Mathematics 1174 SOLUTIONS TO SELECTED EXERCISES FROM MATH 1174 TEXTBOOK November 7, 2023 Minh Phuong Bui & Pichmony Anhaouy Langara College Mathematics and Statistics department panhaouy@langara.ca Contents 1 2 Limits 5 1.1 1.3 1.4 1.5 1.6 5 8 9 12 15 An Introduction to Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . Finding Limits Analytically . . . . . . . . . . . . . . . . . . . . . . . . . . One Sided Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Limits Involving Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Derivatives 17 2.1 Introduction to Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Basic Differentiation Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 The Product and Quotient Rules . . . . . . . . . . . . . . . . . . . . . . 2.5 Derivative as Rates of Change . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Derivative of Trigonometric Functions . . . . . . . . . . . . . . . . . . 2.7 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9 Derivatives of Exponential and Logarithmic Functions . . . . . . 2.10 Derivatives of Inverse Trigonometric Functions . . . . . . . . . . . . 3 Graphical Behaviour of Functions 3.1 3.3 3.4 4 4.1 4.2 4.3 4.4 4.5 4.6 31 Extreme Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Increasing and Decreasing Functions . . . . . . . . . . . . . . . . . . . Concavity and Second Derivative . . . . . . . . . . . . . . . . . . . . . . Applications of the Derivatives 31 32 34 37 Related Rates Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Applied Optimization Problems . . . . . . . . . . . . . . . . . . . . . . . Economics and Business Applications . . . . . . . . . . . . . . . . . . . Linear Approximation and Differentials . . . . . . . . . . . . . . . . . L’Hopital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Antiderivatives and Indefinite Integration . . . . . . . . . . . . . . . 3 17 19 20 22 24 25 26 28 30 37 41 44 46 48 50 Chapter 1 Limits 1.1 An Introduction to Limits  13. Let f (x) = x2 − x + 1, x ≤ 3, 2x + 1, x > 3. x 2.99 2.999 2.9999 .. . f (x) 6.95010 6.99501 6.99950 .. . 3.01 3.001 3.0001 .. . 7.05010 7.00501 7.00050 .. . So, from the table above and the graph below, we can see that the lim f (x) = 7. x→3 12 10 8 6 4 2 −1 1 2 3 4 5 5 17. Here, we have f (x) = 9x + 0.06, a = −1. From the table below, we see that the limit seems to be exactly 9. −0.1 −0.01 .. . f (−1 + h) − f (−1) h 9 9 .. . 0.01 0.1 .. . 9 9 .. . h 1 19. Here, we have f (x) = , a = 2. From the table below, we see that the limit is approximately x+1 −0.11. −0.1 −0.01 .. . f (2 + h) − f (2) h −0.114943 −0.111483 .. . 0.01 0.1 .. . −0.110742 −0.107527 .. . h 21. Here, we have f (x) = ln x, a = 5. From the table below, we see that the limit is approximately 0.2. −0.1 −0.01 .. . f (5 + h) − f (5) h 0.202027 0.2002 .. . 0.01 0.1 .. . 0.1998 0.198026 .. . h Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 6 23. Here, we have f (x) = cos x, a = π. From the table below, we see that the limit is approximately ±0.005. However, if we continue on by increasing the digit 0, eventually we will get the limit approaching 0. −0.1 −0.01 .. . f (π + h) − f (π) h −0.0499583 −0.0499583 .. . 0.01 0.1 .. . 0.00499996 0.0499583 .. . h Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 7 1.3 Finding Limits Analytically  3 lim f (x) 3f (x) 3·6 7. lim = x→9 = = 6. x→9 g(x) lim g(x) 3  x→9   f (x) 9. lim . This is not possible because as x → 6, the denominator goes to zero. So, we x→6 3 − g(x) can say that the limit in this particular case does not exit. 11. lim f (g(x)). Not possible to know; as x approaches 6, g(x) approaches 3, but we know nothing x→6 of the behaviour of f (x) when x is near 3. h i 13. lim f (x)g(x) − f 2 (x) + g 2 (x) = lim f (x) · lim g(x) − lim f 2 (x) + lim g 2 (x). x→6 x→6 x→6 x→6 x→6 = 9 · 3 − 81 + 9 = −45. 15. lim cos((g(x)) = lim cos(π) = −1. x→10 x→10 17. lim g(5f (x)). x→1 lim 5f (x) = 5 lim f (x) = 10 and lim g(x) = π and g(10) = π. x→1 x→1 x→10 ⇒ lim g(5f (x)) = π. x→1 19. lim x→π  x − 3 7 x−5 =  π − 3 7 π−5 ≈ 0. x2 + 3x + 5 π 2 + 3π + 5 = ≈ 0.6064. x→π 5x2 − 2x − 3 5π 2 − 2π − 3 25. lim x2 − 4x − 12 (x − 6)(x + 2) x+2 27. lim 2 = lim = lim = −8. x→6 x − 13x + 42 x→6 (x − 6)(x − 7) x→6 x − 7 x2 + 6x − 16 (x − 2)(x + 8) x+8 = lim = lim = 10. 