible for a function to be increasing and concave up on (0, ∞) with a horizontal asymptote of y = 1? If so, give a sketch of such a function. 22. f (x) = x4 − 4x3 + 6x2 − 4x + 1 Problems In Exercises 5 – 15, a function f (x) is given. 21. f (x) = −3x4 + 8x3 + 6x2 − 24x + 2 23. f (x) = 1 x2 + 1 24. f (x) = x x2 − 1 25. f (x) = sin x + cos x on (−π, π) 00 (a) Compute f (x). 26. f (x) = x2 ex 00 (b) Graph f and f on the same axes (using technology is permitted) and verify Theorem 34. 27. f (x) = x2 ln x 2 5. f (x) = −7x + 3 28. f (x) = e−x 6. f (x) = −4x2 + 3x − 8 In Exercises 29 – 41, a function f (x) is given. Find the critical points of f and use the Second Derivative Test, when possible, to determine the relative extrema. (Note: these are the same functions as in Exercises 16 – 28.) 7. f (x) = 4x2 + 3x − 8 8. f (x) = x3 − 3x2 + x − 1 9. f (x) = −x3 + x2 − 2x + 5 29. f (x) = x2 − 2x + 1 30. f (x) = −x2 − 5x + 7 10. f (x) = cos x 31. f (x) = x3 − x + 1 11. f (x) = sin x 32. f (x) = 2x3 − 3x2 + 9x + 5 12. f (x) = tan x 13. f (x) = 1 x2 + 1 33. f (x) = x4 x3 + − 2x + 3 4 3 34. f (x) = −3x4 + 8x3 + 6x2 − 24x + 2 14. f (x) = 1 x 35. f (x) = x4 − 4x3 + 6x2 − 4x + 1 15. f (x) = 1 x2 36. f (x) = 1 x2 + 1 Custom Made for Math 1174 at Langara College by P. Anhaouy Page 175 37. f (x) = x x2 − 1 45. f (x) = 2x3 − 3x2 + 9x + 5 x4 x3 + − 2x + 3 4 3 38. f (x) = sin x + cos x on (−π, π) 46. f (x) = 39. f (x) = x2 ex 47. f (x) = −3x4 + 8x3 + 6x2 − 24x + 2 40. f (x) = x2 ln x 48. f (x) = x4 − 4x3 + 6x2 − 4x + 1 41. f (x) = e−x 2 49. f (x) = 1 x2 + 1 50. f (x) = x x2 − 1 In Exercises 42 – 54, a function f (x) is given. Find the x values where f 0 (x) has a relative maximum or minimum. (Note: these are the same functions as in Exercises 16 – 28.) 51. f (x) = sin x + cos x on (−π, π) 42. f (x) = x2 − 2x + 1 52. f (x) = x2 ex 43. f (x) = −x2 − 5x + 7 53. f (x) = x2 ln x 44. f (x) = x3 − x + 1 54. f (x) = e−x Page 176 Custom Made for Math 1174 at Langara College by P. Anhaouy 2 3.5 3.5 Curve Sketching Curve Sketching We are attempting to understand the behavior of a function f based on the information given by its derivatives. While all of a function’s derivatives relay information about it, it turns out that “most” of the behavior we care about is explained by f 0 and f 00 . Understanding the interactions between the graph of f and f 0 and f 00 is important. To gain this understanding, one might argue that all that is needed is to look at lots of graphs. This is true to a point, but is somewhat similar to stating that one understands how an engine works after looking only at pictures. It is true that the basic ideas will be conveyed, but “hands–on” access increases understanding. The following Key Idea summarizes what we have learned so far that is applicable to sketching graphs of functions and gives a framework for putting that information together. It is followed by several examples. Key Idea 3 Curve Sketching To produce an accurate sketch a given function f , consider the following steps. 1. Find the domain of f . Generally, we assume that the domain is the entire real line then find restrictions, such as where a denominator is 0 or where negatives appear under the radical. 2. Find the critical values of f . 3. Find the possible points of inflection of f . 4. Find the location of any vertical asymptotes of f (usually done in conjunction with item 1 above). 5. Consider the limits lim f (x) and lim f (x) to determine x→−∞ x→∞ the end behavior of the function. (continued) Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 177 Chapter 3 The Graphical Behaviour of Functions Key Idea 3 Curve Sketching – Continued 6. Create a number line that includes all critical points, possible points of inflection, and locations of vertical asymptotes. For each interval created, determine whether f is increasing or decreasing, concave up or down. 7. Evaluate f at each critical point and possible point of inflection. Plot these points on a set of axes. Connect these points with curves exhibiting the proper concavity. Sketch asymptotes and x and y intercepts where applicable. Example 115 Curve sketching Use Key Idea 3 to sketch f (x) = 3x3 − 10x2 + 7x + 5. Solution We follow the steps outlined in the Key Idea. 1. The domain of f is the entire real line; there are no values x for which f (x) is not defined. 2. Find the critical values of f . We compute f 0 (x) = 9x2 − 20x + 7. Use the Quadratic Formula to find the roots of f 0 : p √  20 ± (−20)2 − 4(9)(7) 1 x= = 10 ± 37 ⇒ x ≈ 0.435, 1.787. 2(9) 9 3. Find the possible points of inflection of f . Compute f 00 (x) = 18x − 20. We have f 00 (x) = 0 ⇒ x = 10/9 ≈ 1.111. 4. There are no vertical asymptotes. 5. We determine the end behavior using limits as x approaches ±infinity. lim f (x) = −∞ x→−∞ lim f (x) = ∞. x→∞ We do not have any horizontal asymptotes. √ 6. We place the values x = (10 ± 37)/9 and x = 10/9 on a number line, as shown in Figure 3.38. We mark each subinterval as increasing or decreasing, concave up or down, using the techniques used in Sections 3.3 and 3.4. Notes: Page 178 Custom Made for Math 1174 at Langara College by P. Anhaouy 3.5 f ′ > 0 incr f ′′ < 0 c. down f 1 9 (10 − √ f ′ < 0 decr f ′ < 0 decr f ′ > 0 incr ′′ ′′ f ′′ < 0 c. up 37) ≈ 0.435 < 0 c. down . f 10 9 ≈ 1.111 > 0 c. up 1 9 (10 + Curve Sketching y 10 √ 37) ≈ 1.787 5 Figure 3.38: Number line for f in Example 115. 7. We plot the appropriate points on axes as shown in Figure 3.39(a) and connect the points with straight lines. In Figure 3.39(b) we adjust these lines to demonstrate the proper concavity. Our curve crosses the y axis at y = 5 and crosses the x axis near x = −0.424. In Figure 3.39(c) we show a graph of f drawn with a computer program, verifying the accuracy of our sketch. x −1 . 1 2 3 2 3 2 3 −5 (a) y 10 Example 116 Curve sketching x2 − x − 2 . Sketch f (x) = 2 x −x−6 Solution 5 We again follow the steps outlined in Key Idea 3. 1. In determining the domain, we assume it is all real numbers and looks for restrictions. We find that at x = −2 and x = 3, f (x) is not defined. So the domain of f is D = {real numbers x | x 6= −2, 3}. x −1 . 1 −5 2. To find the critical values of f , we first find f 0 (x). Using the Quotient Rule, we find (b) y −8x + 4 −8x + 4 f (x) = 2 = . (x + x − 6)2 (x − 3)2 (x + 2)2 10 f 0 (x) = 0 when x = 1/2, and f 0 is undefined when x = −2, 3. Since f 0 is undefined only when f is, these are not critical values. The only critical value is x = 1/2. 5 0 00 3. To find the possible points of inflection, we find f (x), again employing the Quotient Rule: 24x2 − 24x + 56 f 00 (x) = . (x − 3)3 (x + 2)3 00 We find that f (x) is never 0 (setting the numerator equal to 0 and solving for x, we find the only roots to this quadratic are imaginary) and f 00 is undefined when x = −2, 3. Thus concavity will possibly only change at x = −2 and x = 3. x −1 . 1 −5 (c) Figure 3.39: Sketching f in Example 115. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 179 Chapter 3 The Graphical Behaviour of Functions 4. The vertical asymptotes of f are at x = −2 and x = 3, the places where f is undefined. y 5 5. There is a horizontal asymptote of y = 1, as lim f (x) = 1 and x→−∞ lim f (x) = 1. x→∞ x −4 −2 2 4 6. We place the values x = 1/2, x = −2 and x = 3 on a number line as shown in Figure 3.40. We mark in each interval whether f is increasing or decreasing, concave up or down. We see that f has a relative maximum at x = 1/2; concavity changes only at the vertical asymptotes. −5 . (a) f ′ > 0 incr y f ′′ f ′ > 0 incr > 0 c. up f ′′ < 0 c. down . 5 −2 −2 2 4 f ′′ > 0 c. up < 0 c. down 3 1 2 7. In Figure 3.41(a), we plot the points from the number line on a set of axes and connect the points with straight lines to get a general idea of what the function looks like (these lines effectively only convey increasing/decreasing information). In Figure 3.41(b), we adjust the graph with the appropriate concavity. We also show f crossing the x axis at x = −1 and x = 2. −5 .. f ′ < 0 decr Figure 3.40: Number line for f in Example 116. x −4 f f ′ < 0 decr ′′ (b) Figure 3.41(c) shows a computer generated graph of f , which verifies the accuracy of our sketch. y 5 Example 117 x −4 −2 2 Curve sketching 5(x − 2)(x + 1) Sketch f (x) = . x2 + 2x + 4 4 Solution We again follow Key Idea 3. (c) 1. We assume that the domain of f is all real numbers and consider restrictions. The only restrictions come when the denominator is 0, but this never occurs. Therefore the domain of f is all real numbers, R. Figure 3.41: Sketching f in Example 116. 2. We find the critical values of f by setting f 0 (x) = 0 and solving for x. We find . . −5 f 0 (x) = 15x(x + 4) (x2 + 2x + 4)2 Notes: Page 180 Custom Made for Math 1174 at Langara College by P. Anhaouy ⇒ f 0 (x) = 0 when x = −4, 0. 3.5 3. We find the possible points of inflection by solving f 00 (x) = 0 for x. We find 30x3 + 180x2 − 240 . f 00 (x) = − (x2 + 2x + 4)3 Curve Sketching y 5 The cubic in the numerator does not factor very “nicely.” We instead approximate the roots at x = −5.759, x = −1.305 and x = 1.064. 4. There are no vertical asymptotes. x 5. We have a horizontal asymptote of y = 5, as lim f (x) = lim f (x) = x→−∞ x→∞ 5. −5 . (a) 6. We place the critical points and possible points on a number line as shown in Figure 3.42 and mark each interval as increasing/decreasing, concave up/down appropriately. f ′ > 0 incr f ′ > 0 incr f ′ < 0 decr f ′ < 0 decr f ′ > 0 incr f ′ > 0 decr f ′′ > 0 c. up f ′′ < 0 c. down f ′′ < 0 c. down f ′′ > 0 c. up f ′′ > 0 c. up f ′′ < 0 c. down −5.579 −4 . −1.305 0 5 y 5 1.064 x Figure 3.42: Number line for f in Example 117. 7. In Figure 3.43(a) we plot the significant points from the number line as well as the two roots of f , x = −1 and x = 2, and connect the points with straight lines to get a general impression about the graph. In Figure 3.43(b), we add concavity. Figure 3.43(c) shows a computer generated graph of f , affirming our results. In each of our examples, we found a few, significant points on the graph of f that corresponded to changes in increasing/decreasing or concavity. We connected these points with straight lines, then adjusted for concavity, and finished by showing a very accurate, computer generated graph. Why are computer graphics so good? It is not because computers are “smarter” than we are. Rather, it is largely because computers are much faster at computing than we are. In general, computers graph functions much like most students do when first learning to draw graphs: they plot equally spaced points, then connect the dots using lines. By using lots of points, the connecting lines are short and the graph looks smooth. This does a fine job of graphing in most cases (in fact, this is the method used for many graphs in this text). However, in regions where the graph is very “curvy,” this can generate noticeable sharp edges on the graph unless a large number of points are used. High quality computer −5 5 . (b) y 5 x −5 5 . (c) Figure 3.43: Sketching f in Example 117. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 181 Chapter 3 The Graphical Behaviour of Functions algebra systems, such as Mathematica, use special algorithms to plot lots of points only where the graph is “curvy.” In Figure 3.44, a graph of y = sin x is given, generated by Mathematica. The small points represent each of the places Mathematica sampled the function. Notice how at the “bends” of sin x, lots of points are used; where sin x is relatively straight, fewer points are used. (Many points are also used at the endpoints to ensure the “end behavior” is accurate.) 1.0 0.5 1 2 3 4 5 6 -0.5 -1.0 Figure 3.44: A graph of y = sin x generated by Mathematica. How does Mathematica know where the graph is “curvy”? Calculus. When we study curvature in a later chapter, we will see how the first and second derivatives of a function work together to provide a measurement of “curviness.” Mathematica employs algorithms to determine regions of “high curvature” and plots extra points there. Again, the goal of this section is not “How to graph a function when there is no computer to help.” Rather, the goal is “Understand that the shape of the graph of a function is largely determined by understanding the behavior of the function at a few key places.” In Example 117, we were able to accurately sketch a complicated graph using only 5 points and knowledge of asymptotes! There are many applications of our understanding of derivatives beyond curve sketching. The next chapter explores some of these applications, demonstrating just a few kinds of problems that can be solved with a basic knowledge of differentiation. Notes: Page 182 Custom Made for Math 1174 at Langara College by P. Anhaouy Exercises 3.5 Terms and Concepts 14. f (x) = x3 + 3x2 + 3x + 1 1. Why is sketching curves by hand beneficial even though technology is ubiquitous? 15. f (x) = x3 − x2 − x + 1 16. f (x) = (x − 2) ln(x − 2) 2. What does “ubiquitous” mean? 3. T/F: When sketching graphs of functions, it is useful to find the critical points. 17. f (x) = (x − 2)2 ln(x − 2) 18. f (x) = x2 − 4 x2 19. f (x) = x2 − 4x + 3 x2 − 6x + 8 20. f (x) = x2 − 2x + 1 x2 − 6x + 8 4. T/F: When sketching graphs of functions, it is useful to find the possible points of inflection. 5. T/F: When sketching graphs of functions, it is useful to find the horizontal and vertical asymptotes. Problems √ 21. f (x) = x x + 1 In Exercises 6 – 11, practice using Key Idea 3 by applying the principles to the given functions with familiar graphs. 22. f (x) = x2 ex 23. f (x) = sin x cos x on [−π, π] 6. f (x) = 2x + 4 24. f (x) = (x − 3)2/3 + 2 2 7. f (x) = −x + 1 (x − 1)2/3 x 8. f (x) = sin x 25. f (x) = 9. f (x) = ex In Exercises 26 – 28, a function with the parameters a and b are given. Describe the critical points and possible points of inflection of f in terms of a and b. 10. f (x) = 1 x 11. f (x) = 1 x2 26. f (x) = In Exercises 12 – 25, sketch a graph of the given function using Key Idea 3. Show all work; check your answer with technology. 12. f (x) = x3 − 2x2 + 4x + 1 13. f (x) = −x3 + 5x2 − 3x + 2 a x2 + b2 27. f (x) = sin(ax + b) 28. f (x) = (x − a)(x − b) 29. Given x2 + y 2 = 1, use implicit differentiation to find dy d2 y and dx 2 . Use this information to justify the sketch dx of the unit circle. Custom Made for Math 1174 at Langara College by P. Anhaouy Page 183 4: Applications of the Derivative 4.1 Related Rates When two quantities are related by an equation, knowing the value of one quantity can determine the value of the other. For instance, the circumference and radius of a circle are related by C = 2πr; knowing that C = 6πin determines the radius must be 3in. The topic of related rates takes this one step further: knowing the rate at which one quantity is changing can determine the rate at which the other changes. We demonstrate the concepts of related rates through examples. Example 118 Understanding related rates The radius of a circle is growing at a rate of 5in/hr. At what rate is the circumference growing? The circumference and radius of a circle are related by Solution C = 2πr. We are given information about how the length of r changes with respect to time; that is, we are told dr dt = 5in/hr. We want to know how the length of C changes with respect to time, i.e., we want to know dC dt . Implicitly differentiate both sides of C = 2πr with respect to t: C = 2πr   d d C = 2πr dt dt dC dr = 2π . dt dt As we know dr dt = 5in/hr, we know dC = 2π5 = 10π ≈ 31.4in/hr. dt Consider another, similar example. Example 119 Finding related rates Water streams out of a faucet at a rate of 2in3 /s onto a flat surface at a constant rate, forming a circular puddle that is 1/8in deep. Note: This section relies heavily on implicit differentiation, so referring back to Section 2.8 may help. Chapter 4 Applications of the Derivative 1. At what rate is the area of the puddle growing? 2. At what rate is the radius of the circle growing? Solution 1. We can answer this question two ways: using “common sense” or related rates. The common sense method states that the volume of the puddle is growing by 2in3 /s, where volume of puddle = area of circle × depth. Since the depth is constant at 1/8in, the area must be growing by 16in2 /s. This approach reveals the underlying related–rates principle. Let V and A represent the Volume and Area of the puddle. We know V = A × 81 . Take the derivative of both sides with respect to t, employing implicit differentiation. 