2 x→2 x − 3x + 2 x→2 (x − 2)(x − 1) x→2 x − 1 29. lim x2 − 5x − 14 (x − 7)(x + 2) x−7 3 = lim = lim =− . 2 x→−2 x + 10x + 16 x→−2 (x + 2)(x + 8) x→−2 x + 8 2 31. lim Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 8 1.4 5. One Sided Limits a) lim− f (x) = 2. x→1 b) lim+ f (x) = 2. x→1 c) lim f (x) = 2. x→1 d) f (1) = 1. e) lim− f (x). As f is not defined for x < 0, this limit is not defined. x→0 f) lim+ f (x) = 1. x→0 7. a) lim− f (x) = +∞. Note: For now, we can also say the limit does not exist. x→1 b) lim+ f (x) = +∞. Note: For now, we can also say the limit does not exist. x→1 c) lim f (x) = +∞. Note: For now, we can also say the limit does not exist. x→1 d) f (1) is not defined. e) lim− f (x) = 0. x→2 f) lim+ f (x) = 0. x→0 9. a) lim− f (x) = 2. x→1 b) lim+ f (x) = 2. x→1 c) lim f (x) = 2. x→1 d) f (1) = 2. 11. a) b) Math 1174 lim f (x) = 2. x→−2− lim f (x) = 2. x→−2+ Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 9 c) lim f (x) = 2. x→−2 d) f (−2) = 0. e) lim− f (x) = 2. x→2 f) lim+ f (x) = 2. x→2 g) lim f (x) = 2. x→2 h) f (2) is not defined. 13. a) lim− f (x) = lim− (x + 1) = 2. x→1 x→1 b) lim+ f (x) = lim+ (x2 − 5) = −4. x→1 x→1 c) lim f (x) = DN E. x→1 d) f (1) = 2. 15. a) b) lim f (x) = lim − (x2 − 1) = 0. x→−1− x→−1 lim f (x) = lim − (x3 + 1) = 0. x→−1+ x→−1 c) lim f (x) = 0. x→−1 d) f (−1) = 0. e) lim− f (x) = lim− (x3 + 1) = 2. x→1 x→1 f) lim+ f (x) = lim+ (x2 + 1) = 2. x→1 x→1 g) lim f (x) = 2. x→1 h) f (1) = 2. Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 10 17. a) lim− f (x) = lim− (1 − cos2 x) = 1 − cos2 a. x→a x→a b) lim+ f (x) = lim+ sin2 x = sin2 a = 1 − cos2 a. x→a x→a c) lim f (x) = 1 − cos2 a. x→a d) f (a) = sin2 a = 1 − cos2 a. 19. a) lim− f (x) = lim− x2 = 4. x→2 x→2 b) lim+ f (x) = lim+ (−x2 + 2x + 4) = 4. x→2 x→2 c) lim f (x) = 4. x→2 d) f (2) = 2 + 1 = 3. 21. a) lim− f (x) = lim− −x = −1. x b) lim+ f (x) = lim+ x = 1. x x→0 x→0 x→0 x→0 c) lim f (x) = DN E. x→0 d) f (0) = 0. Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 11 1.5 Limits Involving Infinity 9. f (x) = a) b) 1 . (x + 1)2 lim f (x) = +∞. x→−1− lim f (x) = +∞. x→−1+ 11. f (x) = 1 ex + 1 . a) lim f (x) = 1. x→−∞ b) lim f (x) = 0. x→+∞ 1 c) lim− f (x) = . x→0 2 1 d) lim+ f (x) = . x→0 2 13. f (x) = cos x. a) lim f (x) does not exist because cos x is bouncing between −1 and 1 for all x. x→−∞ b) lim f (x) does not exist because cos x is bouncing between −1 and 1 for all x. x→+∞ 15. f (x) = x2 − 1 (x − 1)(x + 1) = . 2 x −x−6 (x − 3)(x + 2) a) lim− f (x) = x→3 (2)(4) 8 = − = −∞. Or we can try table of numbers like below − (0 )(5) 0 x 2.9 2.99 2.999 .. . b) lim+ f (x) = x→3 f (x) −15.1224 −159.12 −1599.12 .. . (2)(4) 8 = + = ∞. Or we can try table of numbers like below + (0 )(5) 0 x 3.1 3.01 3.001 .. . f (x) 16.8824 160.88 1600.88 .. . c) It seems that the limit lim f (x) does not exist. x→3 Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 12 17. f (x) = (x − 6)(x − 5) x2 − 11x + 30 = . 3 2 x − 4x − 3x + 18 (x − 2)(x − 3)2 a) lim− f (x) = x→3 (−3)(−2) 6 = + = ∞. Or we can try table of numbers like below − 2 (1)(0 ) 0 x 2.9 2.99 .. . b) lim+ f (x) = x→3 f (x) 132.857 12124.4 .. . (−3)(−2) 6 = + = ∞. Or we can try table of numbers like below + 2 (1)(0 ) 0 x 3.1 3.01 .. . f (x) 108.039 11876.4 .. . c) So, it seems that the limit lim f (x) = ∞. x→3 19. f (x) = 2x2 − 2x − 4 (x − 2)(x + 1) = . 2 x + x − 20 (x − 4)(x + 5) 2x2 = 2. So horizontal asymptote is y = 2. x→∞ x→∞ x2 (2)(5) 10 (−7)(−4) 10 lim− f (x) = − = − = −∞ and lim − f (x) = = − = +∞. + x→4 x→−5 (0 )(9) 0 (−9)(0 ) 0 So vertical asymptotes are x = 4 and x = −5. lim f (x) ≈ lim x2 + x − 12 (x − 3)(x + 4) x+4 21. f (x) = 3 = = . 2 7x − 14x − 21x x(x − 3)(x + 1) x(x + 1) x2 1 = lim = 0. So horizontal asymptote is y = 0. 3 x→∞ x→∞ 7x x→∞ x 4 4 3 4 lim− f (x) = − = − = −∞ and lim − f (x) = = + = +∞. − x→0 x→−1 (0 )(1) 0 (−1)(0 ) 0 So vertical asymptotes are x = 0 and x = −1. lim f (x) ≈ lim 23. f (x) = x2 − 9 (x − 3)(x + 3) x−3 = = , x 6= −3. 