1 A 8    d 1 d V = A dt dt 8 dV 1 dA = dt 8 dt V = 1 dA dA As dV dt = 2, we know 2 = 8 dt , and hence dt = 16. Thus the area 2 is growing by 16in /s. 2. To start, we need an equation that relates what we know to the radius. We just learned something about the surface area of the circular puddle, and we know A = πr2 . We should be able to learn about the rate at which the radius is growing with this information. Implicitly derive both sides of A = πr2 with respect to t: A = πr2   d d A = πr2 dt dt dA dr = 2πr dt dt Notes: Page 186 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.1 Related Rates dr 2 Our work above told us that dA dt = 16in /s. Solving for dt , we have dr 8 = . dt πr Note how our answer is not a number, but rather a function of r. In other words, the rate at which the radius is growing depends on how big the circle already is. If the circle is very large, adding 2in3 of water will not make the circle much bigger at all. If the circle dime–sized, adding the same amount of water will make a radical change in the radius of the circle. In some ways, our problem was (intentionally) ill–posed. We need to specify a current radius in order to know a rate of change. When the puddle has a radius of 10in, the radius is growing at a rate of 8 4 dr = = ≈ 0.25in/s. dt 10π 5π Using the diagram in Figure 4.1, let’s label what we know about the situation. As both the police officer and other driver are 1/2 mile from the intersection, we √ have A = 1/2, B = 1/2, and through the Pythagorean Theorem, C = 1/ 2 ≈ 0.707. Solution N B = 1/2 . Car A = 1/2 Example 120 Studying related rates Radar guns measure the rate of distance change between the gun and the object it is measuring. For instance, a reading of “55mph” means the object is moving away from the gun at a rate of 55 miles per hour, whereas a measurement of “−25mph” would mean that the object is approaching the gun at a rate of 25 miles per hour. If the radar gun is moving (say, attached to a police car) then radar readouts are only immediately understandable if the gun and the object are moving along the same line. If a police officer is traveling 60mph and gets a readout of 15mph, he knows that the car ahead of him is moving away at a rate of 15 miles an hour, meaning the car is traveling 75mph. (This straight–line principle is one reason officers park on the side of the highway and try to shoot straight back down the road. It gives the most accurate reading.) Suppose an officer is driving due north at 30 mph and sees a car moving due east, as shown in Figure 4.1. Using his radar gun, he measures a reading of 20mph. By using landmarks, he believes both he and the other car are about 1/2 mile from the intersection of their two roads. If the speed limit on the other road is 55mph, is the other driver speeding? E C Officer Figure 4.1: A sketch of a police car (at bottom) attempting to measure the speed of a car (at right) in Example 120. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 187 Chapter 4 Applications of the Derivative Note: Example 120 is both interesting and impractical. It highlights the difficulty in using radar in a non– linear fashion, and explains why “in real life” the police officer would follow the other driver to determine their speed, and not pull out pencil and paper. The principles here are important, though. Many automated vehicles make judgments about other moving objects based on perceived distances, radar–like measurements and the concepts of related rates. We know the police officer is traveling at 30mph; that is, dA dt = −30. The reason this rate of change is negative is that A is getting smaller; the distance between the officer and the intersection is shrinking. The radar dB measurement is dC dt = 20. We want to find dt . We need an equation that relates B to A and/or C. The Pythagorean Theorem is a good choice: A2 + B 2 = C 2 . Differentiate both sides with respect to t: A2 + B 2 = C 2   d d A2 + B 2 = C2 dt dt dA dB dC 2A + 2B = 2C dt dt dt We have values for everything except dB dt . Solving for this we have C dC − A dA dB dt = dt ≈ 58.28mph. dt B The other driver appears to be speeding slightly. 100mph . x 10 θ Figure 4.2: Tracking a speeding car (at left) with a rotating camera. Example 121 Studying related rates A camera is placed on a tripod 10ft from the side of a road. The camera is to turn to track a car that is to drive by at 100mph for a promotional video. The video’s planners want to know what kind of motor the tripod should be equipped with in order to properly track the car as it passes by. Figure 4.2 shows the proposed setup. How fast must the camera be able to turn to track the car? We seek information about how fast the camera is to Solution turn; therefore, we need an equation that will relate an angle θ to the position of the camera and the speed and position of the car. Figure 4.2 suggests we use a trigonometric equation. Letting x represent the distance the car is from the point on the road directly in front of the camera, we have x tan θ = . (4.1) 10 As the car is moving at 100mph, we have dx dt = −100mph (as in the last example, since x is getting smaller as the car travels, dx dt is negative). We need to convert the measurements so they use the same units; rewrite -100mph in terms of ft/s: dx m m ft 1 hr = −100 = −100 · 5280 · = −146.6ft/s. dt hr hr m 3600 s Notes: Page 188 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.1 Related Rates Now take the derivative of both sides of Equation (4.1) using implicit differentiation: x 10  d d x tan θ = dt dt 10 dθ 1 dx sec2 θ = dt 10 dt cos2 θ dx dθ = dt 10 dt tan θ = (4.2) We want to know the fastest the camera has to turn. Common sense tells us this is when the car is directly in front of the camera (i.e., when θ = 0). Our mathematics bears this out. In Equation (4.2) we see this is when cos2 θ is largest; this is when cos θ = 1, or when θ = 0. With dx dt ≈ −146.67ft/s, we have dθ 1rad =− 146.67ft/s = −14.667radians/s. dt 10ft We find that dθ dt is negative; this matches our diagram in Figure 4.2 for θ is getting smaller as the car approaches the camera. What is the practical meaning of −14.667radians/s? Recall that 1 circular revolution goes through 2π radians, thus 14.667rad/s means 14.667/(2π) ≈ 2.33 revolutions per second. The negative sign indicates the camera is rotating in a clockwise fashion. We introduced the derivative as a function that gives the slopes of tangent lines of functions. This chapter emphasizes using the derivative in other ways. Newton’s Method uses the derivative to approximate roots of functions; this section stresses the “rate of change” aspect of the derivative to find a relationship between the rates of change of two related quantities. In the next section we use Extreme Value concepts to optimize quantities. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 189 Exercises 4.1 Terms and Concepts (a) 1 mile away? 1. T/F: Implicit differentiation is often used when solving “related rates” type problems. 2. T/F: A study of related rates is part of the standard police officer training. Problems 3. Water flows onto a flat surface at a rate of 5cm3 /s forming a circular puddle 10mm deep. How fast is the radius growing when the radius is: (b) 1/5 mile away? (c) Directly overhead? 8. An F-22 aircraft is flying at 500mph with an elevation of 100ft on a straight–line path that will take it directly over an anti–aircraft gun as in Exercise 7 (note the lower elevation here). How fast must the gun be able to turn to accurately track the aircraft when the plane is: (a) 1000 feet away? (b) 100 feet away? (c) Directly overhead? (a) 1 cm? (b) 10 cm? 9. A 24ft. ladder is leaning against a house while the base is pulled away at a constant rate of 1ft/s. (c) 100 cm? . 1 /s . (a) 1 cm? At what rate is the top of the ladder sliding down the side of the house when the base is: (b) 10 cm? (a) 1 foot from the house? (c) 100 cm? (b) 10 feet from the house? 5. Consider the traffic situation introduced in Example 120. How fast is the “other car” traveling if the officer and the other car are each 1/2 mile from the intersection, the other car is traveling due west, the officer is traveling north at 50mph, and the radar reading is −80mph? (c) 23 feet from the house? (d) 24 feet from the house? 10. A boat is being pulled into a dock at a constant rate of 30ft/min by a winch located 10ft above the deck of the boat. 6. Consider the traffic situation introduced in Example 120. Calculate how fast the “other car” is traveling in each of the following situations. (a) The officer is traveling due north at 50mph and is 1/2 mile from the intersection, while the other car is 1 mile from the intersection traveling west and the radar reading is −80mph? (b) The officer is traveling due north at 50mph and is 1 mile from the intersection, while the other car is 1/2 mile from the intersection traveling west and the radar reading is −80mph? 7. An F-22 aircraft is flying at 500mph with an elevation of 10,000ft on a straight–line path that will take it directly over an anti–aircraft gun. . 10,000 . 24 4. A circular balloon is inflated with air flowing at a rate of 10cm3 /s. How fast is the radius of the balloon increasing when the radius is: θ . 10 . . At what rate is the boat approaching the dock when the boat is: (a) 50 feet out? (b) 15 feet out? (c) 1 foot from the dock? (d) What happens when the length of rope pulling in the boat is less than 10 feet long? 11. An inverted cylindrical cone, 20ft deep and 10ft across at the top, is being filled with water at a rate of 10ft3 /min. At what rate is the water rising in the tank when the depth of the water is: (a) 1 foot? (b) 10 feet? . x How fast must the gun be able to turn to accurately track the aircraft when the plane is: (c) 19 feet? How long will the tank take to fill when starting at empty? Page 190 Custom Made for Math 1174 at Langara College by P. Anhaouy 30 12. A rope, attached to a weight, goes up through a pulley at the ceiling and back down to a worker. The man holds the rope at the same height as the connection point between rope and weight. (c) How fast is the weight rising after the man has walked 40 feet? (d) How far must the man walk to raise the weight all the way to the pulley? 2 /s . Suppose the man stands directly next to the weight (i.e., a total rope length of 60 ft) and begins to walk away at a rate of 2ft/s. How fast is the weight rising when the man has walked: 14. A hot air balloon lifts off from ground rising vertically. From 100 feet away, a 5’ woman tracks the path of the balloon. When her sightline with the balloon makes a 45◦ angle with the horizontal, she notes the angle is increasing at about 5◦ /min. (a) What is the elevation of the balloon? (a) 10 feet? (b) How fast is it rising? (b) 40 feet? How far must the man walk to raise the weight all the way to the pulley? 13. Consider the situation described in Exercise 12. Suppose the man starts 40ft from the weight and begins to walk away at a rate of 2ft/s. (a) How long is the rope? (b) How fast is the weight rising after the man has walked 10 feet? 15. A company that produces landscaping materials is dumping sand into a conical pile. The sand is being poured at a rate of 5ft3 /sec; the physical properties of the sand, in conjunction with gravity, ensure that the cone’s height is roughly 2/3 the length of the diameter of the circular base. How fast is the cone rising when it has a height of 30 feet? Custom Made for Math 1174 at Langara College by P. Anhaouy Page 191 Chapter 4 Applications of the Derivative 4.2 Applied Optimization Problems In this section we apply the concepts of extreme values to solve “word problems,” i.e., problems stated in terms of situations that require us to create the appropriate mathematical framework in which to solve the problem. We start with a classic example which is followed by a discussion of the topic of optimization. . . . . . . . . . . . . . . . . . . . . . . . Example 122 Optimization: perimeter and area A man has 100 feet of fencing, a large yard, and a small dog. He wants to create a rectangular enclosure for his dog with the fencing that provides the maximal area. What dimensions provide the maximal area? . . y . . . . . . . . . . . . . . One can likely guess the correct answer – that is great. Solution We will proceed to show how calculus can provide this answer in a context that proves this answer is correct. It helps to make a sketch of the situation. Our enclosure is sketched twice in Figure 4.3, either with green grass and nice fence boards or as a simple rectangle. Either way, drawing a rectangle forces us to realize that we need to know the dimensions of this rectangle so we can create an area function – after all, we are trying to maximize the area. We let x and y denote the lengths of the sides of the rectangle. Clearly, Area = xy. . x y We do not yet know how to handle functions with 2 variables; we need to reduce this down to a single variable. We know more about the situation: the man has 100 feet of fencing. By knowing the perimeter of the rectangle must be 100, we can create another equation: Perimeter = 100 = 2x + 2y. x Figure 4.3: A sketch of the enclosure in Example 122. We now have 2 equations and 2 unknowns. In the latter equation, we solve for y: y = 50 − x. Now substitute this expression for y in the area equation: Area = A(x) = x(50 − x). Note we now have an equation of one variable; we can truly call the Area a function of x. This function only makes sense when 0 ≤ x ≤ 50, otherwise we get negative values of area. So we find the extreme values of A(x) on the interval [0, 50]. Notes: Page 192 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.2 Applied Optimization Problems To find the critical points, we take the derivative of A(x) and set it equal to 0, then solve for x. A(x) = x(50 − x) 0 = 50x − x2 A (x) = 50 − 2x We solve 50 − 2x = 0 to find x = 25; this is the only critical point. We evaluate A(x) at the endpoints of our interval and at this critical point to find the extreme values; in this case, all we care about is the maximum. Clearly A(0) = 0 and A(50) = 0, whereas A(25) = 625ft2 . This is the maximum. Since we earlier found y = 50 − x, we find that y is also 25. Thus the dimensions of the rectangular enclosure with perimeter of 100 ft. with maximum area is a square, with sides of length 25 ft. This example is very simplistic and a bit contrived. (After all, most people create a design then buy fencing to meet their needs, and not buy fencing and plan later.) But it models well the necessary process: create equations that describe a situation, reduce an equation to a single variable, then find the needed extreme value. “In real life,” problems are much more complex. The equations are often not reducible to a single variable (hence multi–variable calculus is needed) and the equations themselves may be difficult to form. Understanding the principles here will provide a good foundation for the mathematics you will likely encounter later. We outline here the basic process of solving these optimization problems. Key Idea 4 Solving Optimization Problems 1. Understand the problem. Clearly identify what quantity is to be maximized or minimized. Make a sketch if helpful. 2. Create equations relevant to the context of the problem, using the information given. (One of these should describe the quantity to be optimized. We’ll call this the fundamental equation.) 3. If the fundamental equation defines the quantity to be optimized as a function of more than one variable, reduce it to a single variable function using substitutions derived from the other equations. (continued). . . Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 193 Chapter 4 Applications of the Derivative Key Idea 4 ued Solving Optimization Problems – Contin- 4. Identify the domain of this function, keeping in mind the context of the problem. 5. Find the extreme values of this function on the determined domain. 6. Identify the values of all relevant quantities of the problem. We will use Key Idea 4 in a variety of examples. Example 123 Optimization: perimeter and area Here is another classic calculus problem: A woman has a 100 feet of fencing, a small dog, and a large yard that contains a stream (that is mostly straight). She wants to create a rectangular enclosure with maximal area that uses the stream as one side. (Apparently her dog won’t swim away.) What dimensions provide the maximal area? . . We will follow the steps outlined by Key Idea 4. 1. We are maximizing area. A sketch of the region will help; Figure 4.4 gives two sketches of the proposed enclosed area. A key feature of the sketches is to acknowledge that one side is not fenced. . . . . . . . . . . . . . . . . . . . . . Solution . . . . . y 2. We want to maximize the area; as in the example before, Area = xy. . . This is our fundamental equation. This defines area as a function of two variables, so we need another equation to reduce it to one variable. x x We again appeal to the perimeter; here the perimeter is y Perimeter = 100 = x + 2y. Note how this is different than in our previous example. Figure 4.4: A sketch of the enclosure in Example 123. 3. We now reduce the fundamental equation to a single variable. In the perimeter equation, solve for y: y = 50 − x/2. We can now write Notes: Page 194 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.2 Area as Applied Optimization Problems 1 Area = A(x) = x(50 − x/2) = 50x − x2 . 2 Area is now defined as a function of one variable. 4. We want the area to be nonnegative. Since A(x) = x(50 − x/2), we want x ≥ 0 and 50 − x/2 ≥ 0. The latter inequality implies that x ≤ 100, so 0 ≤ x ≤ 100. 5. We now find the extreme values. At the endpoints, the minimum is found, giving an area of 0. Find the critical points. We have A0 (x) = 50 − x; setting this equal to 0 and solving for x returns x = 50. This gives an area of A(50) = 50(25) = 1250. 6. We earlier set y = 50 − x/2; thus y = 25. Thus our rectangle will have two sides of length 25 and one side of length 50, with a total area of 1250 ft2 . Keep in mind as we do these problems that we are practicing a process; that is, we are learning to turn a situation into a system of equations. These equations allow us to write a certain quantity as a function of one variable, which we then optimize. . Example 124 Optimization: minimizing cost A power line needs to be run from an power station located on the beach to an offshore facility. Figure 4.5 shows the distances between the power station to the facility. It costs $50/ft. to run a power line along the land, and $130/ft. to run a power line under water. How much of the power line should be run along the land to minimize the overall cost? What is the minimal cost? 1000 . . 