9x + 27 9(x + 3) 9 x x2 = lim = ∞. So no horizontal asymptote. x→∞ x→∞ 9x x→∞ 9 The function does not have vertical asymptote as well because the denominator is 9. lim f (x) ≈ lim Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 13 x3 x3 + 2x2 + 1 ≈ lim = lim x2 = ∞. x→∞ x x→∞ x→∞ x−5 25. lim x3 + 2x2 + 1 x3 ≈ lim = lim x = −∞. x→−∞ x→∞ x2 x→−∞ x2 − 5 27. lim Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 14 1.6 Continuity 11. f (x) is not continuous at a = 1 because lim f (x) = lim+ f (x) = 2 but f (1) = 1. x→1− x→1 13. f (x) is not continuous at a = 1 because f (1) is not defined or lim f (x) = ∞ and limx→1+ f (x) = ∞. x→1− 15. f (x) is continuous at a = 1 because lim f (x) = lim+ f (x) = f (1) = 2. x→1− 19. x→1 a) f (x) is continuous at a = 0 because lim f (x) = lim+ f (x) = f (0) = 03 − 0 = 0. x→0− x→0 b) f (x) is not continuous at a = 1 because lim f (x) = lim− x3 − x = 0 and lim+ f (x) = lim+ x − 2 = −1 = f (1). x→1− 21. x→1 x→1 x→1 a) f (x) is continuous at a = 0 because lim− f (x) = lim+ f (x) = f (0) = x→0 x→0 64 02 − 64 =− . 2 0 − 11 · 0 + 24 24 b) f (x) is not continuous at a = 8 because (x − 8)(x + 8) x+8 16 = lim = . x→8 (x − 8)(x − 3) x→8 x − 3 5 lim− f (x) = lim+ f (x) = lim x→8 x→8 but f (8) = 5. √ 23. g(x) = x2 − 4. So the domain of the function is x2 − 4 ≥ 0 ⇒ x ∈ (−∞, −2] ∪ [2, +∞). The function f(x) is continuous at x = −2 and x = 2 because lim + g(x) = 0 and lim− g(x) = 0. x→−2 x→2 Therefore, the continuous interval is (−∞, −2] ∪ [2, +∞). Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 15 25. g(x) = √ 5t2 − 30 = p 5(t2 − 6). √ √ So the domain of the function is x2 − 4 ≥ 0 ⇒ x ∈ (−∞, − 6] ∪ [ 6, +∞). √ √ The function f(x) is continuous at x = − 6, x = 6, because lim √ + x→− 6 g(x) = 0 and lim g(x) = 0. √ − x→ 6 √ √ Therefore, the continuous interval is (−∞, − 6] ∪ [ 6, +∞). 27. g(x) = 1 . 1 + x2 The function g(x) is continuous everywhere, (−∞, +∞). 31. g(x) = 1 − ek . We need 1 − ek ≥ 0 ⇒ 1 ≥ ek ⇒ k ≤ 0. The function g(x) is continuous at x = 0 because lim+ g(x) = 0. x→0 Therefore, the continuous interval is (−∞, 0]. Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 16 Chapter 2 Derivatives 2.1 Introduction to Derivative 7. f (x) = 2x. 2(x + h) − 2x 2h f (x + h) − f (x) = lim = lim = 2. h→0 h→0 h h→0 h h f 0 (x) = lim 9. f (x) = x2 . f (x + h) − f (x) (x + h)2 − x2 x2 + 2xh + h2 − x2 = lim = lim h→0 h→0 h→0 h h h h(2x + h) = lim = lim (2x + h) = 2x. h→0 h→0 h f 0 (x) = lim 11. f (x) = 1 . x 1 − x1 f (x + h) − f (x) x − (x + h) x+h f (x) = lim = lim = lim h→0 h→0 h→0 hx(x + h) h h 1 1 = lim − = − 2. h→0 x(x + h) x 0 15. f (x) = 4 − 3x, at x = 7. f (x + h) − f (x) 4 − 3(x + h) − (4 − 3x) = lim = −3. h→0 h→0 h h At x = 7 ⇒ y = f (7) = −17. Then, f 0 (x) = lim a) the equation of the tangent line is y + 17 = −3(x − 7). 1 b) the equation of the normal line at the same point is y + 17 = (x − 7). 3 Note: No need to simplify the results. 17 17. f (x) = 3x2 − x + 4, at x = −1. f (x + h) − f (x) 3(x + h)2 − (x + h) + 4 − (3x2 − x + 4) = lim . h→0 h→0 h h 6xh + 3h2 − h = lim = 6x − 1. h→0 h At x = −1 ⇒ y = f (−1) = 8, f 0 (−1) = −7. Then, f 0 (x) = lim a) the equation of the tangent line is y − 8 = −7(x + 1). 1 b) the equation of the normal line at the same point is y − 8 = (x + 1). 7 Note: No need to simplify the results. 19. f (x) = 1 , at x = 3. x−2 1 1 − x−2 f (x + h) − f (x) x−2−x−h+2 x+h−2 f (x) = lim = lim = lim . h→0 h→0 h→0 h(x − 2)(x + h − 2) h h −1 1 = lim = lim − . h→0 (x − 2)(x + h − 2) h→0 (x − 2)2 0 At x = 3 ⇒ y = f (3) = 1, f 0 (3) = −1. Then, a) the equation of the tangent line is y − 1 = −1 · (x − 3). b) the equation of the normal line at the same point is y − 1 = 1 · (x − 3). Note: No need to simplify the results. 23. f (x) = 1 . x+1 a) At (0, 1), f 0 (0) ≈ −1. At (1, 0.5), f 0 (1) ≈ −0.25. 1 − 1 f (x + h) − f (x) = lim x+h+1 x+1 h→0 h→0 h h x+1−x−h−1 −1 1 = lim = lim =− . h→0 h(x + 1)(x + h + 1) h→0 (x + 1)(x + h + 1) (x + 1)2 b) f 0 (x) = lim c) At (0, 1), f 0 (0) = −1. Then, the equation of the tangent line is y − 1 = −1(x − 0) ⇒ y = −x + 1. At (1, 0.5), f 0 (1) = −0.25. Then, the equation of the tangent line is y − 0.5 = −0.25(x − 1) ⇒ y = −0.25x + 0.75. Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 18 2.3 Basic Differentiation Rules 9. f (x) = 7x2 − 5x + 7 ⇒ f 0 (x) = 14x − 5. 1 3 11. m(t) = 9t5 − t3 + 3t − 8 ⇒ m0 (t) = 45t4 − t2 + 3. 8 8 1 1 1 13. p(s) = s4 + s3 + s2 + s + 1 ⇒ p0 (s) = s3 + s2 + s + 1. 4 3 2 15. Note that we will have a faster and better way of finding derivative of this type, but for now we expand the function to get the following: g(x) = (2x − 5)3 = (2x − 5)2 (2x − 5) = (4x2 − 20x + 25)(2x − 5) = 8x3 − 60x2 + 150x + 125. Thus, g 0 (x) = 24x2 − 120x + 150. 17. Note: The same as the previous equation. f (x) = (2−3x)2 = 4−12x+9x2 ⇒ f 0 (x) = −12+18x. 19. First 4 derivatives of h(t) = t2 − et . ⇒ h0 (t) = 2t − et ⇒ h00 (t) = 2 − et ⇒ h000 (t) = −et ⇒ h(4) (t) = −et . 21. f (x) = x3 − x ⇒ f (1) = 0. Also, f 0 (x) = 3x2 − 1 ⇒ f 0 (1) = 3 − 1 = 2. Thus, the equation of the tangent line at (1, 0) is y − 0 = 2(x − 1) ⇒ y = 2x − 2. 1 The slope of the normal line is m = − . So, the equation of the normal line at (1, 0) is 2 1 1 1 y − 0 = − (x − 1) ⇒ y = − x + . 2 2 2 23. f (x) = 2x + 3 ⇒ f (5) = 13. Also, f 0 (x) = 2 ⇒ f 0 (5) = 2. Thus, the equation of the tangent line at (5, 13) is ⇒ y − 13 = 2(x − 5) ⇒ y = 2x + 3. Note that since the function is linear, the tangent line at any point is the line itself. 1 The slope of the normal line is m = − . So, the equation of the normal line at (5, 13) is 2 1 1 31 ⇒ y − 13 = − (x − 5) ⇒ y = − x + . 2 2 2 Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 19 2.4 The Product and Quotient Rules 5. f (x) = x(x2 + 3x). a) f (x) = x(x2 + 3x) ⇒ f 0 (x) = 1(x2 + 3x) + x(2x + 3) = 3x2 + 6x. b) f (x) = x(x2 + 3x) = x3 + 3x2 ⇒ f 0 (x) = 3x2 + 6x. c) We can see that they are equal. 7. h(s) = (2s − 1)(s + 4). a) h(s) = (2s − 1)(s + 4) ⇒ h0 (s) = 2(s + 4) + (2s − 1) = 4s − 7. b) h(s) = (2s − 1)(s + 4) = 2s2 − 7s − 4 ⇒ h0 (s) = 4s − 7. c) We can see that they are equal. 9. f (x) = x2 + 3 . x x2 + 3 2x(x) − (x2 + 3) x2 − 3 ⇒ f 0 (x) = = = 1 − 3x−2 . 2 2 x x x 2 x +3 = x + 3x−1 ⇒ f 0 (x) = 1 − 3x−2 . b) f (x) = x c) We can see that they are equal. a) f (x) = 11. h(s) = 3 4s3 3 0 − 3(12s2 ) 3 9 0 ⇒ h (s) = = − 4 = − s−4 . 3 6 4s 16s 4s 4 3 3 9 b) h(s) = 3 = s−3 ⇒ h0 (s) = − s−4 . 4s 4 4 c) We can see that they are equal. a) h(s) = 13. g(x) = x+7 (x − 5) − (x + 7) 12 ⇒ g 0 (x) = =− . 2 x−5 (x − 5) (x − 5)2 15. f (x) = x4 + 2x3 (4x3 + 6x2 )(x − 2) − (x4 + 2x3 ) ⇒ f 0 (x) = . Note: No need to simplify the x+2 (x + 2)2 result. Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 20 17. g(s) = es (s2 + 2) ⇒ g 0 (s) = es (s2 + 2) + es (2s) ⇒ g 0 (0) = 2. Thus, the equation of the tangent line at (0, 2) is ⇒ y − 2 = 2(s − 0) ⇒ y = 2s + 2. 1 The slope of the normal line is m = − . So, the equation of the normal line at (0, 2) is 2 1 1 ⇒ y − 2 = − (x − 0) ⇒ y = − x + 2. 2 2 19. The graph of f (x)has a horizontal tangent line at a point when f 0 (x) = 0. Here, we have 75 3 f (x) = 6x2 − 18x − 24 ⇒ f 0 (x) = 12x − 18 = 0 ⇒ x = ⇒ y = − . 2 2 21. Same as the previous question. x2 2x(x + 1) − x2 x2 + 2x x(x + 2) f (x) = ⇒ f 0 (x) = = = = 0 ⇒ x = 0, −2. 2 2 x+1 (x + 1) (x + 1) (x + 1)2 x = 0 ⇒ y = 0. x = −2 ⇒ y = −4. Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 21 2.5 Derivative as Rates of Change 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. 3. s(t) = 100t − 5t2 ⇒ v(t) = 100 − 10t = 10(10 − t). s(4) − s(1) = 75 m/s. 4−1 b) v = 0 ⇒ t = 10 s. So, smax = s(10) = 500 m. c) v(2) = 80 m/s and v(15) = −50 m/s d) stotal = |smax − s(5)| + |s(12) − smax | = |125| + |20| = 145 m. a) v̄ = 5. h(t) = 490 − 12gt2 . 19.6 − 372.4 h(2) − h(1) = = −352.8. 2−1 1 h(t + ) − h(t) b) h(t + ) = 490 − 12g(t + )2 ⇒ v(t) = . t+−t 490 − 12gt2 − 24gt + 12g2 − 490 + 12gt2 −24gt + 12g2 = = .   = −24gt + 12g r 490 2 c) h(t) = 0 ⇒ 490 − 12gt = 0 ⇒ t = = 2.04. 12 · 9.8 a) v(t) = 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. 9. C(x) = 4200 + 5.40x − 0.001x2 + 0.000002x3 . a) The average cost is given as C̄(x) = C(x) 4200 = + 5.40 − 0.001x + 0.000002x2 . Then, x x 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. Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 22 11. p(x) = −3x2 + 600x and C(x) = 1800 + 357x2 . We have R(x) = p·x = (−3x2 +600x)·x = −3x3 +600x2 ⇒ P (x) = R(x)−C(x) = −3x3 +243x2 −1800 ⇒ 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. Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 23 2.6 Derivative of Trigonometric Functions 3. f (θ) = 9 sin θ + 10 cos θ ⇒ f 0 (θ) = 9 cos θ − 10 sin θ. 5. h(t) = et − sin t − cos t ⇒ h0 (t) = et − cos t + sin t. 7. f (t) = 1 2 1 (csc t − 4) ⇒ f 0 (t) = − 3 (csc t − 4) − 2 csc t tan t. 2 t t t 9. h(x) = cot x − ex ⇒ h0 (x) = − csc2 x − ex . 11. f (x) = cos x(cos x + 3) − sin x(− sin x) sin x 1 + 3 cos x ⇒ f 0 (x) = = . 2 cos x + 3 (cos x + 3) (cos x + 3)2 13. g(t) = 4t3 et − sin t cos t ⇒ g 0 (t) = (12t2 et + 4t3 et ) − (cos2 t − sin2 t). 15. f (x) = x2 ex tan x ⇒ f 0 (x) = (2xex + x2 ex ) tan x + (x2 ex · sec2 x). 17. f (x) = 4 sin x ⇒ f π  2 0 = 4, and also f (x) = 4 cos x ⇒ f The tangent line at the point 0 π  2 = 0.  π , 4 is y = 4 and the normal line to the line y = 4 is x = . 2 2 π 19. g(t) = t sin t ⇒ g 0 (t) = sin t + t cos t ⇒ g 0  3π  = −1.  3π 3π  So, the tangent line at the point is y + = −1 x − ⇒ y = −x. 2 2 The normal line at the point is y + 2  3π 3π  =1 x− ⇒ y = x − 3π. 2 2 21. f (x) = x sin x ⇒ f 0 (x) = sin x + x cos x = 0 ⇒ x = 0; y = 0. 23. f (x) = x sin x ⇒ f 0 (x) = sin x + x cos x ⇒ f 00 (x) = cos x + (cos x − x sin x). ⇒ f (3) (x) = −2 sin x − (sin x + x cos x). ⇒ f (4) (x) = −3 cos x − (cos x − x sin x) = −4 cos x + x sin x. Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 24 2.7 The Chain Rule 5. f (x) = (4x3 − x)10 ⇒ f 0 (x) = 10(4x3 − x)9 (12x2 − 1). 7. g(θ) = (sin θ + cos θ)3 ⇒ g 0 (θ) = 3(sin θ + cos θ)2 (cos θ − sin θ).  9. f (x) = 1 x+ x 4  1 3 1 ⇒ f (x) = 4 x + ) 1 − 2 . x x 0 11. g(x) = tan (5x) ⇒ g 0 (x) = 5 sec2 (5x).  13. p(t) = cos3 (t2 + 3t + 1) ⇒ p0 (t) = 3 cos2 (t2 + 3t + 1) · − sin(t2 + 3t + 1) · (2t + 3). 15. g(t) = cos (t2 + 3t)·sin (5t − 7) ⇒ g 0 (t) = −(2t+3) sin(t2 +3t) sin(5t−7)+5 cos(t2 +3t) cos(5t− 7). 17. f (x) = (4x3 − x)10 ⇒ f (0) = 0. Also, f 0 (x) = 10(4x3 − x)9 (12x2 − 1) ⇒ f 0 (0) = 0. So the tangent line equation at the point (0, 0) is y = 0 and the normal line equation is x = 0. 0 2 19. g (θ) = 3(sin θ + cos θ) (cos θ − sin θ) ⇒ g So the tangent line equation at the point The normal line equation is y − 1 = Math 1174 π  2 =1⇒g 0 π  2 = −3.   π , 1 is y − 1 = −3 x − . 2 2 π 1 π x− . 3 2 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 25 2.8 Implicit Differentiation 13. x4 + y 2 + y = 7 ⇒ 4x3 + 2y dy dy dy 4x3 + =0⇒ =− . dx dx dx 2y + 1 15. cos x + sin y = 1 ⇒ − sin x + cos y 17. dy dy sin x =0⇒ = . dx dx cos y y y0x − y y dy = 10 ⇒ = . =0⇒ 2 x x dx x 19. x2 tan y = 50 ⇒ 2x tan y + x2 sec2 y dy 2x tan y 2 sin y cos y dy =0⇒ = 2 2 =− . dx dx x sec y x 21. (y 2 + 2y − x)2 = 200 ⇒ 2(y 2 + 2y − x)(2yy 0 + 2y 0 − 1) = 0. ⇒ y 0 (2y + 2)(y 2 + 2y − x) = y 2 + 2y − x ⇒ y 0 = 23. 1 . 2y + 2 (cos x + y 0 )(cos y + x) − (sin x + y)(1 − sin yy 0 ) sin x + y =1⇒ = 0. cos y + x (cos y + x)2 ⇒ y0 = sin x + y − cos x(cos y + x) cos y + x + sin y(sin x + y) 25. x0.4 + y 0.4 = 1. x−0.6 y 0.6 a) 0.4x−0.6 + 0.4y −0.6 · y 0 = 0 ⇒ y 0 = − −0.6 = − 0.6 . y x At (1, 0), y 0 = 0, so the tangent line is y − 0 = 0(x − 1) ⇒ y = 0. b) y 0 = −1.859 ⇒ y − 0.281 = −1.859(x − 0.1) ⇒ y = −1.859(x − 0.1) + 0.281. 27. (x2 + y 2 − 4)3 = 108y 2 .  dy  dy  dy dy 2 = 216y ⇒ 3(0 + 16 − 4) 8 = 864 , at (0, 4), y 0 = 0. a) 3(x + y − 4) 2x + 2y dx dx dx dx 2 2 2  ⇒ The tangent line is y − 4 = 0(x − 0) ⇒ y = 4. √ b) At (2, − 4 108), y 0 = 0.93. Then, the equation of the tangent line at the √ 4 point is y + 108 = 0.93(x − 2). Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 26 29. (x − 2)2 + (y − 3)2 = 9. 7   6 + 3√3  dy dy =0⇒2 −2 +2 −3 = 0. a) 2(x − 2) + 2(y − 3) dx 2 2 dx At √ ! 7 6+3 3 dy 1 , ⇒ = − √ . Then, the equation of the tangent line at the 2 2 dx 3 √ 1  7 6+3 3 = −√ x − . point is y − 2 2 3   4 + 3 √3 3 dy − 2 + 2( − 3) = 0, At b) 2 2 2 dx ! √ 4+3 3 3 dy √ , ⇒ = 3. 