5000 Figure 4.5: Running a power line from the power station to an offshore facility with minimal cost in Example 124. We will follow the strategy of Key Idea 4 implicitly, Solution without specifically numbering steps. There are two immediate solutions that we could consider, each of which we will reject through “common sense.” First, we could minimize the distance by directly connecting the two locations with a straight line. However, this requires that all the wire be laid underwater, the most costly option. Second, we could minimize the underwater length by running a wire all 5000 ft. along the beach, directly across from the offshore facility. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 195 Chapter 4 Applications of the Derivative This has the undesired effect of having the longest distance of all, probably ensuring a non–minimal cost. The optimal solution likely has the line being run along the ground for a while, then underwater, as the figure implies. We need to label our unknown distances – the distance run along the ground and the distance run underwater. Recognizing that the underwater distance can be measured as the hypotenuse of a right triangle, we choose to label the distances as shown in Figure 4.6. By choosing x as we did, we make the expression under the square root simple. We now create the cost function. Cost = land cost $50 × land distance 50(5000 − x) + water cost + $130 ×√water distance + 130 x2 + 10002 . √ So we have c(x) = 50(5000 − x) + 130 x2 + 10002 . This function only makes sense on the interval [0, 5000]. While we are fairly certain the endpoints will not give a minimal cost, we still evaluate c(x) at each to verify. c(0) = 380, 000 c(5000) ≈ 662, 873. We now find the critical values of c(x). We compute c 0 (x) as . c 0 (x) = −50 + √ 2 0 √ x2 + 0 10 1000 130x x2 + 10002 . . . 5000 − x x Recognize that this is never undefined. Setting c 0 (x) = 0 and solving Figure 4.6: Labeling unknown distances in Example 124. Notes: Page 196 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.2 Applied Optimization Problems for x, we have: 130x −50 + √ x2 + 10002 √ x2 + 10002 130x 1302 x2 x2 + 10002 =0 = 50 = 502 1302 x2 = 502 (x2 + 10002 ) 1302 x2 − 502 x2 = 502 · 10002 (1302 − 502 )x2 = 50, 0002 50, 0002 1302 − 502 50, 000 x= √ 1302 − 502 50, 000 2 x= = 416 ≈ 416.67. 120 3 x2 = Evaluating c(x) at x = 416.67 gives a cost of about $370,000. The distance the power line is laid along land is 5000 − 416.67 = 4583.33 ft., √ and the underwater distance is 416.672 + 10002 ≈ 1083 ft. In the exercises you will see a variety of situations that require you to combine problem–solving skills with calculus. Focus on the process; learn how to form equations from situations that can be manipulated into what you need. Eschew memorizing how to do “this kind of problem” as opposed to “that kind of problem.” Learning a process will benefit one far longer than memorizing a specific technique. The next section introduces our applications of the derivatives to economics and business problems such as elasticity, optimal harvest, and inventory cost control. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 197 Exercises 4.2 Terms and Concepts 1. T/F: An “optimization problem” is essentially an “extreme values” problem in a “story problem” setting. 2. T/F: This section teaches one to find the extreme values of function that have more than one variable. What is the maximum volume of a package with a square cross section (w = h) that does not exceed the 108” standard? 12. The strength S of a wooden beam is directly proportional to its cross sectional width w and the square of its height h; that is, S = kwh2 for some constant k. w Problems 3. Find the maximum product of two numbers (not necessarily integers) that have a sum of 100. 4. Find the minimum sum of two numbers whose product is 500. 5. Find the maximum sum of two numbers whose product is 500. 6. Find the maximum sum of two numbers, each of which is in [0, 300] whose product is 500. 7. Find the maximal area of a right triangle with hypotenuse of length 1. 8. A rancher has 1000 feet of fencing in which to construct adjacent, equally sized rectangular pens. What dimensions should these pens have to maximize the enclosed area? 9. A standard soda can is roughly cylindrical and holds 355cm3 of liquid. What dimensions should the cylinder be to minimize the material needed to produce the can? Based on your dimensions, determine whether or not the standard can is produced to minimize the material costs. 10. Find the dimensions of a cylindrical can with a volume of 206in3 that minimizes the surface area. The “#10 can”is a standard sized can used by the restaurant industry that holds about 206in3 with a diameter of 6 2/16in and height of 7in. Does it seem these dimensions where chosen with minimization in mind? 11. The United States Postal Service charges more for boxes whose combined length and girth exceeds 108” (the “length” of a package is the length of its longest side; the girth is the perimeter of the cross section, i.e., 2w + 2h). 12 h Given a circular log with diameter of 12 inches, what sized beam can be cut from the log with maximum strength? 13. A power line is to be run to an offshore facility in the manner described in Example 124. The offshore facility is 2 miles at sea and 5 miles along the shoreline from the power plant. It costs $50,000 per mile to lay a power line underground and $80,000 to run the line underwater. How much of the power line should be run underground to minimize the overall costs? 14. A power line is to be run to an offshore facility in the manner described in Example 124. The offshore facility is 5 miles at sea and 2 miles along the shoreline from the power plant. It costs $50,000 per mile to lay a power line underground and $80,000 to run the line underwater. How much of the power line should be run underground to minimize the overall costs? 15. A woman throws a stick into a lake for her dog to fetch; the stick is 20 feet down the shore line and 15 feet into the water from there. The dog may jump directly into the water and swim, or run along the shore line to get closer to the stick before swimming. The dog runs about 22ft/s and swims about 1.5ft/s. How far along the shore should the dog run to minimize the time it takes to get to the stick? (Hint: the figure from Example 124 can be useful.) 16. A woman throws a stick into a lake for her dog to fetch; the stick is 15 feet down the shore line and 30 feet into the water from there. The dog may jump directly into the water and swim, or run along the shore line to get closer to the stick before swimming. The dog runs about 22ft/s and swims about 1.5ft/s. How far along the shore should the dog run to minimize the time it takes to get to the stick? (Google “calculus dog” to learn more about a dog’s ability to minimize times.) 17. What are the dimensions of the rectangle with largest area that can be drawn inside the unit circle? Page 198 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.3 4.3 Economics and Business Applications 4.3.1 Elasticity of Demand Problems Definition 22 Economics and Business Applications General Elasticity The elasticity, E ∗ (p), of a demand curve at a price level p is defined as relative rate of change of the demand E ∗ (p) = . relative rate of change of price Theorem 37 If price and demand are related by a demand equation x = f (p), or q = f (p), then the elasticity of the demand is given by E ∗ (p) = f 0 (p) x0 q0 · p, or E ∗ (p) = · p, or E ∗ (p) = · p, f (p) x q if f (p) is a differentiable function of p. Notice that here, x0 = dx f 0 (p) d x0 . The quotient = [ln f (p)], or = dp f (p) dp x d [ln x], defines the rate of growth of the function x = f (p). This ratio is dp sometimes called the the relative rate of change of the function x = f (p). This leads to elasticity as f 0 (p) d [ln f (p)] relative rate of change of f (p) f (p) dp = = . E ∗ (p) = 1 d relative rate of change of p [ln p] p dp rate of growth of f (p) or . rate of growth of p Remarks: • When |E ∗ | > 1, the demand curve is called elastic at price p. It means that the demand is sensitive to changes in price. A percentage change in price produces a large percentage change in demand. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 199 Chapter 4 Applications of the Derivative • When |E ∗ | < 1, the demand curve is called inelastic at price p. This means that the demand is not sensitive to changes in price. A percentage change in price produces smaller percentage change in demand. • When |E ∗ | = 1 the demand curve is perfect elastic or unitary meaning that the percentage change in demand and price are relatively equal. The figures below show two different demand curves relating the price p at which an item sells to the quantity x = f (p) of that item consumers will purchase in a given period of time. Note that because of the steepness of the graph of the curve f1 , a small percentage change in price will result in a substantial percentage change in demand. Economists refer to such demand curve as elastic. Because the slope of the demand curve f2 is relatively small, a small percentage change in p will cause only a small percentage change in demand. Demand curves with this property are called inelastic. x x x = f (p) : elastic x = f(p) : inelastic p Figure 1 (a) p Figure 2 (b) Figure 4.7: Price-Demand Case: Elastic and Inelastic Curves. Special Case: For the Price-Demand Case shown in Figure4.7. We define the elasticity as follow. Notes: Page 200 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.3 Definition 23 Function Economics and Business Applications Special Case - Elasticity of Price Demand The elasticity, E(p), of the demand curve for x = f (p) at price level p is defined as E=− relative rate of change of the demand . relative rate of change of price , Theorem 38 Special Case If price and demand are related by a demand equation x = f (p), or q = f (p), then the elasticity of the demand is given by E(p) = − f 0 (p) x0 q0 · p, or E(p) = − · p, or E(p) = − · p, f (p) x q if f (p) is a differentiable function of p. Remarks: For the Special Case, • The minus sign is to make sure the E(p) is always a positive quantity. Another way is to use absolute value: E = E(p) . • When E > 1, the demand curve is called elastic at any price p. It means that the demand is sensitive to changes in price. A percentage change in price produces a large percentage change in demand. When 0 < E < 1, the demand curve is called inelastic at any price p. This means that the demand is not sensitive to changes in price. A percentage change in price produces smaller percentage change in demand. When E = 1 the demand curve is perfect elastic meaning or unitary meaning that the percentage change in demand and price are relatively equal. • In this case, we may interpret elasticity as the (negative of the) ratio of the relative change in demand to the relative change in price at any price level p. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 201 Chapter 4 Applications of the Derivative Recall R(p) = x · p = p · f (p) ⇒ R0 (p) = p · f 0 (p) + f (p) =  that 0 p · f (p) +1 f (p) f (p) ⇒ R0 (p) = f (p)[1 − E(p)] ⇒ M R(p) = f (p) (1  − E).  1 Note: It can be shown also that M R(x) = p 1 − . E So, we can say (see Figure 4.8) 1. Elastic: E > 1 : p ↑ ⇒ R(p) ↓ or p ↓ ⇒ R(p) ↑. 2. Inelastic: E < 1 : p ↑ ⇒ R(p) ↑ or p ↓ ⇒ R(p) ↓. Inelastic increasing revenue Elastic decreasing revenue R(p) E(p) < 1, R0 (p) > 0 R" p# p" R# E(p) > 1, R0 (p) < 0 R" p" p# R# p Figure 4.8: PriceFigure Demand3 Case: Revenue Curve. Notice that the following examples are based on the price-demand case only. Notes: Page 202 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.3 Example 125 Find the elasticity of the demand curve with equation x = Economics and Business Applications 500 at (p + 1)2 price level p = 4. 1000 − (p+1) 3 1000 2p x =− . So, E(p) = − 500 × p = ⇒ Solution (p + 1)3 p +1 (p+1)2 (2)(4) 8 E(4) = = > 1. So, we say that this demand curve is elastic at 4+1 5 price p = 4. 0 Example 126 The demand for an item is found to be roughly x = 200 − p, for prices between p = 25 and p = 150 dollars. For which prices is this demand curve (a) elastic? (b) inelastic ? −(−1) p ×p = . We need 200 − p 200 − p to solve E > 1 for elastic and E < 1 for inelastic. E = 1 ⇒ p = 100. After using test points, we can see that we have inelastic curve for p ∈ [25, 100) and elastic curve for p ∈ (100, 150]. Example 127 Skytrain used to charge $2 per person (for a one-zone trip). Say the demand function is given by x2 + 450p = 1000, where x is the number of riders (in thousands) and p is the price of a ticket in dollars. (a) Compute the elasticity of the demand at p = $2. (b) Should the ticket price be raised or lowered to increase the revenue from the Skytrain? (c) What price will maximize the revenue? What will be the demand at this price? Solution x0 = −1. So, E(p) = − 0 x (a) 2xx0 +450 = 0 ⇒ x0 = − 450 2x . Since E = − x ×p, at 45 45 p = 2 we have x = 10 and x0 = − and E(2) = > 1. So, it is elastic. 2 10 (b) We should lower the price of the ticket in order to increase revenue. 1 1 (c) R(x) = px = (1000x − x3 ) ⇒ R0 (x) = (1000 − 3x2 ) = 0 ⇒ 450 450 r 1000 6 x= . R00 (x) = − x < 0 for x above. So, R(x) has maximum 3 450 at x above. Si, the price should be p = 2000 1350 = 1.48 and the demand is q x = 1000 3 . Solution OR: p x = 1000 − 450p, 0 < p < 2.2 (radicand is greater than zero). This gives Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 203 Chapter 4 Applications of the Derivative 225p . Since M R(p) = f (p)(1−E), to maximize we set this 1000 − 450p 225p ⇒p= to be zero and get E = 1 (perfect elastic). So, 1 = 1000 p− 450p 1.48 (same answer). Of couse, we have to check R(p) = 1000 − 450p · p at critical number p = 1.48 and other two end points for the absolute max. values. E(p) = Example 128 Suppose that you sell books for $7 each. On average 50 books are sold each week. If the price of the book is dropped to $5 each then each week you sell 58. Assume that the demand is linear function of the price. (a) Find the price that will maximize revenues from selling the books. (b) Assuming that each book costs $3 to make, find the elasticity of demand at the profit maximizing price. 58 − 50 · (p − 7) = 50 − 4(p − 7) ⇒ 5−7 2 x = −4p + 78. Then, R(p) = x · p = −4p + 78p ⇒ M R(p) = −8p + 78. M R(p) = 0 gives p = $9.75. Solution (a) x = f (p) = 50 + (b) P = R − C = −4p2 + 78p − 3x = −4p2 + 78p + 12p − 3 · 78 ⇒ P 0 (p) = x0 4p dx = −4 ⇒ E(p) = − · p = . −8p + 90 = 0 ⇒ p = 11.25. x0 = dp x 78 − 4p When p = 11.25, we have E(11.15) = 1.36 > 1. So, it is elastic. 4.3.2 Optimal Harvest Problems Let’s consider the following example. Suppose that the owner of a stand of timber estimates that the timber is currently worth $5000 and that√ it t will increase in value with time according to the formula V (t) = 5000e 2 , where t is the time in years. But, the prevailing interest rates expected to remain at 12%, compounded continuously, and the owner worries that by waiting longer to convert the timber to cash considerable interest income will be forgone. When should the owner sell the timber so as to maximize revenue? Since the money received in the future has different values today depending on when it is to be received, we must convert the revenue to be received in the future into its present value using the formula, A = P ert ⇒ P = e−rt A, where P is present value, and A is future value. In this problem, Notes: Page 204 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.3 √ Economics and Business Applications t the future value is V (t) = 5000e 2 , and r = 0.12, the present value of the revenue in selling the timber after t years is given by   √ √ √ t t t 1 P (t) = e−0.12t 5000e 2 = 5000e 2 −0.12t ⇒ P 0 (t) = 5000e 2 −0.12t √ − 0.12 . 4 t 1 To find the maximum value of P (t), we set P 0 (t) = 0 ⇒ √ − 0.12 = 4 t  2 √ 1 1 1 0⇒ t= = . So, we have t = = 4.34 years. It is 4(0.12) 0.48 0.48 about 4 years and 4 months. The above example is a typical of many situations in which the value of an asset V (t) is growing in time. The question of when to sell the asset is called an optimal harvest problem because it applies harvesting fish from a fishery, trees from a woodland, wine from an ageing facility, and so on. What makes this question of when to harvest interesting is that while the value of the asset V (t) is growing in time, the owner is missing the opportunity to invest these dollars at the prevailing interest rate r, which usually is compounded continuously. This means that V (t) must be discounted by the factor e−rt used to calculate present value. In this case, the present value of the revenue received from harvesting after t years is P (t) = V (t)e−rt . To determine the optimal harvest time, we maximize P (t) by setting P 0 (t) = 0. P 0 (t) = V 0 (t)e−rt +V (t)e−rt (−r) = [V 0 (t) − rV (t)] e−rt = 0 ⇒ V 0 (t)−rV (t) = 0 V 0 (t) , which is the rate of growth of V (t). V (t) Therefore, under a constant prevailing rate of interest r, the optimal time to harvest the asset whose value is given by the function V (t) is when the rate of growth of V (t) equals to the √prevailing interest rate. ⇒r= t In the above example, V (t) = 5000e 2 and r = 0.12,, when we harvest we must have  √  t 1 √ 2 5000e 0 V (t) 1 1 4 t √ 0.12 = = = √ ⇒ 0.12 = √ ⇒ t = 4.34 years . t V (t) 4 t 4 t 5000e 2 In general, the interest rate r(t) depends on the time t. We can write it as follow V 0 (t) d r(t) = , or r(t) = [ln V (t)] . V (t) dt Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 205 Chapter 4 Applications of the Derivative d [ln V (tc )]. It can be dt 00 0 shown that P (tc ) = r (tc )P (tc ). So, the present value P (tc ) is a maximum when r0 (tc ) < 0 and a minimum when r0 (tc ) > 0. Note: We use the second derivative test. When harvest time tc , we must have r = r(tc ) = Example 129 √ The value at time t of an art object is given by V (t) = 1000 × 10 t . 