2 2 dx Then, the equation of the tangent line at the point is √ 3 √  4 + 3 3 ⇒y− = 3 x− . 2 2 y 0.6 31. x0.4 + y 0.4 = 1 ⇒ 0.4x−0.6 + 0.4y −0.6 · y 0 = 0 ⇒ y 0 = − 0.6 . x y 00 = − Math 1174 0.6y −0.4 y 0 x0.6 − 0.6y 0.6 x−0.4 3y 0.6 3 = + . 1.2 1.6 x 5x 5yx1.2 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 27 2.9 Derivatives of Exponential and Logarithmic Functions 1 1 9. y = f (x) = 5x + 10 ⇒ x = 5y + 10 ⇒ y = x − 2 ⇒ f −1 (x) = x − 2. Note that we don’t 5 5 need to find the inverse in this problem. Just want to show how to find one. 1 1 f 0 (x) = 5 ⇒ (f −1 )0 (20) = 0 = . f (2) 5 √ 11. f (x) = sin 2x ⇒ f 0 (x) = 2 cos 2x ⇒ (f −1 )0 ( 3/2) = 13. f (x) = f0 1  = π 6 1 1 = = 1. 2 cos (π/3) 2 · 21 1 2x 1 1 = = −2. ⇒ f 0 (x) = − ⇒ (f −1 )0 (1/2) = 0 2 2 2 1+x (1 + x ) f (1) −2/4 15. f (x) = ln(cos x) ⇒ f 0 (x) = − 17. f (x) = 2 ln(x) ⇒ f 0 (x) = sin x = − tan x. cos x 2 . x 19. g(t) = 5cos t ⇒ g 0 (t) = − ln 5 · 5cos t · sin t. 21. g(t) = 152 ⇒ g 0 (t) = 0. 23. m(w) = 25. h(t) = 27. ln 3 · 3w · 2w − ln 2 · 2w (3w + 1) 3w + 1 0 ⇒ m (w) = . 2w 22w 2t + 3 ln 2 · 2t (3t + 2) − ln 3 · 3t (2t + 3) 0 ⇒ h (t) = . 3t + 2 (3t + 2)2 a) d 1 k (ln(xk )) = kxk−1 k = . dx x x b) d d k (ln(xk )) = (k ln(x)) = . dx dx x 2 29. y = (2x)x ⇒ y(1) = 2. Taking natural log both sides, simplifying, and then implicit y0 2x2 differentiating gives ln(y) = x2 ln(2x) ⇒ = 2x ln(2x) + . y 2x ⇒ y 0 = y(2x ln(2x) + x) ⇒ y 0 (1) = 2(2 ln(2) + 1) = 4 ln(2) + 2. The equation of the tangent line is y − 2 = (4 ln(2) + 2)(x − 1). Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 28 31. y = xsin x+2 ⇒ y (sin x + 2) ln(x). π  2 = (π)3 . Taking natural log both sides and simplifying gives ln(y) = 8  y0 sin x + 2 ⇒ y 0 = y cos x ln(x)+ Then implicit differentiating both sides gives = cos x ln(x)+ y x  π  (π)3 6 2 sin x + 2  3(π) ⇒ y0 = · = . x 2 8 π 4 The equation of the tangent line is y − 33. y = (π)3 3(π)2  π = x− . 8 4 2 (x + 1)(x + 2) 1 ⇒ y(0) = . Now, Taking natural log both sides and (x + 3)(x + 4) 6 simplifying gives ln y = ln(x + 1) + ln(x + 2) − ln(x + 3) − ln(x + 4). Then implicit differentiating both sides gives y0 1 1 1 1 = + − − y x +1 x + 2 x + 3 x + 4   1 1 1 1 1 1 1 1  1 0 0 + − − ⇒y 0 = · + − − . ⇒y =y x+1 x+2 x+3 x+4 6 1 2 3 4 11 1 11 ⇒ y 0 (0) = ⇒ the equation of the tangent line is y − = (x − 0) 72 6 72 1 11 ⇒y = x+ . 72 6 Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 29 2.10 Derivatives of Inverse Trigonometric Functions 2 1 · 2 ⇒ h0 (t) = √ . 1. h(t) = sin−1 (2t) ⇒ h0 (t) = √ 1 − 4t2 1 − 4t2 3. g(x) = tan−1 (2x) ⇒ g 0 (x) = 1 2 0 · 2 ⇒ g (x) = . 1 + 4x2 1 + 4x2 sin t . 5. g(t) = sin t cos−1 t ⇒ g 0 (t) = cos t cos−1 t − √ 1 − t2 √ 7. g(x) = tan−1 ( x) ⇒ g 0 (x) = 1 1 1 · √ ⇒ g 0 (x) = √ . 1+x 2 x 2 x(1 + x) 1 9. f (x) = sin(sin−1 x) ⇒ f 0 (x) = cos(sin−1 x) · √ 1 − x2 −1 cos(sin x) ⇒ f 0 (x) = √ . Note that from precalculus, we can simplify this to f (x) = 1 − x2 sin(sin−1 x) = x ⇒ f 0 (x) = 1. 11. f (x) = tan−1 (tan x). a) f (x) = x ⇒ f 0 (x) = 1. b) f 0 (x) = 13. f (x) = sin −1 sec2 x 1 2 · sec x = = 1. 1 + tan2 x sec2 x  √2  π  √2  √ 1 0 x ⇒ f (x) = √ ⇒f = ⇒f = 2. 2 4 2 1 − x2 0 √ √  2 π √  2 So, the equation of the tangent line at x = is y − = 2 x − . 2 4 2 Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 30 Chapter 3 Graphical Behaviour of Functions 3.1 Extreme Values 7. Absolute maximum is the point C. Absolute minimum is the point A. Relative maximums are B and D. Relative minimum is E. √ x3 ⇒ f 0 (2) = 0. 9. f 0 (x) = 2x( 6 − x2 ) − √ 2 6 − x2 13. f 0 (0) is not defined. 1  1  15 15. f (x) = x2 + x + 4 ⇒ f 0 (x) = 2x + 1 = 0 ⇒ x = − . f − = ; f (−1) = 4; f (2) = 10. 2 2 4 1 The absolute minimum is at x = − and the absolute maximum is at x = 2. 2 π   2π  3√3 3 17. f (x) = 3 sin x ⇒ f = √ ;f = 4 3 2 2   π  π π 3 ⇒ f 0 (x) = 3 cos x = 0 ⇒ x = ⇒ f = 3 (max). And f = √ is min. 2 2 4 2 3 28 ⇒ f (1) = 4; f (5) = . x 5 √ √ √ √ 3 28 f 0 (x) = 1 − 2 = 0 ⇒ x = 3; − 3 ⇒ f ( 3) = 2.732; f (− 3) = −0.732 (min). f (5) = x 5 is max. 19. f (x) = x + 21. f (x) = ex cos x ⇒ f (0) = 1; f (π) = −eπ . π  π f 0 (x) = ex (cos x − sin x) = 0 ⇒ x = ⇒ f = 1.551 (max). And f (π) = −eπ is min. 4 4 ln x ⇒ f (1) = 0; f (4) = 0.347. x 1 − ln x f 0 (x) = = 0 ⇒ x = e. Thus, f (e) = 0.368 (max). And f (1) = 0 is min. x2 23. f (x) = 31 3.3 Increasing and Decreasing Functions 15. f (x) = x3 + 3x2 + 3. a) D = (−∞, +∞). b) f 0 (x) = 3x2 + 6x = 0 ⇒ x = 0, −2. f 0 (x) is defined on (−∞, +∞). So critical numbers are x = 0, −2. c) Put these two x values on signs chart (or numbers line). Pick some test values in the intervals and plug them in the f 0 (x) above to see if it is positive or negative. After doing that we get f 0 (x) > 0 on (−∞, −2) & (0, +∞), on which the function is increasing; and f 0 (x) < 0 on (−2, 0), on which the the function is decreasing. d) f 0 (x) changes from positive to negative at x = −2 so the relative maximum is at x = −2. f 0 (x) changes from negative to positive at x = 0 so the relative minimum is at x = 0. 17. f (x) = x3 − 3x2 + 3x − 1. a) D = (−∞, +∞). b) f 0 (x) = 3x2 − 6x + 3 = 3(x2 − 2x + 1) = 3(x − 1)2 = 0 ⇒ x = 1. f 0 (x) is defined on (−∞, +∞). So, a critical number is x = 1. c) We can see from b) that f 0 (x) > 0 ∀x. So, the function is always increase for all values of x. d) No sign change so there is no maximum and minimum. x2 − 4 19. f (x) = 2 . x −1 a) D = (−∞, −1) ∪ (−1, 1) ∪ (1, +∞). 6x b) f 0 (x) = 2 = 0 ⇒ x = 0. f 0 (x) is undefined at x = −1, 1. Thus, the critical (x − 1)2 numbers are x = −1, 0, 1 c) Put these three x values on signs chart (or numbers line). Pick some test values in the intervals and plug them in the f 0 (x) above to see if it is positive or negative. After doing that we get f 0 (x) < 0 on (−∞, 0), on which the function is decreasing; and f 0 (x) > 0 on (0, +∞), on which the function is increasing. d) f 0 (x) changes from negative to positive at x = 0 so the relative minimum is at x = 0. Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 32 2 (x − 2) 3 . 21. f (x) = x a) D = (−∞, 0) ∪ (0, +∞). b) f 0 (x) = 6−x 1 3x2 (x − 2) 3 = 0 ⇒ x = 6. f 0 (x) is undefined at x = 0, 2. So the critical numbers are x = 0, 2, 6. c) Put these three x values on signs chart (or numbers line). Pick some test values in the intervals and plug them in the f 0 (x) above to see if it is positive or negative. After doing that we get f 0 (x) < 0 on (−∞, 2) & (6, +∞), on which the function is decreasing; and f 0 (x) > 0 on (2, 6), on which the function is increasing. d) f 0 (x) changes from negative to positive at x = 2 so the relative minimum is at x = 2. f 0 (x) changes from positive to negative at x = 6 so the relative maximum is at x = 6. 23. f (x) = x5 − 5x. a) D = (−∞, +∞). b) f 0 (x) = 5x4 − 5 = 5(x − 1)(x + 1)(x2 + 1) = 0 ⇒ x = −1, 1. f 0 (x) is defined ∀x. So the critical points are at x = −1, 1. c) Put these two x values on signs chart (or numbers line). Pick some test values in the intervals and plug them in the f 0 (x) above to see if it is positive or negative. After doing that we get f 0 (x) > 0 on (−∞, −1) & (1, +∞), on which the function is increasing; and f 0 (x) < 0 on (−1, 1), on which the function is decreasing. d) f 0 (x) changes from negative to positive at x = 1 so the relative minimum is at x = 1. f 0 (x) changes from positive to negative at x = −1 so the relative maximum is at x = −1. Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 33 3.4 Concavity and Second Derivative 17. f (x) = −x2 − 5x + 7. 5 a) f 0 (x) = −2x − 5 = 0 ⇒ x = − . 2 f 00 (x) = −2 < 0 ∀x, thus there is no inflection point. b) f 00 (x) = −2 < 0 ∀x, thus f (x) is concave down everywhere. 19. f (x) = 2x3 − 3x2 + 9x + 5. a) f 0 (x) = 6x2 − 6x + 9 = 3(2x2 − 2x + 3) > 0 ∀x. 1 f 00 (x) = 12x − 6 = 0 ⇒ x = . f 00 (x) is defined everywhere. Thus the possible inflection 2 1 point is at x = . 2 b) Put this x value on signs chart (or numbers line). Pick some test values in the intervals and plug them in the f 00 (x) above to see if it is positive or negative. After doing that we get 1  1 , +∞ . Note and f 00 (x) > 0 (concave up) on 2 2 1 that since the concavity changes direction at x = , we can say that the inflection point 2 1 is at x = . 