1. Find r(t), the rate of growth of the value at time t. 2. If the cost-of-capital (applicable interest rate) is r = 10% (continuous p.a.), when should the art object be sold to maximize present value? 3. If the art object was sold after 50 years to maximize present value, what was the interest rate used? Solution 1. r(t) = V (t) = 1000×10 √ t 0 ⇒ V (t) = 1000×(ln 10)×10 √ t ·  1 √ 2 t  . V 0 (t) ln 10 = √ . V (t) 2 t ln 10 2. 0.10 = r(t) = √ ⇒ t = 2 t  ln 10 2(0.01) 2 = 132.5 years. ln 10 3. r(50) = √ = 0.1628. That is r = 16.28% 2 50 Example 130 A poor math teacher is hoping to make some money by selling some antique plates she had just received from her grandma. The value √ of the plates is appreciating in time following a model V (t) = 3000 2t + 1. When is the right time to sell the plate if we know the interest rate is 4%? √ 1 V (t) = 3000 2t + 1 ⇒ V 0 (t) = 3000 √ · (2) = 2 2t + 1 3000 √ 0 1 3000 V (t) 1 2t + 1 √ √ = ⇒ 0.04 = ⇒t= . r(t) = = V (t) 2t + 1 2t + 1 2t + 1 3000 2t + 1 12 years. Solution Notes: Page 206 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.3 Economics and Business Applications Example 131 A 12th century collection of artefacts has its ”nominal” market value (in dollars) modelled by V (t) = 100te−0.05t , where t is measured in years from today. 1. How long will it take until V will achieve its maximum nominal value? 2. The owner can invest her savings at an interest rate of 5% per year compounded continuously. When should she sell her collection in order to maximize her wealth? Show that your answer gives the maximum (not the minimum) value. V (t) = 100te−0.05t , where t is measured in years from Solution today. 1. V 0 (t) = 100e−0.05t + 100te−0.05t (−0.05) = 100e−0.05t (1 − 0.05t) = 0 ⇒ 0.05t = 1 ⇒ t = 20 years. 100e−0.05t (1 − 0.05t) 1 − 0.05t V 0 (t) ⇒ 0.05 = = ⇒ 0.05t = V (t) 100te−0.05t t 1 − 0.05t ⇒ t = 10 years.   d V 0 (t) To be a maximum, we needs r0 (r) < 0, i.e. < 0. Here dt V (t)  0    d 1 − 0.05t 1 d V (t) = = − 2 < 0 for all t. dt V (t) dt t t 2. r(t) = 4.3.3 Inventory Costs Control Suppose we run a small business selling stuff. We must know how to order if we want to keep an inventory available all the time during the year. One way we can do is to make order every year. However, doing this requires that we have to stock it somewhere and we have our capital sitting in a storage. These are called carrying costs. Another way, we can order lots of small shipments over the year. Here, we have to pay for shipping and labor to check the shipments arrived and fill out order forms. So, our decision is to be between these two approaches so that we can minimize the total cost. Our approach follows the model called the Economic Order Quantity (EOQ). The model makes the assumptions that 1. x be the order quantity, r be the cost of storing one unit for one year, and s be the cost for placing an order. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 207 Chapter 4 Applications of the Derivative 2. The inventory is drawn down at a fixed continuous rate Q units per year. 3. The inventory slowly declines to zero at the exact time that the next order arrives. This means that it starts at x units and decreases to 0 units, then instantly goes up to x. From this assumptions, we can see that Q Q • the annual ordering costs is Co (x) = · s, where is the number x x of orders placed per year, x x • the annual storage cost is Cs (x) = · r ,where is the average size 2 2 of the inventory through the year, and Q • the total annual inventory costs are C(x) = Co (x) + Cs (x) = · s + x x · r, x > 0. 2 Example 132 Show that C above is minimized when x = √ total cost is 2Qrs. r 2Qs , and show that the r Qs r C (x) = − 2 + = 0 ⇒ x = x 2 r 2Qs . Since C 00 (x) = r r 2Qs 2Qs > 0 for all x > 0, C(x) has local min. at x = . Now, 3 x r lim C(x) = ∞, and lim C(x) = ∞. C(x) is in fact minimized when x→∞ x→0+ q r 2Qs 2Qs 2Qs r r x= . The minimum value is C = q . Simplify this + r 2 2Qs Solution 0 r gives the answer. Example 133 Suppose that a liquor store manager wants to have 1200 cases of beers over the year. He realizes that the cost to stock a case of beer is $8 per year and each delivery the order costs him $75. What size of order should he place in order to minimize the total cost? How often should he order a year? Notes: Page 208 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.3 Economics and Business Applications 1200 x 90000 Here, C(x) = · 75 + · 8 = + 4x ⇒ C 0 (x) = x 2 x 90000 180000 . Setting C 0 (x) = 0 and solving it gives − 2 + 4, C 00 (x) = x x3 x = 150 (take the positive x only). Since C 00 (150) > 0, C(x) has a minimum at x = 150 (no need to show that this is a global minimum because it has been shown in the previous equation). Thus, the manage 1200 = 8 times a year should order 150 cases of beers and he should order 150 in order to minimize the total cost. Solution Example 134 A merchant sells chairs rather uniformly at a rate of about 1600 chairs a year. Accounting procedures determine that the cost of carrying a chair in stock is $2 per year. When ordering chairs, the merchant experiences a fixed cost of $25 plus a variable order of 10 cents per chair. Assuming that orders can be placed so that delivery occurs precisely when stock is depleted, how often should the merchant order chair so that to minimize yearly cost? 1600 Q ·s = · (25 + 0.10x) = 25 · Here C0 (x) = x x 1600 x x + 160. Cs (x) = · r = · 2 = x. The total cost function is then x 2 2 25 · 1600 25 · 1600 C(x) = + 160 + x. C 0 (x) = − + 1 = 0 ⇒ x = 200. x x2 50 · 1600 Since C 00 (x) = > 0 for all x > 0, C(x) has a min. at x = 200. x3 1600 The number of times that the merchant should order is = 8 in order 200 minimize the total cost. Solution Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 209 Exercises 4.3 Problems In Exercises 1 – 10, solve the elasticity of demand problems below. 1. Suppose that a store selling electric-bikes has a demand function x = 500 − 10p, where x is the number of bikes and p is the unit price. (a) Find the elasticity of this function. (b) Is the demand elastic or inelastic when the price is $30? (c) What is the percentage change in the demand if the price is $30, but changing by 4.5%? 2. Suppose that we have a demand function x = 100 − p . 4 (a) Find the elasticity of this function. (a) If currently, p = $200, should we increase or the decrease the revenue if the price is lowered slightly? (b) Find the price that maximize the revenue. 8. Suppose that the demand and price per unit is related in the following equation xp + 30p + 50x = 15840. (a) If currently, p = $150, should we increase or the decrease the revenue if the price is lowered slightly? (b) Find the price that maximize the revenue. In Exercises 9 – 16, solve the optimal harvest problems below. (c) Find the intervals of elasticity in term of the demand x. 9. An asset has value at time t of V (t) = 100(60 + t2 ). An investor can borrow and lend (deposit in the bank) at interest rate r = 6.25% per annum, continuously compounded. When should this investor acquire the asset and when should she sell it? 3. Suppose that the demand and price per unit is related in the following equation x2 + p2 = 4, where x = f (p). 10. The value at time t of an art object is given by V (t) = 1000(t2 + 3t + 4). (b) Is the demand elastic or inelastic when the price is $20? (a) Find the elasticity of the demand curve. (b) Is the demand elastic or inelastic when x = 1? (c) Find the intervals of elasticity in term of the demand x. 4. Suppose that the demand and price per unit is related in the following equation p2 (100 + x) = 250, where x = f (p). (a) Find the elasticity of the demand curve. (b) Is the demand elastic or inelastic when x = 25? 5. Suppose that the demand and price per unit is related  x 2 in the following equation p = 100 − . Currently, 10 the price per unit is $100. (a) Find the elasticity of demand curve. (b) Should the unit price be increased or decreased in order to increase the revenue? 6. Suppose that the demand and price per unit is related  2 x − 3000 in the following equation p = . Currently, 600 the price per unit is $1800. (a) Find the elasticity of demand curve. (b) Should the unit price be increased or decreased in order to increase the revenue? 7. Suppose that the demand and price per unit is related in the following equation ln x − 2 ln p + 0.02p = 7, where p ≥ 100. (a) Find r(t), the rate of growth of the value at time t. (b) If the cost-of-capital (applicable interest rate) is r = 10%, when should the art object be sold to maximize the present value of the sale’s proceeds? 11. A work of art has its value t years in the future modelled by V (t) = 250(5 + 0.2t)3/2 , t ≥ 0. It is expected that the applicable interest rate will remain at r = 4%, continuously compounded. When should the holder of the art piece sell it, to maximize their wealth? How much is the value then? 12. A work of art has its value t years in the future mod√ 3 t elled by V (t) = 100 · 7 , t ≥ 0. It is expected that the applicable interest rate will remain at r = 15%, continuously compounded. When should the holder of the art piece sell it, to maximize their wealth? 13. A collection of memorabilia has its nominal market value (in dollars) modeled by V (t) = 50te−0.10t , where t is measured in years. (a) Find the time tc when the nominal value V is at its maximum. Explain or show V (tc ) is the absolute maximum of V for t ≥ 0. (b) The owner can invest her savings at interest rate of r = 2.5% per year compounded continuously. When should she sell her collection in order to maximize her wealth? Page 210 Custom Made for Math 1174 at Langara College by P. Anhaouy 14. A collection of films has its nominal market value (in dollars) modeled by V (t) = 100(60t − t2 ), where t is measured in years. The owner can invest her savings at interest rate of r = 3.5% per year compounded continuously. When should she sell her collection in order to maximize her wealth? 15. A collection of films has its nominal market value (in dollars) modeled by V (t) = 100(60t − t2 ), where t is measured in years. Find the time tc when the nominal value V is at its maximum. Explain or show V (tc ) is the absolute maximum of V for t ≥ 0. 16. A collection of films has its nominal market value (in dollars) modeled by V (t) = 0.20te−0.20t , where t is measured in years. Find the time tc when the nominal value V is at its maximum. Explain or show V (tc ) is the absolute maximum of V for t ≥ 0. The owner can invest her savings at interest rate of r = 0.5% per year compounded continuously. When should she sell her collection in order to maximize her wealth? In Exercises 17 – 18, solve the inventory control problems below. 17. A merchant sells chairs rather uniformly at a rate of about 10, 000 chairs a year. Accounting procedures determine that the cost of carrying a chair in stock is $5.76 per chair per year. When ordering chairs, the merchant experiences a fixed cost of $4.50 per order. Assuming that orders can be placed so that delivery occurs precisely when stock is depleted, how often should the merchant order chair so that to minimize yearly cost? Find the minimum total cost. 18. By allowing production runs, the EOQ model becomes Lot-Size model,. The total cost then becomes C(x) = r(1 − d/p) Q ·s+ · x, where d is the daily demand and x 2 p is the daily production. Showsthat the minimum cost 2s Q in this case happen when x = · . r (1 − d/p) Custom Made for Math 1174 at Langara College by P. Anhaouy Page 211 Chapter 4 Applications of the Derivative 4.4 Linear Approximation and Differentials In Section 2.2 we explored the meaning and use of the derivative. One idea was about using the tangent line equation at a point to approximate a value of a function near by that point. This section starts by revisiting this idea. 4.4.1 Linear Approximation (Tangent Line Approximation Revisited) Suppose that the function y = f (x) is differentiable at x = a. Then, f 0 (a) is the slope of the tangent line at (a, f (a)), and it is given by f 0 (a) = lim x→a f (x) − f (a) f (x) − f (a) ⇒ f 0 (a) ≈ . x−a x−a This means the same as the slope of the tangent line is approximately the same as the slope of the secant line as x approaches a. Multiplying both sides by x − a and rearranging terms gives f (x) ≈ f (a) + f 0 (a)(x − a). So, basically, we can use this to approximate the value of f (x) for any x close to a. Of course, in any approximation problem, we should have some errors involved. In this case, we let the approximation error to be E(x), which defined as the difference between the sides of the above formula and as x approaches a, E(x) goes to zero even faster. Theorem 39 Linear Approximation If the function y = f (x) is a differentiable function at x = a, then f (x) ≈ f (a) + f 0 (a)(x − a). Moreover, the error E(x) = f (x) − f (a) − f 0 (a)(x − a) satisfies E(x) lim = 0. x→a x − a Remarks: • We can show that the error goes to zero as follow: E(x) f (x) − f (a) − f 0 (a)(x − a) f (x) − f (a) = = + f 0 (a). x−a x−a x−a Notes: Page 212 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.4 Linear Approximation and Differentials Then, taking the limits both sides, we obtain   E(x) f (x) − f (a) f (x) − f (a) 0 0 lim = lim − f (a) = lim −f (a) = x→a x − a x→a x→a x−a x−a 0 0 f (a) − f (a) = 0. • The linear function on the right of the approximation above is called the linear approximation or the tangent line approximation or the linearization of the function f (x) near the point a, and it is denoted by L(x) and we write it L(x) = f (a) + f 0 (a)(x − a). • So, we can say that, near the point a, f (x) ≈ L(x). • Graphically, see the figure on the left below y y y = f(x) y = f(x) y = L(x) f(x) E(x) ∆y L(x) dy y = L(x) a x x x ∆x = dx x + dx (a) x (b) Figure 4.9: Linear Approximations Example 135 Find the linearization, L(x), of f (x) = x2 − x at x = 2 and compare f (2.01) and L(2.01). f 0 (x) = 2x − 1. Then, L(x) = f (2) + f 0 (2)(x − 2) = 2 + 3(x − 2) = 3x − 4. So, f (2.01) = 2.0301, and L(2.01) = 2.03. Solution Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 213 Chapter 4 Applications of the Derivative Example 136 π π Approximate the value of tan 46o , using a = and ∆x = . 4 180 π . Let f (x) = tan x. Here, f (460 ) = f (45o ) + 4  π f 0 (45o ) x− . Since f 0 (x) = sec2 x, we have f (460 ) ≈ tan 45o +sec2 45o · 4 π ≈ 1.035. 180 Solution 45o = Example 137 √ √ How much bigger than 100 is 103? Solution Let f (x) = √ √ √ 1 x ⇒ f 0 (x) = √ . So, 103 − 100 ≈ 2 x 1 (3) = 0.15. f 0 (100)(103 − 100) = √ 2 100 Example 138 √ √ 3 Use the function f (x) = 3 x to approximate 25. √ 3 1 . Use a = 27, we have x2/3 √ 1 79 3 25 = f (25) ≈ f (27) + f 0 (27)(x − 27) = 3 + (25 − 27) = . 27 27 Solution f (x) = x ⇒ f 0 (x) = Example 139 A manufacturer company estimates the cost of hiring and training x new employees is √ C(x) = 500(x − x). 1. Find C 0 (16). 2. The company is trying to decide whether to hire 16 or 17 new employees. Approximate the difference in cost using linear approximation. Solution 1 1 1. C 0 (x) = 500(1 − √ ⇒ C 0 (16) = 500(1 − √ = $437.50. 2 x 2 16 2. C(17) − C(16) ≈ C 0 (16).(1) ≈ 437.50. Notes: Page 214 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.4 4.4.2 Linear Approximation and Differentials Differentials Recall that the derivative of a function f can be used to find the slopes of lines tangent to the graph of f . At x = a, the tangent line to the graph of f has equation y = f (a) + f 0 (a)(x − a). The tangent line can be used to find good approximations of f (x) for values of x near a as we have seen in linear approximation. For instance, we can approximate sin 1.1 using the tangent line√to the graph of f (x) = sin x at x = π/3 ≈ 1.05. Recall that sin(π/3) = 3/2 ≈ 0.866, and cos(π/3) = 1/2. Thus the tangent line to f (x) = sin x at x = π/3 is: 1 `(x) = (x − π/3) + 0.866. 2 In Figure 4.10(a), we see a graph of f (x) = sin x graphed along with its tangent line at x = π/3. The small rectangle shows the region that is displayed in Figure 4.10(b). In this figure, we see how we are approximating sin 1.1 with the tangent line, evaluated at 1.1. Together, the two figures show how close these values are. Using this line to approximate sin 1.1, we have: y 1 √ 3 2 0.5 x . π 3 (a) 1 (1.1 − π/3) + 0.866 2 1 = (0.053) + 0.866 = 0.8925. 2 `(1.1) = (We leave it to the reader to see how good of an approximation this is.) We now generalize this concept. Given f (x) and an x–value a, the tangent line is `(x) = f (a) + f 0 (a)(x − a). Clearly, f (a) = `(a). Let ∆x be a small number, representing a small change in x value. We assert that: y ℓ(1.1) ≈ sin 1.1 0.89 0.88 sin 1.1 0.87 √ 3 2 f (a + ∆x) ≈ `(a + ∆x), since the tangent. line to a function approximates well the values of that function near x = a. As the x value changes from a to a + ∆x, the y value of f changes from f (a) to f (a + ∆x). We call this change of y value ∆y. That is: ∆y = f (a + ∆x) − f (a). . x π 3 1.1 (b) Figure 4.10: Graphing f (x) = sin x and its tangent line at x = π/3 in order to estimate sin 1.1. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 215 Chapter 4 Applications of the Derivative Replacing f (a + ∆x) with its tangent line approximation, we have ∆y ≈ `(a + ∆x) − f (a)  = f 0 (a) (a + ∆x) − a + f (a) − f (a) = f 0 (a)∆x (4.3) This final equation is important; we’ll come back to it in Key Idea 5. We introduce two new variables, dx and dy in the context of a formal definition. Definition 24 Differentials of x and y. Let y = f (x) be differentiable. The differential of x, denoted dx, is any nonzero real number (usually taken to be a small number). The differential of y, denoted dy, is dy = f 0 (x)dx. We can solve for f 0 (x) in the above equation: f 0 (x) = dy/dx. This states that the derivative of f with respect to x is the differential of y divided by the differential of x; this is not the alternate notation for the dy derivative, dx . This latter notation was chosen because of the fraction–like qualities of the derivative, but again, it is one symbol and not a fraction. It is helpful to organize our new concepts and notations in one place. Key Idea 5 Differential Notation Let y = f (x) be a differentiable function. 1. ∆x represents a small, nonzero change in x value. 2. dx represents a small, nonzero change in x value (i.e., ∆x = dx). 3. ∆y is the change in y value as x changes by ∆x; hence ∆y = f (x + ∆x) − f (x). 4. dy = f 0 (x)dx which, by Equation (4.3), is an approximation of the change in y value as x changes by ∆x; dy ≈ ∆y. Notes: Page 216 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.4 Linear Approximation and Differentials What is the value of differentials? Like many mathematical concepts, differentials provide both practical and theoretical benefits. We explore both here. Example 140 Finding and using differentials Consider f (x) = x2 . Knowing f (3) = 9, approximate f (3.1). The x value is changing from x = 3 to x = 3.1; thereSolution fore, we see that dx = 0.1. If we know how much the y value changes from f (3) to f (3.1) (i.e., if we know ∆y), we will know exactly what f (3.1) is (since we already know f (3)). We can approximate ∆y with dy. ∆y ≈ dy = f 0 (3)dx = 2 · 3 · 0.1 = 0.6. We expect the y value to change by about 0.6, so we approximate f (3.1) ≈ 9.6. We leave it to the reader to verify this, but the preceding discussion links the differential to the tangent line of f (x) at x = 3. One can verify that the tangent line, evaluated at x = 3.1, also gives y = 9.6. Of course, it is easy to compute the actual answer (by hand or with a calculator): 3.12 = 9.61. (Before we get too cynical and say “Then why bother?”, note our approximation is really good!) So why bother? In “most” real life situations, we do not know the function that describes a particular behavior. Instead, we can only take measurements of how things change – measurements of the derivative. Imagine water flowing down a winding channel. It is easy to measure the speed and direction (i.e., the velocity) of water at any location. It is very hard to create a function that describes the overall flow, hence it is hard to predict where a floating object placed at the beginning of the channel will end up. However, we can approximate the path of an object using differentials. Over small intervals, the path taken by a floating object is essentially linear. Differentials allow us to approximate the true path by piecing together lots of short, linear paths. This technique is called Euler’s Method, studied in introductory Differential Equations courses. We use differentials once more to approximate the value of a function. Even though calculators are very accessible, it is neat to see how these techniques can sometimes be used to easily compute something that looks rather hard. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 217 Chapter 4 Applications of the Derivative Example 141√ Using differentials to approximate a function value Approximate 4.5. √ We expect 4.5 ≈ 2, yet we can do better. Let √ f (x) = √ Solution x, and let c = 4. Thus f (4) = 2. We can compute f 0 (x) = 1/(2 x), so f 0 (4) = 1/4. We approximate the difference between f (4.5) and f (4) using differentials, with dx = 0.5: f (4.5) − f (4) = ∆y ≈ dy = f 0 (4) · dx = 1/4 · 1/2 = 1/8 = 0.125. The approximate change in f from x = 4 to x = 4.5 is 0.125, so we ap√ proximate 4.5 ≈ 2.125. Differentials are important when we discuss integration. When we study that topic, we will use notation such as Z f (x) dx quite often. While we don’t discuss here what all of that notation means, note the existence of the differential dx. Proper handling of integrals comes with proper handling of differentials. In light of that, we practice finding differentials in general. Example 142 Finding differentials In each of the following, find the differential dy. 1. y = sin x 2. y = ex (x2 + 2) 3. y = √ x2 + 3x − 1 Solution 1. y = sin x: As f (x) = sin x, f 0 (x) = cos x. Thus dy = cos(x)dx. 2. y = ex (x2 + 2): Let f (x) = ex (x2 + 2). We need f 0 (x), requiring the Product Rule. We have f 0 (x) = ex (x2 + 2) + 2xex , so  dy = ex (x2 + 2) + 2xex dx. Notes: Page 218 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.4 Linear Approximation and Differentials √ √ 3. y = x2 + 3x − 1: Let f (x) = x2 + 3x − 1; we need f 0 (x), requiring the Chain Rule. We have f 0 (x) = 1 2x + 3 1 2 (x + 3x − 1)− 2 (2x + 3) = √ . Thus 2 2 x2 + 3x − 1 (2x + 3)dx . dy = √ 2 x2 + 3x − 1 Finding the differential dy of y = f (x) is really no harder than finding the derivative of f ; we just multiply f 0 (x) by dx. It is important to remember that we are not simply adding the symbol “dx” at the end. We have seen a practical use of differentials as they offer a good method of making certain approximations. Another use is error propagation. Suppose a length is measured to be x, although the actual value is x + ∆x (where we hope ∆x is small). This measurement of x may be used to compute some other value; we can think of this as f (x) for some function f . As the true length is x + ∆x, one really should have computed f (x + ∆x). The difference between f (x) and f (x + ∆x) is the propagated error. How close are f (x) and f (x + ∆x)? This is a difference in “y” values; f (x + ∆x) − f (x) = ∆y ≈ dy. We can approximate the propagated error using differentials. Example 143 Using differentials to approximate propagated error A steel ball bearing is to be manufactured with a diameter of 2cm. The manufacturing process has a tolerance of ±0.1mm in the diameter. Given that the density of steel is about 7.85g/cm3 , estimate the propagated error in the mass of the ball bearing. The mass of a ball bearing is found using the equation Solution “mass = volume × density.” In this situation the mass function is a product of the radius of the ball bearing, hence it is m = 7.85 43 πr3 . The differential of the mass is dm = 31.4πr2 dr. The radius is to be 1cm; the manufacturing tolerance in the radius is Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 219 Chapter 4 Applications of the Derivative ±0.05mm, or ±0.005cm. The propagated error is approximately: ∆m ≈ dm = 31.4π(1)2 (±0.005) = ±0.493g Is this error significant? It certainly depends on the application, but we can get an idea by computing the relative error. The ratio between amount of error to the total mass is dm 0.493 =± m 7.85 43 π 0.493 =± 32.88 = ±0.015, or ±1.5%. We leave it to the reader to confirm this, but if the diameter of the ball was supposed to be 10cm, the same manufacturing tolerance would give a propagated error in mass of ±12.33g, which corresponds to a percent error of ±0.188%. While the amount of error is much greater (12.33 > 0.493), the percent error is much lower. Notes: Page 220 Custom Made for Math 1174 at Langara College by P. Anhaouy Exercises 4.4 Terms and Concepts 22. y = 4 x4 1. T/F: Given a differentiable function y = f (x), we are generally free to choose a value for dx, which then determines the value of dy. 23. y = 2x tan x + 1 24. y = ln(5x) 2. T/F: The symbols “dx” and “∆x” represent the same concept. 3. T/F: The symbols “dy” and “∆y” represent the same concept. 4. T/F: Differentials are important in the study of integration. 5. How are differentials and tangent lines related? Problems In Exercises 6 – 17, use differentials to approximate the given value by hand. 7. 5.932 8. 5.13 13. 28. y = 3x ln x 29. y = x ln x − x 30. A set of plastic spheres are to be made with a diameter of 1cm. If the manufacturing process is accurate to 1mm, what is the propagated error in volume of the spheres? 32. What is the propagated error in the measurement of the cross sectional area of a circular log if the diameter is measured at 1500 , accurate to 1/400 ? √ 11. 24 √ 3 x+1 x+2 (b) 5 seconds? √ 16.5 √ 3 27. y = (a) 2 seconds? 9. 6.83 12. 26. y = cos(sin x) 31. The distance, in feet, a stone drops in t seconds is given by d(t) = 16t2 . The depth of a hole is to be approximated by dropping a rock and listening for it to hit the bottom. What is the propagated error if the time measurement is accurate to 2/10ths of a second and the measured time is: 6. 2.052 10. 25. y = ex sin x 63 33. A wall is to be painted that is 80 high and is measured to be 100 , 700 long. Find the propagated error in the measurement of the wall’s surface area if the measurement is accurate to 1/200 . 8.5 14. sin 3 Exercises 34 – 38 explore some issues related to surveying in which distances are approximated using other measured distances and measured angles. (Hint: Convert all angles to radians before computing.) 15. cos 1.5 16. e0.1 In Exercises 17 – 29, compute the differential dy. 17. y = x2 + 3x − 5 34. The length l of a long wall is to be approximated. The angle θ, as shown in the diagram (not to scale), is measured to be 85.2◦ , accurate to 1◦ . Assume that the triangle formed is a right triangle. 18. y = x7 − x5 19. y = 1 4x2 20. y = (2x + sin x)2 2 3x 21. y = x e l =? θ 250 (a) What is the measured length l of the wall? Custom Made for Math 1174 at Langara College by P. Anhaouy Page 221 (b) What is the propagated error? (c) What is the percent error? 35. Answer the questions of Exercise 34, but with a measured angle of 71.5◦ , accurate to 1◦ , measured from a point 1000 from the wall. 36. The length l of a long wall is to be calculated by measuring the angle θ shown in the diagram (not to scale). Assume the formed triangle is an isosceles triangle. The measured angle is 143◦ , accurate to 1◦ . θ 500 (a) What is the measured length of the wall? (b) What is the propagated error? (c) What is the percent error? 37. The length of the walls in Exercises 34 – 36 are essentially the same. Which setup gives the most accurate result? 38. Consider the setup in Exercises 36. This time, assume the angle measurement of 143◦ is exact but the measured 500 from the wall is accurate to 600 . What is the approximate percent error? l =? Page 222 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.5 4.5 L’Hôpital’s Rule L’Hôpital’s Rule This section focuses on another application of derivative. It is concerned with a technique of evaluating certain limits. Our treatment of limits exposed us to “0/0”, an indeterminate form. If lim f (x) = 0 and lim g(x) = 0, we do not conclude that lim f (x)/g(x) is x→c x→c x→c 0/0; rather, we use 0/0 as notation to describe the fact that both the numerator and denominator approach 0. The expression 0/0 has no numeric value; other work must be done to evaluate the limit. Other indeterminate forms exist; they are: ∞/∞, 0 · ∞, ∞ − ∞, 00 , 1∞ and ∞0 . Just as “0/0” does not mean “divide 0 by 0,” the expression “∞/∞” does not mean “divide infinity by infinity.” Instead, it means “a quantity is growing without bound and is being divided by another quantity that is growing without bound.” We cannot determine from such a statement what value, if any, results in the limit. Likewise, “0 · ∞” does not mean “multiply zero by infinity.” Instead, it means “one quantity is shrinking to zero, and is being multiplied by a quantity that is growing without bound.” We cannot determine from such a description what the result of such a limit will be. This section introduces l’Hôpital’s Rule, a method of resolving limits that produce the indeterminate forms 0/0 and ∞/∞. We’ll also show how algebraic manipulation can be used to convert other indeterminate expressions into one of these two forms so that our new rule can be applied. Theorem 40 L’Hôpital’s Rule, Part 1 Let lim f (x) = 0 and lim g(x) = 0, where f and g are differentiable x→c x→c functions on an open interval I containing c, and g 0 (x) 6= 0 on I except possibly at c. Then f (x) f 0 (x) = lim 0 . x→c g(x) x→c g (x) lim We demonstrate the use of l’Hôpital’s Rule in the following examples; we will often use “LHR” as an abbreviation of “l’Hôpital’s Rule.” Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 223 Chapter 4 Applications of the Derivative Example 144 Using l’Hôpital’s Rule Evaluate the following limits, using l’Hôpital’s Rule as needed. sin x x→0 x √ x+3−2 2. lim x→1 1−x x2 x→0 1 − cos x 1. lim 3. lim x2 + x − 6 x→2 x2 − 3x + 2 4. lim Solution 1. We proved this limit is 1 in Example 13 using the Squeeze Theorem. Here we use l’Hôpital’s Rule to show its power. by LHR sin x x→0 x lim 2. 3. lim x→1 √ x+3−2 1−x cos x = 1. x→0 1 = lim 1 by LHR lim 2 = x→1 (x + 3)−1/2 1 =− . −1 4 by LHR x2 2x = lim . x→0 1 − cos x x→0 sin x This latter limit also evaluates to the 0/0 indeterminate form. To evaluate it, we apply l’Hôpital’s Rule again. lim 2x x→0 sin x lim by LHR = 2 = 2. cos x x2 = 2. x→0 1 − cos x Thus lim 4. We already know how to evaluate this limit; first factor the numerator and denominator. We then have: x2 + x − 6 (x − 2)(x + 3) x+3 = lim = lim = 5. x→2 x2 − 3x + 2 x→2 (x − 2)(x − 1) x→2 x − 1 lim We now show how to solve this using l’Hôpital’s Rule. x2 + x − 6 x→2 x2 − 3x + 2 lim by LHR = lim 2x + 1 x→2 2x − 3 = 5. Note that at each step where l’Hôpital’s Rule was applied, it was needed : the initial limit returned the indeterminate form of “0/0.” If Notes: Page 224 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.5 L’Hôpital’s Rule the initial limit returns, for example, 1/2, then l’Hôpital’s Rule does not apply. The following theorem extends our initial version of l’Hôpital’s Rule in two ways. It allows the technique to be applied to the indeterminate form ∞/∞ and to limits where x approaches ±∞. Theorem 41 L’Hôpital’s Rule, Part 2 1. Let lim f (x) = ±∞ and lim g(x) = ±∞, where f and g are x→a x→a differentiable on an open interval I containing a. Then f (x) f 0 (x) = lim 0 . x→a g(x) x→a g (x) lim 2. Let f and g be differentiable functions on the open interval (a, ∞) for some value a, where g 0 (x) 6= 0 on (a, ∞) and f (x) returns either 0/0 or ∞/∞. Then lim x→∞ g(x) f (x) f 0 (x) = lim 0 . x→∞ g(x) x→∞ g (x) lim A similar statement can be made for limits where x approaches −∞. Example 145 Using l’Hôpital’s Rule with limits involving ∞ Evaluate the following limits. 3x2 − 100x + 2 x→∞ 4x2 + 5x − 1000 ex . x→∞ x3 1. lim 2. lim Solution 1. We can evaluate this limit already using Theorem 8; the answer is 3/4. We apply l’Hôpital’s Rule to demonstrate its applicability. 3x2 − 100x + 2 x→∞ 4x2 + 5x − 1000 lim ex 2. lim 3 x→∞ x ∞. by LHR = by LHR = ex x→∞ 3x2 lim 6x − 100 x→∞ 8x + 5 by LHR ex x→∞ 6x by LHR lim by LHR = lim = = lim 6 x→∞ 8 = 3 . 4 ex = x→∞ 6 lim Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 225 Chapter 4 Applications of the Derivative Recall that this means that the limit does not exist; as x approaches ∞, the expression ex /x3 grows without bound. We can infer from this that ex grows “faster” than x3 ; as x gets large, ex is far larger than x3 . (This has important implications in computing when considering efficiency of algorithms.) Indeterminate Forms 0 · ∞ and ∞ − ∞ L’Hôpital’s Rule can only be applied to ratios of functions. When faced with an indeterminate form such as 0 · ∞ or ∞ − ∞, we can sometimes apply algebra to rewrite the limit so that l’Hôpital’s Rule can be applied. We demonstrate the general idea in the next example. Example 146 Applying l’Hôpital’s Rule to other indeterminate forms Evaluate the following limits. 1. lim+ x · e1/x 3. lim x→0 2. lim x · e x→∞ 1/x h ln(x + 1) − ln x 4. lim (x2 − ex ) x→0− i x→∞ Solution 1. As x → 0+ , x → 0 and e1/x → ∞. Thus we have the indeterminate e1/x form 0 · ∞. We rewrite the expression x · e1/x as ; now, as 1/x x → 0+ , we get the indeterminate form ∞/∞ to which l’Hôpital’s Rule can be applied. lim+ x·e1/x = lim+ x→0 x→0 e1/x 1/x by LHR = lim+ x→0 (−1/x2 )e1/x = lim+ e1/x = ∞. −1/x2 x→0 Interpretation: e1/x grows “faster” than x shrinks to zero, meaning their product grows without bound. 2. As x → 0− , x → 0 and e1/x → e−∞ → 0. The the limit evaluates to 0 · 0 which is not an indeterminate form. We conclude then that lim x · e1/x = 0. x→0− 3. This limit initially evaluates to the indeterminate form ∞ − ∞. By applying a logarithmic rule, we can rewrite the limit as   h i x+1 lim ln(x + 1) − ln x = lim ln . x→∞ x→∞ x Notes: Page 226 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.5 L’Hôpital’s Rule As x → ∞, the argument of the ln term approaches ∞/∞, to which we can apply l’Hôpital’s Rule. x+1 x→∞ x lim Since x → ∞ implies x→∞ by LHR = 1 = 1. 1 x+1 → 1, it follows that x   x+1 implies ln → ln 1 = 0. x Thus   x+1 lim ln(x + 1) − ln x = lim ln = 0. x→∞ x→∞ x Interpretation: since this limit evaluates to 0, it means that for large x, there is essentially no difference between ln(x + 1) and ln x; their difference is essentially 0. 4. The limit lim (x2 − ex ) initially returns the indeterminate form x→∞ ∞−∞. Wecan rewrite the expression by factoring out x2 ; x2 −ex =  ex x2 1 − 2 . We need to evaluate how ex /x2 behaves as x → ∞: x ex by LHR ex by LHR ex = lim = lim = ∞. x→∞ x2 x→∞ 2x x→∞ 2  ex  Thus lim x2 1 − 2 evaluates to ∞ · (−∞), which is not an inx→∞ x determinate form; rather, ∞ · (−∞) evaluates to −∞. We conclude that lim (x2 − ex ) = −∞. lim x→∞ Interpretation: as x gets large, the difference between x2 and ex grows very large. Indeterminate Forms 00 , 1∞ and ∞0 Key Idea 6 Evaluating Limits Involving Indeterminate Forms 00 , 1∞ and ∞0  If lim ln f (x) = L, then lim f (x) = lim eln(f (x)) = e L . x→c x→c x→c Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 227 Chapter 4 Applications of the Derivative Example 147 Using l’Hôpital’s Rule with indeterminate forms involving exponents Evaluate the following limits. x  1 2. lim+ xx . 1. lim 1 + x→∞ x x→0 Solution 1. This equivalent to a special limit given in Theorem 3; these limits have important applications within mathematics and finance. Note that the exponent approaches ∞ while the base approaches 1, leading to the indeterminate form 1∞ . Let f (x) = (1 + 1/x)x ; the prob lem asks to evaluate lim f (x). Let’s first evaluate lim ln f (x) . x→∞ x→∞ x 1 lim ln f (x) = lim ln 1 + x→∞ x→∞ x   1 = lim x ln 1 + x→∞ x  1 ln 1 + x = lim x→∞ 1/x   This produces the indeterminate form 0/0, so we apply l’Hôpital’s Rule. 1 2 1+1/x · (−1/x ) x→∞ (−1/x2 ) = lim = lim 1 x→∞ 1 + 1/x = 1.  Thus lim ln f (x) = 1. We return to the original limit and apply x→∞ Key Idea 6. lim x→∞  1+ 1 x x = lim f (x) = lim eln(f (x)) = e1 = e. x→∞ x→∞ 2. This limit leads to the indeterminate form 00 . Let f (x) = xx and Notes: Page 228 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.5  consider first lim+ ln f (x) . L’Hôpital’s Rule y x→0  lim+ ln f (x) = lim+ ln (xx ) x→0 x→0 = lim+ x ln x x→0 ln x . x→0+ 1/x = lim This produces the indeterminate form −∞/∞ so we apply l’Hôpital’s Rule. 1/x = lim+ x→0 −1/x2 = lim −x 4 3 2 1 f(x) = xx . x 1 2 Figure 4.11: A graph of f (x) = xx supporting the fact that as x → 0+ , f (x) → 1. x→0+ = 0.  Thus lim+ ln f (x) = 0. We return to the original limit and apply x→0 Key Idea 6. lim xx = lim f (x) = lim eln(f (x)) = e0 = 1. x→0+ x→0+ x→0+ This result is supported by the graph of f (x) = xx given in Figure 4.11. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 229 Exercises 4.5 Terms and Concepts 19. lim x4 x→∞ ex 1. List the different indeterminate forms described in this section. 2. T/F: l’Hôpital’s Rule provides a faster method of computing derivatives. 3. T/F: l’Hôpital’s Rule states that   f 0 (x) d f (x) . = 0 dx g(x) g (x) √ x x→∞ ex 20. lim ex 21. lim √ x→∞ x ex x→∞ 2x 22. lim 4. Explain what the indeterminate form “1∞ ” means. f (x) g(x) ; l’Hôpital’s Rule is applied when . ex x→∞ 3x 23. lim 5. Fill in the blanks: The Quotient Rule is applied to when taking taking certain 6. Create (but do not evaluate!) a limit that returns “∞0 ”. x3 − 5x2 + 3x + 9 x→3 x3 − 7x2 + 15x − 9 24. lim 25. 7. Create a function f (x) such that lim f (x) returns “00 ”. x→−2 x3 + 7x2 + 16x + 12 x→1 26. lim ln x x 27. lim ln(x2 ) x x→∞ Problems x→∞ In Exercises 8 – 52, evaluate the given limit. x2 + x − 2 x→1 x−1 8. lim 2 x +x−6 9. lim x→2 x2 − 7x + 10  ln x 28. lim 2 x x→∞ 29. lim x · ln x x→0+ √ x · ln x sin x x→π x − π 30. lim sin x − cos x x→π/4 cos(2x) 31. lim xe1/x 10. lim 11. x3 + 4x2 + 4x lim lim sin(5x) 12. lim x→0 x 13. lim x→0 14. lim sin(2x) x+2 sin(2x) x→0+ x→0+ 32. lim x3 − x2 x→∞ 33. lim x→∞ 34. lim xex x→−∞ x→0 sin(3x) 35. lim 15. lim sin(ax) x→0 sin(bx) x e −1 16. lim + x2 x→0 √ x − ln x 1 x→0+ x 2 e−1/x 36. lim (1 + x)1/x x→0+ 37. lim (2x)x x→0+ ex − x − 1 17. lim + x2 x→0 18. lim x→0+ x − sin x x3 − x2 38. lim (2/x)x x→0+ 39. lim (sin x)x x→0+ Page 230 Custom Made for Math 1174 at Langara College by P. Anhaouy Hint: use the Squeeze Theorem. 40. lim (1 − x)1−x x→1+ 41. lim (x)1/x 47. lim tan x sin(2x) x→π/2 48. lim x→∞ 1 x→1+ ln x 42. lim (1/x)x x→∞ 49. lim 43. lim (ln x)1−x x→3+ x − 5 2 −9 1 x−1 − x x−3 x→1+ 50. lim x tan(1/x) 1/x x→∞ 44. lim (1 + x) x→∞ (ln x)3 x→∞ x 2 1/x 51. lim lim tan x cos x 52. lim 45. lim (1 + x ) x→∞ 46. x→π/2 x2 + x − 2 x→1 ln x Custom Made for Math 1174 at Langara College by P. Anhaouy Page 231 Chapter 4 Applications of the Derivative 4.6 Antiderivatives and Indefinite Integration We have spent considerable time considering the derivatives of a function and their applications. In the following chapters, we are going to starting thinking in “the other direction.” That is, given a function f (x), we are going to consider functions F (x) such that F 0 (x) = f (x). There are numerous reasons this will prove to be useful: these functions will help us compute areas, volumes, mass, force, pressure, work, and much more. Given a function y = f (x), a differential equation is one that incorporates y, x, and the derivatives of y. For instance, a simple differential equation is: y 0 = 2x. Solving a differential equation amounts to finding a function y that satisfies the given equation. Take a moment and consider that equation; can you find a function y such that y 0 = 2x? Can you find another? And yet another? Hopefully one was able to come up with at least one solution: y = x2 . “Finding another” may have seemed impossible until one realizes that a function like y = x2 + 1 also has a derivative of 2x. Once that discovery is made, finding “yet another” is not difficult; the function y = x2 + 123, 456, 789 also has a derivative of 2x. The differential equation y 0 = 2x has many solutions. This leads us to some definitions. Definition 25 Antiderivatives and Indefinite Integrals Let a function f (x) be given. An antiderivative of f (x) is a function F (x) such that F 0 (x) = f (x). The set of all antiderivatives of f (x) is the indefinite integral of f , denoted by Z f (x) dx. Make a note about our definition: we refer to an antiderivative of f , as opposed to the antiderivative of f , since there is always an infinite number of them. We often use upper-case letters to denote antiderivatives. Notes: Page 232 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.6 Antiderivatives and Indefinite Integration Knowing one antiderivative of f allows us to find infinitely more, simply by adding a constant. Not only does this give us more antiderivatives, it gives us all of them. Theorem 42 Antiderivative Forms Let F (x) and G(x) be antiderivatives of f (x). Then there exists a constant C such that G(x) = F (x) + C. Given a function f and one of its antiderivatives F , we know all antiderivatives of f have the form F (x) + C for some constant C. Using Definition 25, we can say that Z f (x) dx = F (x) + C. Let’s analyze this indefinite integral notation. Integra on symbol ∫ Differen al of x Constant of integra on f(x) dx = F(x) + C Integrand One an deriva ve Figure 4.12: Understanding the indefinite integral notation. . Figure 4.12 showsR the typical notation of the indefinite integral. The integration symbol, , is in reality an “elongated S,” representing “take the sum.” We will later see how sums and antiderivatives are related. The function we want to find an antiderivative of is called the integrand. It contains R the differential of the variable we are integrating with respect to. The symbol and the differential dx are notR“bookends” with a function sandwiched in between; rather, the symbol means “find all antiderivatives of what follows,” and the function f (x) and dx are multiplied together; the dx does not “just sit there.” Let’s practice using this notation. ExampleZ 148 Evaluating indefinite integrals Evaluate sin x dx. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 233 Chapter 4 Applications of the Derivative We are asked to find all functions F (x) such that F 0 (x) = sin x. Some thought will lead us to one solution: F (x) = − cos x, because d dx (− cos x) = sin x. The indefinite integral of sin x is thus − cos x, plus a constant of integration. So: Z sin x dx = − cos x + C. Solution A commonly asked question is “What happened to the dx?” The unenlightened response is “Don’t worry about it. It just goes away.” A full understanding includes the following. This process of antidifferentiation is really solving a differential question. The integral Z sin x dx presents us with a differential, dy = sin x dx. It is asking: “What is y?” We found lots of solutions, all of the form y = − cos x + C. Letting dy = sin x dx, rewrite Z Z sin x dx as dy. This is asking: “What functions have a differential of the form dy?” The answer is “Functions of the form y + C, where C is a constant.” What is y? We have lots of choices, all differing by a constant; the simplest choice is y = − cos x. Understanding all of this is more important later as we try to find antiderivatives of more complicated functions. In this section, we will simply explore the rules of indefinite integration, and one can succeed for now with answering “What happened to the dx?” with “It went away.” Let’s practice once more before stating integration rules. ExampleZ 149 Evaluating indefinite integrals 2 Evaluate (3x + 4x + 5) dx. We seek a function F (x) whose derivative is 3x2 +4x+5. When taking derivatives, we can consider functions term–by–term, so we can likely do that here. What functions have a derivative of 3x2 ? Some thought will lead us to a cubic, specifically x3 + C1 , where C1 is a constant. What functions have a derivative of 4x? Here the x term is raised to the first power, so we likely seek a quadratic. Some thought should lead us to 2x2 + C2 , where C2 is a constant. Solution Notes: Page 234 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.6 Antiderivatives and Indefinite Integration Finally, what functions have a derivative of 5? Functions of the form 5x + C3 , where C3 is a constant. Our answer appears to be Z (3x2 + 4x + 5) dx = x3 + C1 + 2x2 + C2 + 5x + C3 . We do not need three separate constants of integration; combine them as one constant, giving the final answer of Z (3x2 + 4x + 5) dx = x3 + 2x2 + 5x + C. It is easy to verify our answer; take the derivative of x3 + 2x3 + 5x + C and see we indeed get 3x2 + 4x + 5. This final step of “verifying our answer” is important both practically and theoretically. In general, taking derivatives is easier than finding antiderivatives so checking our work is easy and vital as we learn. We also see that taking the derivative of our answer returns the function in the integrand. Thus we can say that: d dx Z f (x) dx  = f (x). Differentiation “undoes” the work done by antidifferentiation. We have seen a list of the derivatives of common functions we had learned at that point. We restate part of that list here to stress the relationship between derivatives and antiderivatives. This list will also be useful as a glossary of common antiderivatives as we learn. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 235 Chapter 4 Applications of the Derivative Theorem 43 Derivatives and Antiderivatives (n 6= −1) Common Differentiation Rules Common Indefinite Integral Rules Z   d 1. c1 f (x) ± c2 g(x) dx = 1. c1 f (x) ± c2 g(x) = dx Z Z c1 f 0 (x) ± c2 g 0 (x) c1 f (x) dx ± c2 g(x) dx  d C =0 dx d  3. x =1 dx d n 4. x = n · xn−1 dx 1 d 1 =− 2 5. dx x x d √  1 6. x = √ dx 2 x 2.  d sin x = cos x dx  d 8. cos x = − sin x dx  d 9. tan x = sec2 x dx  d 10. csc x = − csc x cot x dx  d 11. sec x = sec x tan x dx  d 12. cot x = − csc2 x dx d x e = ex 13. dx d x 14. a = ln a · ax dx  1 d ln x = 15. dx x 7. Notes: Page 236 Custom Made for Math 1174 at Langara College by P. Anhaouy 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. Z Z Z 0 dx = C 1 dx = Z xn dx = dx = x + C 1 xn+1 + C n+1 Z 1 1 dx = − + C x2 x Z cos x dx = sin x + C Z Z Z Z Z Z Z Z Z √ 1 √ dx = 2 x + C x sin x dx = − cos x + C sec2 x dx = tan x + C csc x cot x dx = − csc x + C sec x tan x dx = sec x + C csc2 x dx = − cot x + C ex dx = ex + C ax dx = 1 · ax + C ln a 1 dx = ln |x| + C x 4.6 Antiderivatives and Indefinite Integration We highlight a few important points from Theorem 43: R R • Rule #1 states c · f (x) dx = c · f (x) dx. This is the Constant Multiple Rule: we can temporarily ignore constants when finding antiderivatives, just as we did when computing derivatives (i.e.,   d d 2 x2 ). An example: is just as easy to compute as dx dx 3x Z Z 5 cos x dx = 5 · cos x dx = 5 · (sin x + C) = 5 sin x + C. In the last step we can consider the constant as also being multiplied by 5, but “5 times a constant” is still a constant, so we just write “C ”. • Rule #2 is the Sum/Difference Rule: we can split integrals apart when the integrand contains terms that are added/subtracted, as we did in Example 149. So: Z Z Z Z 2 2 (3x + 4x + 5) dx = 3x dx + 4x dx + 5 dx Z Z Z = 3 x2 dx + 4 x dx + 5 dx 1 1 = 3 · x3 + 4 · x2 + 5x + C 3 2 = x3 + 2x2 + 5x + C In practice we generally do not write out all these steps, but we demonstrate them here for completeness. • Rule #5 is the Power Rule of indefinite integration. There are two important things to keep in mind: R 1. Notice the restriction that n 6= −1. This is important: x1 dx 6= “ 10 x0 + C”; rather, see Rule #14. 2. We are presenting antidifferentiation as the “inverse operation” of differentiation. Here is a useful quote to remember: “Inverse operations do the opposite things in the opposite order.” When taking a derivative using the Power Rule, we first multiply by the power, then second subtract 1 from the power. To find the antiderivative, do the opposite things in the opposite order: first add one to the power, then second divide by the power. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 237 Chapter 4 Applications of the Derivative • Note that Rule #14 incorporates the absolute value of x. The exercises will work the reader through why this is the case; for now, know the absolute value is important and cannot be ignored. Initial Value Problems In differential calculus we see that the derivative of a position function gave a velocity function, and the derivative of a velocity function describes acceleration. We can now go “the other way:” the antiderivative of an acceleration function gives a velocity function, etc. While there is just one derivative of a given function, there are infinite antiderivatives. Therefore we cannot ask “What is the velocity of an object whose acceleration is −32ft/s2 ?”, since there is more than one answer. We can find the answer if we provide more information with the question, as done in the following example. Often the additional information comes in the form of an initial value, a value of the function that one knows beforehand. Example 150 Solving initial value problems The acceleration due to gravity of a falling object is −32 ft/s2 . At time t = 3, a falling object had a velocity of −10 ft/s. Find the equation of the object’s velocity. Solution We want to know a velocity function, v(t). We know two things: • The acceleration, i.e., v 0 (t) = −32, and • the velocity at a specific time, i.e., v(3) = −10. Using the first piece of information, we know that v(t) is an antiderivative of v 0 (t) = −32. So we begin by finding the indefinite integral of −32: Z (−32) dt = −32t + C = v(t). Now we use the fact that v(3) = −10 to find C: v(t) = −32t + C v(3) = −10 −32(3) + C = −10 C = 86 Notes: Page 238 Custom Made for Math 1174 at Langara College by P. Anhaouy 4.6 Antiderivatives and Indefinite Integration Thus v(t) = −32t + 86. We can use this equation to understand the motion of the object: when t = 0, the object had a velocity of v(0) = 86 ft/s. Since the velocity is positive, the object was moving upward. When did the object begin moving down? Immediately after v(t) = 0: 43 ≈ 2.69s. 16 Recognize that we are able to determine quite a bit about the path of the object knowing just its acceleration and its velocity at a single point in time. −32t + 86 = 0 ⇒ t= Example 151 Solving initial value problems Find f (t), given that f 00 (t) = cos t, f 0 (0) = 3 and f (0) = 5. We start by finding f 0 (t), which is an antiderivative of Solution 00 f (t): Z f 00 (t) dt = Z cos t dt = sin t + C = f 0 (t). So f 0 (t) = sin t + C for the correct value of C. We are given that f (0) = 3, so: 0 f 0 (0) = 3 ⇒ sin 0 + C = 3 0 ⇒ C = 3. Using the initial value, we have found f (t) = sin t + 3. We now find f (t) by integrating again. Z Z f (t) = f 0 (t) dt = (sin t + 3) dt = − cos t + 3t + C. We are given that f (0) = 5, so − cos 0 + 3(0) + C = 5 −1 + C = 5 C=6 Thus f (t) = − cos t + 3t + 6. This section introduced antiderivatives and the indefinite integral. We found they are needed when finding a function given information about its derivative(s). For instance, we found a position function given a velocity function. In the next section, we will see how position and velocity are unexpectedly related by the areas of certain regions on a graph of the velocity function. Then, in Section ??, we will see how areas and antiderivatives are closely tied together. Notes: Custom Made for Math 1174 at Langara College by P. Anhaouy Page 239 Exercises 4.6 Terms and Concepts Z 5eθ dθ Z 3t dt Z 5t dt 2 Z (2t + 3)2 dt Z (t2 + 3)(t3 − 2t) dt Z x2 x3 dx Z eπ dx 19. 1. Define the term “antiderivative” in your own words. 20. 2. Is it more accurate to refer to “the” antiderivative of f (x) or “an” antiderivative of f (x)? 21. 3. Use your own words to define the indefinite integral of f (x). 22. 4. Fill in the blanks: “Inverse operations do the things in the order.” 23. 5. What is an “initial value problem”? 6. The derivative of a position function is a tion. func- 24. 7. The antiderivative of an acceleration function is a function. 25. Z Problems In Exercises 8 – 26, evaluate the given indefinite integral. 26. a dx 27. Z This problem investigates why Theorem 43 states that 1 dx = ln |x| + C. x Z 3x3 dx (a) What is the domain of y = ln x?  d (b) Find dx ln x . Z x8 dx (c) What is the domain of y = ln(−x)?  d (d) Find dx ln(−x) . Z (10x2 − 2) dx 8. 9. 10. Z 11. dt (e) You should find that 1/x has two types of antiderivatives, depending on whether x > 0 or x Z < 0. In one expression, give a formula for 1 dx that takes these different domains into x account, and explain your answer. Z 1 ds 12. Z 13. 1 dt 3t2 In Exercises 28 – 38, find f (x) described by the given initial value problem. 28. f 0 (x) = sin x and f (0) = 2 29. f 0 (x) = 5ex and f (0) = 10 Z 3 dt t2 Z 1 √ dx x 31. f 0 (x) = sec2 x and f (π/4) = 5 sec2 θ dθ 32. f 0 (x) = 7x and f (2) = 1 14. 30. f 0 (x) = 4x3 − 3x2 and f (−1) = 9 15. Z 16. 33. f 00 (x) = 5 and f 0 (0) = 7, f (0) = 3 Z sin θ dθ 17. 34. f 00 (x) = 7x and f 0 (1) = −1, f (1) = 10 Z 18. (sec x tan x + csc x cot x) dx 35. f 00 (x) = 5ex and f 0 (0) = 3, f (0) = 5 Page 240 Custom Made for Math 1174 at Langara College by P. Anhaouy 36. f 00 (θ) = sin θ and f 0 (π) = 2, f (π) = 4 Review 37. f 00 (x) = 24x2 + 2x − cos x and f 0 (0) = 5, f (0) = 0 39. Use information gained from the first and second deriva1 . tives to sketch f (x) = x e +1 38. f 00 (x) = 0 and f 0 (1) = 3, f (1) = 1 40. Given y = x2 ex cos x, find dy. Custom Made for Math 1174 at Langara College by P. Anhaouy Page 241 A: Solutions To Selected Problems Chapter 1 7. Let  > 0 be given. We wish to find δ > 0 such that when |x − 4| < δ, |f (x) − 15| < . Consider |f (x) − 15| < , keeping in mind we want to make a statement about |x − 4|: Section 1.1 1. Answers will vary. |f (x) − 15| <  3. F |x2 + x − 5 − 15| <  5. Answers will vary. |x2 + x − 20| <  7. −5 |x − 4| · |x + 5| <  9. 2 |x − 4| < /|x + 5| 11. Limit does not exist. 13. 7 Since x is near 4, we can safely assume that, for instance, 3 < x < 5. Thus 15. Limit does not exist. h −0.1 −0.01 0.01 0.1 f (a+h)−f (a) h h −0.1 −0.01 0.01 0.1 f (a+h)−f (a) h f (a+h)−f (a) h 21. h −0.1 −0.01 0.01 0.1 f (a+h)−f (a) h 23. h −0.1 −0.01 0.01 0.1 17. 19. 9 9 9 9 3+5 0 be given. We wish to find δ > 0 such that when |x − 5| < δ, |f (x) − (−2)| < . Consider |f (x) − (−2)| < : |f (x) + 2| <  |(3 − x) + 2| <  |5 − x| <  − < 5 − x <  − < x − 5 < . This implies we can let δ = . Then: |x − 5| < δ 9. Let  > 0 be given. We wish to find δ > 0 such that when |x − 2| < δ, |f (x) − 5| < . However, since f (x) = 5, a constant function, the latter inequality is simply |5 − 5| < , which is always true. Thus we can choose any δ we like; we arbitrarily choose δ = . 11. Let  > 0 be given. We wish to find δ > 0 such that when |x − 0| < δ, |f (x) − 0| < . In simpler terms, we want to show that when |x| < δ, | sin x| < . Set δ = . We start with assuming that |x| < δ. Using the hint, we have that | sin x| < |x| < δ = . Hence if |x| < δ, we know immediately that | sin x| < . Section 1.3 −δ < x − 5 < δ − < x − 5 <  1. Answers will vary. − < (x − 3) − 2 <  3. Answers will vary. − < (−x + 3) − (−2) <  |3 − x − (−2)| < , which is what we wanted to prove. 5. As x is near 1, both f and g are near 0, but f is approximately twice the size of g. (I.e., f (x) ≈ 2g(x).) 7. 6 11. 9. Limit does not exist. (a) 2 (b) 2 11. Not possible to know; as x approaches 6, g(x) approaches 3, but we know nothing of the behavior of f (x) when x is near 3. (c) 2 (d) 0 13. −45 (e) 2 15. −1 (f) 2 17. π (g) 2 19. −0.000000015 ≈ 0 (h) Not defined 21. Limit does not exist 13. 23. 2 25. (a) 2 (b) −4 π 2 +3π+5 ≈ 0.6064 5π 2 −2π−3 (c) Does not exist. 27. −8 (d) 2 29. 10 15. 31. −3/2 (a) 0 (b) 0 33. 0 (c) 0 35. 1 (d) 0 37. 3 (e) 2 39. 1 (f) 2 41. (g) 2 (a) Apply Part 1 of Theorem 1. (b) (h) 2 x Apply Theorem 6; g(x) = x is the same as g(x) = 1 everywhere except at x = 0. Thus lim g(x) = lim 1 = 1. x→0 x→0  (c) The function f (x) is always 0, so g f (x) is never defined as g(x) is not defined at x = 0. Therefore the limit does not exist. 17. (d) The Composition Rule requires that lim g(x) be 19. (b) sin2 a (c) sin2 a (d) sin2 a x→0 (c) 4 (d) 3 21. 5. (a) −1 (b) 1 1. The function approaches different values from the left and right; the function grows without bound; the function oscillates. 3. F (a) 4 (b) 4 equal to g(0). They are not equal, so the conditions of the Composition Rule are not satisfied, and hence the rule is not violated. Section 1.4 (a) 1 − cos2 a = sin2 a (c) Does not exist (d) 0 23. 2/3 (a) 2 25. −9 (b) 2 (c) 2 (d) 1 1. F (e) As f is not defined for x < 0, this limit is not defined. 3. F (f) 1 7. Section 1.5 (a) Does not exist. (b) Does not exist. 5. T 7. Answers will vary. 9. (c) Does not exist. (d) Not defined. 11. (a) 1 (b) 0 (e) 0 9. (a) ∞ (b) ∞ (f) 0 (c) 1/2 (a) 2 (d) 1/2 (b) 2 (c) 2 (d) 2 13. (a) Limit does not exist (b) Limit does not exist 15. Tables will vary. Page A.2 Custom Made for Math 1174 at Langara College by P. Anhaouy (a) x f (x) 2.9 −15.1224 It seems 2.99 −159.12 2.999 −1599.12 limx→3− f (x) = −∞. (b) x 3.1 3.01 3.001 f (x) 16.8824 160.88 1600.88 35. We cannot say; the Intermediate Value Theorem only applies to function values between −10 and 10; as 11 is outside this range, we do not know. It seems limx→3+ f (x) = ∞. 39. Approximate root is x = 0.69. The intervals used are: [0.65, 0.7] [0.675, 0.7] [0.6875, 0.7] [0.6875, 0.69375] [0.690625, 0.69375] (c) It seems limx→3 f (x) does not exist. 17. Tables will vary. (a) (b) 37. Approximate root is x = 1.23. The intervals used are: [1, 1.5] [1, 1.25] [1.125, 1.25] [1.1875, 1.25] [1.21875, 1.25] [1.234375, 1.25] [1.234375, 1.2421875] [1.234375, 1.2382813] 41. (a) 20 (b) 25 x 2.9 2.99 f (x) 132.857 12124.4 It seems limx→3− f (x) = ∞. x 3.1 3.01 f (x) 108.039 11876.4 It seems limx→3+ f (x) = ∞. (c) Limit does not exist (d) 25 43. Answers will vary. Chapter 2 (c) It seems limx→3 f (x) = ∞. 19. Horizontal asymptote at y = 2; vertical asymptotes at x = −5, 4. 21. Horizontal asymptote at y = 0; vertical asymptotes at x = −1, 0. Section 2.1 1. T 3. Answers will vary. 5. Answers will vary. 23. No horizontal or vertical asymptotes. 7. f 0 (x) = 2 25. ∞ 9. g 0 (x) = 2x 27. −∞ 11. r 0 (x) = −1 x2 29. Solution omitted. (a) y = 6 13. 31. Yes. The only “questionable” place is at x = 3, but the left and right limits agree. (b) x = −2 (a) y = −3x + 4 15. Section 1.6 (b) y = 1/3(x − 7) − 17 (a) y = −7(x + 1) + 8 17. 1. Answers will vary. (b) y = 1/7(x + 1) + 8 3. A root of a function f is a value c such that f (c) = 0. (a) y = −1(x − 3) + 1 19. 5. F (b) y = 1(x − 3) + 1 7. T 21. y = −0.099(x − 9) + 1 9. F 23. 11. No; lim f (x) = 2, while f (1) = 1. (a) Approximations will vary; they should match (c) closely. x→1 13. No; f (1) does not exist. (b) f 0 (x) = −1/(x + 1)2 15. Yes (c) At (0, 1), slope is −1. At (1, 0.5), slope is −1/4. 17. y (a) No; lim f (x) 6= f (−2) x→−2 (b) Yes 2 (c) No; f (2) is not defined. 19. x (a) Yes −6 (b) No; the left and right hand limits at 1 are not equal. 21. (a) Yes −4 −2 2 −2 25. . y (b) No. limx→8 f (x) = 16/5 6= f (8) = 5. . 1 23. (−∞, −2] ∪ [2, ∞) √ √ 25. (−∞, − 6] ∪ [ 6, ∞) 0.5 27. (−∞, ∞) x −2π 29. (0, ∞) 2π −0.5 31. (−∞, 0] 33. Yes, by the Intermediate Value Theorem. π −π 27. . −1 Custom Made for Math 1174 at Langara College by P. Anhaouy Page A.3 29. Approximately 24. 31. 9. x(2x)−(x2 +3)1 x2 f 0 (x) = 1 − x32 (a) f 0 (x) = (a) 1 (b) (b) 3 (c) They are equal. (c) Does not exist (d) (−∞, −3) ∪ (3, ∞) 11. (b) Section 2.2 1. Velocity 3. Linear functions. 5. −17 7. f (10.1) is likely most accurate, as accuracy is lost the farther from x = 10 we go. 4s3 (0)−3(12s2 ) 16s6 h0 (s) = −9/4s−4 (a) h0 (s) = (c) They are equal. 13. 15. g 0 (x) = −12 (x−5)2 (a) f 0 (x) = (x+2)(4x3 +6x2 )−(x4 +2x3 )(1) (x+2)2 (b) f (x) = x3 when x 6= −2, so f 0 (x) = 3x2 . (c) They are equal. 11. ft/s2 17. Tangent line: y = 2x + 2 Normal line: y = −1/2x + 2 13. (a) thousands of dollars per car 19. x = 3/2 (b) It is likely that P (0) < 0. That is, negative profit for not producing any cars. 21. x = −2, 0 9. 6 Section 2.5 15. f (x) = g 0 (x) 3. One answer is f (x) = 10. 1. We have s(t) = 2t3 − 9t2 + 12t, where s is in cm and t is in s. Its velocity is v(t) = 6t2 − 18t + 12 = 6(t − 1)(t − 2). The particle is at rest when v = 0 ⇒ t = 1, 2. Here, we know that the particle changes its direction in its motion. When v > 0, it moves to the right, and v < 0, it moves to the left. Solving these inequalities, we get it moves to the right until t = 1; then moves to the left until t = 2; then it moves to the right again for t > 2. Note: When a > 0, it’s accelerating to the right, and a < 0, it accelerates to the left. But, it speeds up when v and a have the same signs; it slows down when they have different signs. 5. One possible answer is f (x) = 17x − 205. 3. s(t) = 100t − 5t2 ⇒ v(t) = 100 − 10t = 10(10 − t). 17. Either g(x) = f 0 (x) or f (x) = g 0 (x) is acceptable. The actual answer is g(x) = f 0 (x), but is very hard to show that f (x) 6= g 0 (x) given the level of detail given in the graph. 19. f 0 (x) = 10x Section 2.3 1. Power Rule. 7. f 0 (x) is a velocity function, and f 00 (x) is acceleration. 9. f 0 (x) = 14x − 5 11. m0 (t) = 45t4 − 38 t2 + 3 13. p0 (s) = s3 + s2 + s + 1 15. g 0 (x) = 24x2 − 120x + 150 17. f 0 (x) = 18x − 12 19. h0 (t) = 2t − et h00 (t) = 2 − et h000 (t) = −et h(4) (t) = −et 21. Tangent line: y = 2(x − 1) Normal line: y = −1/2(x − 1) 23. Tangent line: y = 2x + 3 Normal line: y = −1/2(x − 5) + 13 25. The tangent line to f (x) = x4 at x = 3 is y = 108(x − 3) + 81; thus (3.01)4 ≈ y(3.01) = 108(.01) + 81 = 82.08. Section 2.4 3. F (a) f 0 (x) = (x2 + 3x) + x(2x + 3) (b) f 0 (x) = 3x2 + 6x (c) They are equal. 7. (c) v(2) = 80 m/s and v(15) = −50 m/s (d) stotal = |smax − s(5)| + |s(12) − smax | = |125| + |20| = 145 m. 5. h(t) = 490 − 12gt2 ⇒ v(t) = 490 − 24gt. (a) −14.7 m/s. g (b) −gt − . 2 (c) 10 s. 7. R(x) = p · x = (−3x2 + 600x) · x = −3x3 + 600x2 ⇒ R0 (x) = −9x2 + 1200x. (a) M R(100) = 30, 000. This means that if the store increases sale from 100 to 101 cameras, revenue will increasing by $30, 000. (b) M R(300) = −450, 000. This means that if the store increases sale from 300 to 301 cameras, revenue will decreasing by $450, 000. 1. T 5. s(4) − s(1) = 75 m/s. 4−1 (b) v = 0 ⇒ t = 10 s. So, smax = s(10) = 500 m. (a) v̄ = (a) h0 (s) = 2(s + 4) + (2s − 1)(1) (b) h0 (s) = 4s + 7 (c) They are equal. 9. C̄(x) = C(x) 4200 = + 5.40 − 0.001x + 0.000002x2 . x x (a) C̄(1000) = $10.60 per unit. (b) M C(x) = 5.40 − 0.002x + 0.000006x2 ⇒ M C(1000) = $9.40 per unit. (c) C̄(1001) = $10.5988 per unit. The average cost will decrease by $0.0012 per unit. Page A.4 Custom Made for Math 1174 at Langara College by P. Anhaouy 11. R(x) = p·x = (−3x2 +600x)·x = −3x3 +600x2 ⇒ P (x) = R(x) − C(x) = −3x3 + 243x2 ⇒ M P (x) = −9x2 + 486x. (a) M P (10) = 3960. This means that if we increase the production from 10 to 11, we will increase the profit by $3960. (b) M P (100) = −41, 400. This means that if we increase the production from 100 to 101, we will decrease the profit by $41, 400. 7. f 0 (t) = √ −t 2 1−t √ 9. h0 (x) = 1.5x0.5 = 1.5 x 11. g 0 (x) = √ x(1)−(x+7)(1/2x−1/2 ) 1 = 2√ − √7 3 x x 2 x 13. 3 dy = −4x dx 2y+1 15. dy = sin(x) sec(y) dx 17. dy y = x dx 21. 2 sin(y) cos(y) x 1 2y+2 23. − cos(x)(x+cos(y))+sin(x)+y sin(y)(sin(x)+y)+x+cos(y) 19. − Section 2.6 1. F 3. f 0 (θ) = 9 cos θ − 10 sin θ (a) y = 0 25. 5. h0 (t) = et − cos t + sin t (b) y = −1.859(x − 0.1) + 0.281 (csc t − 4) + t12 (− csc t cot t) 7. f 0 (t) = −2 t3 9. 27. (a) y = 4 √ (b) y = 0.93(x − 2) + 4 108 29. (a) y = − √1 (x − 72 ) + 6+32 3 3 √ √ (b) y = 3(x − 4+32 3 ) + 32 h0 (x) = − csc2 x − ex 11. f 0 (x) = √ sin2 (x)+cos2 (x)+3 cos(x) (cos(x)+3)2 13. g 0 (t) = 12t2 et + 4t3 et − cos2 t + sin2 t 15. f 0 (x) = 2xex tan x = x2 ex tan x + x2 ex sec2 x 31. 3/5 d2 y = 35 y 8/5 + 53 16/5 dx2 x yx 17. Tangent line: y = 4 Normal line: x = π/2 33. d2 y =0 dx2 Section 2.9 19. Tangent line: y = −(x − 3π ) − 3π = −x 2 2 Normal line: y = (x − 3π ) − 3π = x − 3π 2 2 1. F 3. The point (10, 1) lies on the graph of y = f −1 (x) (assuming f is invertible). 21. x = 0 23. f (4) (x) = −4 cos x + x sin x 5. Compose f (g(x)) and g(f (x)) to confirm that each equals x. Section 2.7 1. T 3. T 5. f 0 (x) = 10(4x3 − x)9 · (12x2 − 1) = (120x2 − 10)(4x3 − x)9 7. g 0 (θ) = 3(sin θ + cos θ)2 (cos θ − sin θ)  3 9. f 0 (x) = 4 x + x1 1 − x12 11. 7. Compose f (g(x)) and g(f (x)) to confirm that each equals x. 0 9. f −1 (20) = f 01(2) = 1/5 0 √ 1 11. f −1 ( 3/2) = f 0 (π/6) =1  0 1 −1 13. f (1/2) = f 0 (1) = −2 15. f 0 (x) = − tan x g 0 (x) = 5 sec2 (5x) 17. f 0 (x) = 2/x 13. p0 (t) = −3 cos2 (t2 + 3t + 1) sin(t2 + 3t + 1)(2t + 3) 19. g 0 (t) = − ln 5 · 5cos t sin t 15. g 0 (t) = 5 cos(t2 + 3t) cos(5t − 7) − (2t + 3) sin(t2 + 3t) sin(5t − 7) 21. g 0 (t) = 0 17. Tangent line: y = 0 Normal line: x = 0 19. Tangent line: y = −3(θ − π/2) + 1 Normal line: y = 1/3(θ − π/2) + 1 21. (a) ◦ F/mph (b) The sign would be negative; when the wind is blowing at 10 mph, any increase in wind speed will make it feel colder, i.e., a lower number on the Fahrenheit scale.  23. m0 (w) = 2w ln 3·3w −ln 2·(3w +1) 22w 25. f 0 (x) = (3t +2) (ln 2)2t −(2t +3) (ln 3)3t (3t +2)2   27. In both cases the derivative is the same: k/x.  2 29. y 0 = (2x)x 2x ln(2x) + x Tangent line: y = (2 + 4 ln 2)(x − 1) + 2  31. y 0 = xsin(x)+2 cos x ln x + sin xx+2 Tangent line: y = (3π 2 /4)(x − π/2) + (π/2)3  (x+1)(x+2) 1 1 1 1 + x+2 − x+3 − x+4 33. y 0 = (x+3)(x+4) x+1 Tangent line: y = 11/72x + 1/6 Section 2.8 35. y = −4/5(x − 1) + 2 1. Answers will vary. Section 2.10 3. T 1 5. f 0 (x) = 21 x−1/2 − 12 x−3/2 = 2√ − x 2 √1 x3 1. h0 (t) = √ 2 1−4t2 Custom Made for Math 1174 at Langara College by P. Anhaouy Page A.5 2 3. g 0 (x) = 1+4x 2 1. Answers will vary. 3. Answers will vary. sin(t) 5. g 0 (t) = cos−1 (t) cos(t) − √ 1−t2 5. Increasing 1 7. g 0 (x) = √x(2x+2) 7. Graph and verify. (a) f (x) = x, so f 0 (x) = 1 9. 9. Graph and verify. (b) f 0 (x) = cos(sin−1 x) √ 1 1−x2 = 1. (a) f (x) = x, so f 0 (x) = 1 11. f 0 (x) = 1 sec2 x = 1 1+tan2 x (b) √ √ 13. y = 2(x − 2/2) + π/4 Chapter 3 Section 3.1 1. Answers will vary. 3. Answers will vary. 5. F 7. A: abs. min B: none C: abs. max D: none E: none 9. f 0 (0) = 0 f 0 (2) = 0 11. f 0 (0) = 0 f 0 (3.2) = 0 f 0 (4) is undefined 13. f 0 (0) is not defined 15. min: (−0.5, 3.75) max: (2, 10) √ 17. min: (π/4, 3 2/2) max: (π/2, 3) √ √ 19. min: ( 3, 2 3) max: (5, 28/5) 21. min: (π, −eπ ) √ max: (π/4, 2eπ/4 ) 2 23. min: (1, 0) max: (e, 1/e) 25. y(y−2x) dy = x(x−2y) dx 27. 3x2 + 1 Section 3.2 11. Graph and verify. 13. Graph and verify. 15. domain=(−∞, ∞) c.p. at c = −2, 0; increasing on (−∞, −2) ∪ (0, ∞); decreasing on (−2, 0); rel. min at x = 0; rel. max at x = −2. 17. domain=(−∞, ∞) c.p. at c = 1; increasing on (−∞, ∞); 19. domain=(−∞, −1) ∪ (−1, 1) ∪ (1, ∞) c.p. at c = 0; decreasing on (−∞, −1) ∪ (−1, 0); increasing on (0, 1) ∪ (1, ∞); rel. min at x = 0; 21. domain=(−∞, 0) ∪ (0, ∞); c.p. at c = 2, 6; decreasing on (−∞, 0) ∪ (0, 2) ∪ (6, ∞); increasing on (2, 6); rel. min at x = 2; rel. max at x = 6. 23. domain = (−∞, ∞); c.p. at c = −1, 1; decreasing on (−1, 1); increasing on (−∞, −1) ∪ (1, ∞); rel. min at x = 1; rel. max at x = −1 25. c = ± cos−1 (2/π) Section 3.4 1. Answers will vary. 3. Yes; Answers will vary. 1. Answers will vary. 3. Any c in [−1, 1] is valid. 5. c = −1/2 7. Rolle’s Thm. does not apply. 5. Graph and verify. 7. Graph and verify. 9. Graph and verify. 11. Graph and verify. 9. Rolle’s Thm. does not apply. 11. c = 0 √ 13. c = 3/ 2 15. The Mean Value Theorem does not apply. √ 17. c = ± sec−1 (2/ π) 19. √ c = 5±76 7 21. Max value of 19 at x = −2 and x = 5; min value of 6.75 at x = 1.5. 23. They are the odd, integer valued multiples of π/2 (such as 0, ±π/2, ±3π/2, ±5π/2, etc.) Section 3.3 13. Graph and verify. 15. Graph and verify. 17. Possible points of inflection: none; concave down on (−∞, ∞) 19. Possible points of inflection: x = 1/2; concave down on (−∞, 1/2); concave up on (1/2, ∞) √ 21. Possible points of√inflection: x =√(1/3)(2 ± 7); concave up on ((1/3)(2 −√ 7), (1/3)(2 + √ 7)); concave down on (−∞, (1/3)(2 − 7)) ∪ ((1/3)(2 + 7), ∞) √ 23. Possible √ points√of inflection: x = ±1/ 3; concave down 3, 1/ 3); √ concave up on on (−1/ √ (−∞, −1/ 3) ∪ (1/ 3, ∞) Page A.6 Custom Made for Math 1174 at Langara College by P. Anhaouy 25. Possible points of inflection: x = −π/4, 3π/4; concave down on (−π/4, 3π/4) concave up on (−π, −π/4) ∪ (3π/4, π) 27. Possible points of inflection: x = 1/e3/2 ; concave down on (0, 1/e3/2 ) concave up on (1/e3/2 , ∞) (c) 1/(40π) ≈ 0.00796 cm/s 5. 63.14mph 7. Due to the height of the plane, the gun does not have to rotate very fast. (a) 0.0573 rad/s 29. min: x = 1 √ √ 31. max: x = −1/ 3 min: x = 1/ 3 (b) 0.0725 rad/s (c) In the limit, rate goes to 0.0733 rad/s 33. min: x = 1 9. 35. min: x = 1 (a) 0.04 ft/s (b) 0.458 ft/s 37. critical values: x = −1, 1; no max/min (c) 3.35 ft/s 39. max: x = −2; min: x = 0 (d) Not defined; as the distance approaches 24, the rates approaches ∞. 41. max: x = 0 11. 43. f 0 has no maximal or minimal value 45. f 0 has a minimal value at x = 1/2 47. f 0 has a relative max at: x = (1/3)(2 + at: x = (1/3)(2 − √ (a) 50.92 ft/min (b) 0.509 ft/min √ (c) 0.141 ft/min As the tank holds about 523.6ft3 , it will take about 52.36 minutes. 7) relative min 7) √ 49. f 0 has √ a relative max at x = −1/ 3; relative min at x = 1/ 3 51. f 0 has a relative min at x = 3π/4; relative max at x = −π/4 √ 53. f 0 has a relative min at x = 1/ e3 = e−3/2 13. (a) The rope is 80ft long. (b) 1.71 ft/sec (c) 1.87 ft/sec (d) About 34 feet. 15. The cone is rising at a rate of 0.003ft/s. Section 3.5 Section 4.2 1. Answers will vary. 1. T 3. T 3. 2500; the two numbers are each 50. 5. T 7. A good sketch will include the x and y intercepts.. 9. Use technology to verify sketch. 11. Use technology to verify sketch. 9. The radius should be about 3.84cm and the height should be 2r = 7.67cm. No, this is not the size of the standard can. 13. Use technology to verify sketch. 15. Use technology to verify sketch. 11. The height and width should be 18 and the length should be 36, giving a volume of 11, 664in3 . √ 13. 5 − 10/ 39 ≈ 3.4 miles should be run underground, giving a minimum cost of $374,899.96. 17. Use technology to verify sketch. 19. Use technology to verify sketch. 21. Use technology to verify sketch. 15. The dog should run about 19 feet along the shore before starting to swim. 23. Use technology to verify sketch. 25. Use technology to verify sketch. nπ/2−b , where n is an odd integer 27. Critical points: x = a Points of inflection: (nπ − b)/a, where n is an integer. 29. dy = −x/y, so the function is increasing in second and dx fourth quadrants, decreasing in the first and third quadrants. d2 y = −1/y − x2 /y 3 , which is positive when y < 0 and is dx2 negative when y > 0. Hence the function is concave down in the first and second quadrants and concave up in the third and fourth quadrants. Chapter 4 Section 4.1 1. T 3. (a) 5/(2π) ≈ 0.796cm/s (b) 1/(4π) ≈ 0.0796 cm/s 5. There is no maximum sum; the fundamental equation has only 1 critical value that corresponds to a minimum. √ 7. Area = 1/4, with sides of length 1/ 2. 17. The largest √ area is 2 formed by a square with sides of length 2. Section 4.3 1. x = 500 − 10p ⇒ x0 = −10. p (a) E(p) = − . 50 − p (b) E(30) = −1.5. Since |E| = 1.5 > 1, the demand is elastic. (c) The percentage change in demand is (−1.5)(4.5%) = −6.75%. p 2 3. x + p2 = 4 ⇒ 2xx0 + 2p = 0 ⇒ x0 = − . x (a) E(p) = − p2 . 4 − p2 √ √ (b) If x = 1, then p = 3. So, E( 3) = −3. Since |E| = 3 > 1, the demand is elastic. Custom Made for Math 1174 at Langara College by P. Anhaouy Page A.7 (c) Now, we can solve for E in term of x from (a). x2 − 4 Here, we get E = . So, elastic on x2 0 < x < 1.42 and inelastic on 1.42 < x < 2.  x 2 5 √ 5. p = 100 − ⇒ x = 1000 − 10 p ⇒ x0 = − √ . 10 p (a) E(p) = − √ 500 √ . 100(1000 − 10 p) (b) Currently, at p = 100, |E| = 0.056 < 1, the demand is inelastic. Therefore, the price per unit should be increased in order to increased the revenue. 7. ln x − 2 ln p + 0.02p = 7 ⇒ 2 x0 = − 0.02. x p 100 − p . Currently, at p = 200, 50 |E| = 2 > 1, the demand is elastic. Therefore, we should increase the revenue. (a) Here, we have E = (b) When |E| = 1 ⇒ 100 − p = 50 ⇒ p = 150. 9. V (t) = 100(60 + t2 ) ⇒ V 0 (t) = 100(0 + 2t) = 200t. Then, V0 2t r= ⇒ 0.0625 = ⇒ t = 2, 30. Now, we check V 60 + t2 2 2(60 − t ) r0 (t) = . Since r0 (2) > 0 and r0 (30) < 0, the (60 + t2 )2 present value is a minimum at t = 2 (good to buy), and is maximum at t = 30, when the asset should be sold. 11. V (t) = 250(5 + 0.2t)3/2 ⇒ V 0 (t) = 75(5 + 0.2t)1/2 . Then, V0 75(5 + 0.2t)1/2 3 r= ⇒ 0.04 = ⇒t= = V 10(5 + 0.2t) 250(5 + 0.2t)3/2 12.5. Sell in about 12.5 years. V (12.5) = $5134.90. 13. V (t) = 50te−0.10t ⇒ V 0 (t) = 50e−0.10t (1 − 0.10t). 3. F 5. Answers will vary. 7. Use y = x2 ; dy = 2x · dx with x = 6 and dx = −0.07. Thus dy = −0.84; knowing 62 = 36, we have 5.932 ≈ 35.16. 9. Use y = x3 ; dy = 3x2 · dx with x = 7 and dx = −0.2. Thus dy = −29.4; knowing 73 = 343, we have 6.83 ≈ 313.6. √ √ 11. Use y = x; dy = 1/(2 x)√· dx with x = 25 and √ dx = −1. Thus dy = −0.1; knowing 25 = 5, we have 24 ≈ 4.9. √ √ 3 13. Use y = 3 x; dy = 1/(3 x2 ) · dx with x = 8 and dx = 0.5. Thus dy√= 1/24 ≈ 1/25 = 0.04; knowing √ 3 8 = 2, we have 3 8.5 ≈ 2.04. 15. Use y = cos x; dy = − sin x · dx with x = π/2 ≈ 1.57 and dx ≈ −0.07. Thus dy = 0.07; knowing cos π/2 = 0, we have cos 1.5 ≈ 0.07. 17. dy = (2x + 3)dx −2 19. dy = 4x 3 dx  21. dy = 2xe3x + 3x2 e3x dx 23. dy = 2(tan x+1)−2x sec2 x dx (tan x+1)2 25. dy = (ex sin x + ex cos x)dx 1 27. dy = (x+2) 2 dx 29. dy = (ln x)dx 31. (a) ±12.8 feet (b) ±32 feet 33. ±48in2 , or 1/3ft2 35. (a) 298.8 feet (b) ±17.3 ft (a) V 0 (t) = 0 ⇒ tc = 10 years. When 0 ≤ t < 10, V 0 > 0 and when t > 10, V 0 < 0. So, V (10) is a local maximum. Because the domain of V is all real numbers, and there is only one critical number tc = 10, this local maximum is the absolute maximum. 37. The isosceles triangle setup works the best with the smallest percent error. V 0 (t) 50e−0.10t (1 − 0.10t) (b) r = ⇒ 0.025 = = V (t) 50te−0.10t 1 − 0.10t ⇒ t = 8 years. 1. 0/0, ∞/∞, 0 · ∞, ∞ − ∞, 00 , 1∞ , ∞0 15. V (t) = 100(60t−t2 ) ⇒ V 0 (t) = 100(60−2t) = 0 ⇒ t = 30, that is, tc = 30. When 0 ≤ t < 30, V 0 > 0 and when t > 30, V 0 < 0. So, V (30) is a local maximum. Because the domain of V is all real numbers, and there is only one critical number tc = 30, this local maximum is the absolute maximum. Q 10, 000 45, 000 17. Here C0 (x) = ·s= · (4.50) = . x x x x x Cs (x) = · r = · 5.76 = 2.88x. The total cost function 2 2 45, 000 is then C(x) = + 2.88x. x 45, 000 C 0 (x) = − + 2.88 = 0 ⇒ x = 125. Since x2 90, 000 C 00 (x) = > 0 for all x > 0, C(x) has a min. at x3 x = 125. The number of times that the merchant should 10, 000 order is = 80 in order minimize the total cost. 125 The minimum total cost is C(125) = 45,000 + 2.88(125) = 720. 125 Section 4.4 1. T (c) ±5.8% Section 4.5 3. F 5. derivatives; limits 7. Answers will vary. 9. −5/3 √ 11. − 2/2 13. 0 15. a/b 17. 1/2 19. 0 21. ∞ 23. 0 25. −2 27. 0 29. 0 31. ∞ 33. ∞ 35. 0 37. 1 Page A.8 Custom Made for Math 1174 at Langara College by P. Anhaouy 39. 1 17. − cos θ + C 41. 1 19. 5eθ + C 43. 1 21. 45. 1 23. t6 /6 + t4 /4 − 3t2 + C 47. 2 25. eπ x + C 49. −∞ 27. 51. 0 (c) x < 0 (d) 1/x 1. Answers will vary. 5. Answers will vary. 7. velocity 9. 1/9x9 + C 11. t + C 13. −1/(3t) + C √ 15. 2 x + C (a) x > 0 (b) 1/x Section 4.6 3. Answers will vary. 5t +C 2 ln 5 (e) ln |x| + C. Explanations will vary. 29. 5ex + 5 31. tan x + 4 33. 5/2x2 + 7x + 3 35. 5ex − 2x 37. 2x4 ln2 (2)+2x +x ln 2)(ln 32−1)+ln2 (2) cos(x)−1−ln2 (2) ln2 (2) 39. No answer provided. Custom Made for Math 1174 at Langara College by P. Anhaouy Page A.9 Index absolute maximum, 147 absolute minimum, 147 acceleration, 81, 101 antiderivative, 232 asymptote horizontal, 42 vertical, 40 average cost function, 47 Bisection Method, 57 Chain Rule, 113 notation, 119 concave down, 169 concave up, 169 concavity, 169 inflection point, 170 test for, 170 Constant Multiple Rule of derivatives, 88 of integration, 237 continuous function, 52 properties, 55 Cost function linear, 47 average, 47 derivative, 103 critical number, 149 critical point, 149 curve sketching, 177 decreasing function, 160 finding intervals, 162 strictly, 160 derivative as a function, 73 at a point, 68 basic rules, 86 Chain Rule, 113, 119 Constant Multiple Rule, 88 Constant Rule, 86 differential, 216 exponential functions, 135 First Deriv. Test, 164 general logarithmic functions, 136 Generalized Power Rule, 114 higher order, 89 interpretation, 90 implicit, 122 interpretation, 79 inverse function, 132 inverse trig., 143 Mean Value Theorem, 156 natural exponential, 86 natural exponential functions, 134 natural logarithmic functions, 136 normal line, 70 notation, 73, 89 Power Rule, 86, 97, 128 Product Rule, 92 Quotient Rule, 95 Reciprocal Rule, 86 Second Deriv. Test, 174 Square Root Rule, 86 Sum/Difference Rule, 88 tangent line, 68 trigonometric functions, 110 differentiable, 68 differential, 216 notation, 216 Elasticity General, 199 of Demand Problems, 199 of price demand function, 201 special case, 201 extrema absolute, 147 and First Deriv. Test, 164 and Second Deriv. Test, 174 finding, 150 relative, 148 Extreme Value Theorem, 147 extreme values, 147 First Derivative Test, 164 floor function, 53 Generalized Power Rule, 114 implicit differentiation, 122 increasing function, 160 finding intervals, 162 strictly, 160 indefinite integral, 232 indeterminate form, 2, 41, 226, 227 inflection point, 170 initial value problem, 238 integration indefinite, 232 notation, 233 Power Rule, 237 Sum/Difference Rule, 237 Intermediate Value Theorem, 57 Inventory Costs Control, 207 L’Hôpital’s Rule, 223, 225 limit at infinity, 42 definition, 11 difference quotient, 7 does not exist, 5, 34 indeterminate form, 2, 41, 226, 227 L’Hôpital’s Rule, 223, 225 left handed, 32 of infinity, 39 one sided, 32 properties, 19 pseudo-definition, 2 right handed, 32 Squeeze Theorem, 24 Linear Approximation formula and error, 212 revisited, 212 Linear motion, 100 Linearizaton of a function, 213 logarithmic differentiation, 138 Marginal Analysis, 102 marginal cost, 103 marginal profit, 103 marginal revenue, 103 maximum absolute, 147 and First Deriv. Test, 164 and Second Deriv. Test, 174 relative/local, 148 Mean Value Theorem of differentiation, 156 minimum absolute, 147 and First Deriv. Test, 164, 174 relative/local, 148 natural exponential differentiation, 86 normal line, 70 Optimal Harvest Probelms, 204 optimization, 192 point of inflection, 170 Power Rule differentiation, 86, 92, 97, 128 integration, 237 Profit function derivative, 103 Quotient Rule, 95 related rates, 185 Revenue and Demand, 103 Revenue function derivative, 103 simple, 48 Rolle’s Theorem, 156 Second Derivative Test, 174 Squeeze Theorem, 24 Sum/Difference Rule of derivatives, 88 of integration, 237 tangent line, 68 velocity, 81 average, 100 instantaneous, 100 Differentiation Rules 1. d (cx) = c dx 10. d (ax ) = ln a · ax dx 19.  1 d sin−1 x = √ dx 1 − x2 2. d (u ± v) = u0 ± v 0 dx 11. d 1 (ln x) = dx x 20.  d −1 cos−1 x = √ dx 1 − x2 12. d 1 1 (loga x) = · dx ln a x 21. 13. d (sin x) = cos x dx d (u · v) = uv 0 + dx u0 v d u 4. = dx v vu0 − uv 0 v2 d 5. (u(v)) = u0 (v)v 0 dx 3. 6. d 14. (cos x) = − sin x dx d 15. (csc x) = dx − csc x cot x d 16. (sec x) = sec x tan x dx d (c) = 0 dx d (x) = 1 7. dx 17. d 8. (xn ) = nxn−1 dx 18.  d csc−1 x dx −1 √ |x| x2 − 1  d 22. sec−1 x dx 1 √ |x| x2 − 1  d 1 23. tan−1 x = dx 1 + x2 24. d (tan x) = sec2 x dx  d −1 cot−1 x = dx 1 + x2 25. d (cot x) = − csc2 x dx d (cosh x) = sinh x dx 26. d (sinh x) = cosh x dx 27. d (tanh x) = sech2 x dx d 9. (ex ) = ex dx d (sech x) = dx − sech x tanh x d 29. (csch x) = dx − csch x coth x d 30. (coth x) = − csch2 x dx 28. = = 31.  d 1 cosh−1 x = √ dx x2 − 1 32.  1 d sinh−1 x = √ dx x2 + 1 33.  d −1 sech−1 x = √ dx x 1 − x2  d csch−1 x = dx −1 √ |x| 1 + x2  d 1 35. tanh−1 x = dx 1 − x2 34. 36.  d 1 coth−1 x = dx 1 − x2 Integration Rules Z Z c · f (x) dx = c 1. Z Z f (x) dx tan x dx = − ln | cos x| + C 11. Z Z f (x) ± g(x) dx = Z Z f (x) dx ± g(x) dx Z 3. 0 dx = C 2. 12. sec x dx = ln | sec x + tan x| + C 23. csc x dx = − ln | csc x + cot x| + C 24. a2 − x2 √ dx = sin−1 1 x x2 − a2 dx = x a 1 sec−1 a +C  |x| a  +C Z Z 14. 1 Z Z 13. √ 22. cosh x dx = sinh x + C Z cot x dx = ln | sin x| + C 25. sec2 x dx = tan x + C 26. csc2 x dx = − cot x + C 27. sec x tan x dx = sec x + C 28. csc x cot x dx = − csc x + C 29. sinh x dx = cosh x + C Z 1 dx = x + C 4. Z 5. xn dx = 1 xn+1 + C, n 6= n+1 −1 n 6= −1 Z 6. ex dx = ex + C Z 15. Z 16. Z Z Z 17. 7. Z 8. 1 · ax + C a dx = ln a 1 dx = ln |x| + C x Z 19. Z cos x dx = sin x + C sin x dx = − cos x + C cos2 x dx =  1 1 x + sin 2x + C 2 4  1 1 x − sin 2x + C 2 4 Z x 1 1 dx = tan−1 +C 21. 2 2 x +a a a 20. Z 10. √ Z √ x Z 9. 18. coth x dx = ln | sinh x| + C Z Z Z tanh x dx = ln(cosh x) + C sin2 x dx = 1 x2 − a2 1 x2 + a2 dx = ln x + p x2 − a2 + C dx = ln x + p x2 + a2 + C 1 1 a+x dx = ln +C a2 − x2 2a a−x   Z 1 1 x √ √ 31. dx = ln + a x a2 − x2 a + a2 − x2 C Z 1 1 x √ √ 32. dx = ln +C a x x2 + a2 a + x2 + a2 Z 30. The Unit Circle  √  −1, 3  √ √ 2 2 − 22 , 22 2π/3  √  − 23 , 12 3π/4 120◦ ◦ 5π/6 135 Definitions of the Trigonometric Functions y (0, 1)  π/2 90◦ 1 2, 3  2  √ Unit Circle Definition 2 2 , π/3 π/4 ◦ 60 ◦ 45 150◦ √ π/6 √ 2 2 √  y 3 1 2 , 2 (x, y)  sin θ = y θ y 30◦ π 180◦ 0 270◦ 3π/2 (0, −1) 0 (1, 0) 330◦ 11π/6 315◦ √  ◦ 300 3 7π/4 ,−1 2 5π/3  √ x √ 2 2 2 ,− 2 √  3 1 2,− 2  1 y sec θ = 1 x tan θ = y x cot θ = x y Right Triangle Definition 2 e us en t po Hy θ Adjacent Opposite 210◦ 7π/6 225◦  √  ◦ 5π/4 240 − 23 , − 12  √ 4π/3 √  − 22 , − 22  √  − 12 , − 23 ◦ csc θ = x x (−1, 0) cos θ = x sin θ = O H csc θ = H O cos θ = A H sec θ = H A tan θ = O A cot θ = A O Common Trigonometric Identities Pythagorean Identities sin2 x + cos2 x = 1 2 2 tan x + 1 = sec x 1 + cot2 x = csc2 x Cofunction Identities   π π − x = cos x − x = sec x sin csc 2 2 π π   cos sec − x = sin x − x = csc x 2 2 π  π  tan − x = cot x cot − x = tan x 2 2 Sum to Product Formulas     x+y x−y cos sin x + sin y = 2 sin 2 2     x−y x+y sin x − sin y = 2 sin cos 2 2     x+y x−y cos x + cos y = 2 cos cos 2 2     x+y x−y cos x − cos y = −2 sin sin 2 2 Even/Odd Identities sin(−x) = − sin x cos(−x) = cos x tan(−x) = − tan x csc(−x) = − csc x sec(−x) = sec x cot(−x) = − cot x Power–Reducing Formulas 1 − cos 2x sin2 x = 2 1 + cos 2x cos2 x = 2 1 − cos 2x tan2 x = 1 + cos 2x Double Angle Formulas sin 2x = 2 sin x cos x cos 2x = cos2 x − sin2 x = 2 cos2 x − 1 = 1 − 2 sin2 x 2 tan x tan 2x = 1 − tan2 x Product to Sum Formulas  1 cos(x − y) − cos(x + y) sin x sin y = 2  1 cos(x − y) + cos(x + y) cos x cos y = 2  1 sin x cos y = sin(x + y) + sin(x − y) 2 Angle Sum/Difference Formulas sin(x ± y) = sin x cos y ± cos x sin y cos(x ± y) = cos x cos y ∓ sin x sin y tan x ± tan y tan(x ± y) = 1 ∓ tan x tan y Areas and Volumes Triangles h = a sin θ Right Cone c a h Area = 12 bh θ Law of Cosines: c2 = a2 + b2 − 2ab cos θ b Parallelograms Volume = 13 πr2 h h h r Surface Area = √ πr r2 + h2 + πr2 Right Cylinder Area = bh Circular Circular r 2 Volume = πr h h Surface Area = b 2πrh + 2πr2 Trapezoids Sphere a Area = 12 (a + b)h Volume = 43 πr3 r Surface Area =4πr2 h b Circles General Cone Area = πr2 Area of Base = A r Circumference = 2πr h Volume = 13 Ah A Sectors of Circles θ in radians Area = 12 θr2 s = rθ s General Cylinder Right Area of Base = A θ r h Volume = Ah A Algebra Factors and Zeros of Polynomials Let p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 be a polynomial. If p(a) = 0, then a is a zero of the polynomial and a solution of the equation p(x) = 0. Furthermore, (x − a) is a f actor of the polynomial. Fundamental Theorem of Algebra An nth degree polynomial has n (not necessarily distinct) zeros. Although all of these zeros may be imaginary, a real polynomial of odd degree must have at least one real zero. Quadratic Formula If p(x) = ax2 + bx + c, and 0 ≤ b2 − 4ac, then the real zeros of p are x = (−b ± √ b2 − 4ac)/2a Special Factors x2 − a2 = (x − a)(x + a) x3 − a3 = (x − a)(x2 + ax + a2 ) 3 3 2 2 x + a = (x + a)(x − ax + a ) x4 − a4 = (x2 − a2 )(x2 + a2 ) n(n−1) (x + y)n = xn + nxn−1 y + 2! xn−2 y 2 + · · · + nxy n−1 + y n (x − y)n = xn − nxn−1 y + n(n−1) xn−2 y 2 − · · · ± nxy n−1 ∓ y n 2! Binomial Theorem 2 2 2 (x + y) = x + 2xy + y (x + y)3 = x3 + 3x2 y + 3xy 2 + y 3 (x + y)4 = x4 + 4x3 y + 6x2 y 2 + 4xy 3 + y 4 (x − y)2 = x2 − 2xy + y 2 (x − y)3 = x3 − 3x2 y + 3xy 2 − y 3 (x − y)4 = x4 − 4x3 y + 6x2 y 2 − 4xy 3 + y 4 Rational Zero Theorem If p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 has integer coefficients, then every rational zero of p is of the form x = r/s, where r is a factor of a0 and s is a factor of an . Factoring by Grouping acx3 + adx2 + bcx + bd = ax2 (cs + d) + b(cx + d) = (ax2 + b)(cx + d) Arithmetic Operations ab + ac = a(b + c) a  cb  = d a a d b c =   b ab = c c ad bc c ad + bc a + = b d bd a b c a+b a b = + c c c ac a  = b b c a bc = a−b b−a = c−d d−c ab + ac =b+c a Exponents and Radicals a0 = 1, a 6= 0 (ab)x = ax bx  a x √ n b ax = x b am = am/n ax ay = ax+y −x a 1 = x a √ a = a1/2 √ n √ √ ab = n a n b ax = ax−y ay x y xy (a ) = a √ n a = a1/n r n √ n a a = √ n b b Additional Formulas Summation Formulas: n X n X n(n + 1) 2 i=1  2 n X n(n + 1) 3 i = 2 i=1 c = cn i=1 n X i2 = i=1 n(n + 1)(2n + 1) 6 i= Trapezoidal Rule: Z b f (x) dx ≈ a with Error ≤  ∆x  f (x1 ) + 2f (x2 ) + 2f (x3 ) + ... + 2f (xn ) + f (xn+1 ) 2  (b − a)3  max f 00 (x) 2 12n Simpson’s Rule: Z b a f (x) dx ≈ with Error ≤  ∆x  f (x1 ) + 4f (x2 ) + 2f (x3 ) + 4f (x4 ) + ... + 2f (xn−1 ) + 4f (xn ) + f (xn+1 ) 3  (b − a)5  max f (4) (x) 4 180n Arc Length: Surface of Revolution: Z bp 1 + f 0 (x)2 dx L= S = 2π Z b a a p f (x) 1 + f 0 (x)2 dx (where f (x) ≥ 0) S = 2π Z b p x 1 + f 0 (x)2 dx a (where a, b ≥ 0) Work Done by a Variable Force: W = Z b F (x) dx Force Exerted by a Fluid: F = a Z b w d(y) `(y) dy a Taylor Series Expansion for f (x): pn (x) = f (c) + f 0 (c)(x − c) + f 00 (c) f 000 (c) f (n) (c) (x − c)2 + (x − c)3 + ... + (x − c)n 2! 3! n! Maclaurin Series Expansion for f (x), where c = 0: pn (x) = f (0) + f 0 (0)x + f 00 (0) 2 f 000 (0) 3 f (n) (0) n x + x + ... + x 2! 3! n! Summary of Tests for Series: Test Series ∞ X nth-Term n=1 ∞ X Geometric Series Condition(s) of Convergence an Telescoping Series rn p-Series Integral Test (bn − bn+a ) n=1 ∞ X 1 (an + b)p n=1 ∞ X an Direct Comparison ∞ X an ∞ X an n=0 Limit Comparison n=0 Ratio Test ∞ X an n=0 Root Test ∞ X n=0 lim an 6= 0 This test cannot be used to show convergence. |r| < 1 |r| ≥ 1 lim bn = L Sum = n=1 p≤1 p>1 Z ∞ 1 1−r ! a X bn − L Sum = n→∞ a(n) dn Z ∞ a(n) dn 1 1 is convergent n=0 Comment n→∞ n=0 ∞ X Condition(s) of Divergence ∞ X is divergent ∞ X bn n=0 an = a(n) must be continuous bn n=0 converges and diverges and 0 ≤ an ≤ bn ∞ X bn 0 ≤ bn ≤ an ∞ X bn n=0 converges and n=0 diverges and n→∞ lim an /bn ≥ 0 n→∞ an+1 lim <1 n→∞ an an+1 lim >1 n→∞ an lim an /bn > 0 Also diverges if lim an /bn = ∞ n→∞ {an } must be positive Also diverges if lim an+1 /an = ∞ n→∞ an lim an n→∞ 1/n <1 lim an n→∞ 1/n {an } must be positive >1 Also diverges if 1/n lim an =∞ n→∞ @ Typesetting and Editing by Pichmony Anhaouy, Mathematics and Statistic Department, Langara College, 2023.