2 f 00 (x) < 0 (concave down) on  − ∞, 21. f (x) = −3x4 + 8x3 + 6x2 − 24x + 2. a) f 0 (x) = −12x3 + 24x2 + 12x − 24 = −12(x + 1)(x − 1)(x − 2). √ √ 2 − 7 2 + 7 , . f 00 (x) is defined everywhere. f 00 (x) = −36x2 + 48x + 12 = 0 ⇒ x = 3 √ 3 √ 2− 7 2+ 7 Thus, the possible inflection points are x = , . 3 3 b) Put these x values on signs chart (or numbers line). Pick some test values in the intervals and plug them in the f 00 (x) above to see if it is positive or negative. After doing that we get √  2 + √7  2 − 7 f (x) < 0 (concave down) on −∞, & , +∞ , and f 00 (x) > 0 (concave 3 3  2 − √7 2 + √7  up) on , . Note that since the concavity changes direction at these two 3 3 √ √ 2− 7 2+ 7 points, we can say that the inflection points are at x = , . 3 3 00 Math 1174  Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 34 23. f (x) = 1 . x2 + 1 2(x2 − 13 ) −2x 1 1 00 , and then f (x) = = 0 ⇒ x = − √ , √ . f 00 (x) is defined 2 2 2 3 (x + 1) (x + 1) 3 3 1 1 everywhere because x2 + 1 > 0 ∀x. Thus, the possible inflection points are x = − √ , √ . 3 3 a) f 0 (x) = b) Put these x values on signs chart (or numbers line). Pick some test values in the intervals and plug them in the f 00 (x) above to see if it is positive or negative. After doing that we get 00   1  1  √ √ & − ∞, − , +∞ and f 00 (x) < 0 (concave down) 3 3 f (x) > 0 (concave up) on  1 1  √ on − , √ . Note that since the concavity changes direction at these two points, 3 3 1 1 we can say that the inflection points are at x = − √ , √ . 3 3 25. f (x) = sin x + cos x. π 3π a) f 0 (x) = cos x − sin x, and then f 00 (x) = − sin x − cos x = 0 ⇒ x = − , . f 00 (x) is 4 4 π 3π defined everywhere. Thus, f (x) has possible inflection points are at x = − , . 4 4 b) Put these x values on signs chart (or numbers line). Pick some test values in the intervals and plug them in the f 00 (x) above to see if it is positive or negative. After doing that we get  π 3π   π   3π  f 00 (x) < 0 (concave down) on − , and f 00 (x) > 0 (concave up) on −π, − & ,π . 4 4 4 4 Note that since the concavity changes direction at these two points, we can say that the π 3π inflection points are at x = − , . 4 4 27. f (x) = x2 ln x. a) f 0 (x) = 2x ln x + x, and then f 00 (x) = 2 ln x + 2 + 1 = 2 ln x + 3 = 0 ⇒ x = e−1.5 . f 00 (x) is defined on (0, +∞). Thus, the possible inflection point is at x = e−1.5 . b) Put this x value on signs chart (or numbers line). Pick some test values in the intervals and plug them in the f 00 (x) above to see if it is positive or negative. After doing that we get f 00 (x) < 0 (concave down) on (0, e−1.5 ) and f 00 (x) > 0 (concave up) on (e−1.5 , +∞). Note that since the concavity changes direction at these two points, we can say that the inflection points are at x = e−1.5 . Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 35 31. f (x) = x3 − x + 1.   −1 1 00 00 −1 f (x) = 3x − 1 = 0 ⇒ x = √ , √ , and then f (x) = 6x ⇒ f √ < 0. Thus, f (x) has a 3 3  3 1  −1 1 00 local maximum at x = √ . Also, f √ > 0. Thus, f (x) has a local minimum at x = √ . 3 3 3 0 2 35. f (x) = x4 − 4x3 + 6x2 − 4x + 1. f 0 (x) = 4x3 − 12x2 + 12x − 4 = 4(x − 1)3 = 0 ⇒ x = 1, and then f 00 (x) = 12x2 − 24x + 12 = 12(x − 1)2 ≥ 0 ⇒ f 00 (1) = 0. Thus, f (x) has a local minimum at x = 1. 37. f (x) = x x2 − 1 f 0 (x) = − . x2 + 1 < 0 ∀x 6= −1, 1. Thus, there is no maximum or minimum value. (x2 − 1)2 39. f (x) = x2 ex . f 0 (x) = 2xex +x2 ex = xex (x+2) = 0 ⇒ x = −2, 0, and then f 00 (x) = 2ex +2xex +2xex +x2 ex = ex (x2 + 4x + 2). f 00 (0) > 0; f 00 (−2) < 0. Thus, f (x) has a local minimum at x = 0 and a local maximum at x = −2. 2 41. f (x) = e−x . 2 2 2 f 0 (x) = −2xe−x = 0 ⇒ x = 0, and then f 00 (x) = −2e−x + 4x2 e−x . f 00 (0) < 0. Thus, f (x) has a local maximum at x = 0. 43. f (x) = −x2 − 5x + 7. 5 f 0 (x) = −2x − 5 = 0 ⇒ x = − , and then f 00 (x) = −2 < 0 ∀x. Thus, f 0 (x) does not have 2 maximum or minimum. 45. f (x) = 2x3 − 3x2 + 9x + 5.  3 1 f 0 (x) = 6x2 − 6x + 9 = 6 x2 − x + > 0 ∀x, and then f 00 (x) = 12x − 6 = 0 ⇒ x = . 2 2 1 000 0 f (x) = 12 > 0. Thus, f (x) has a minimum at x = . 2 53. f (x) = x2 ln x. f 0 (x) = 2x ln x + x, and then f 00 (x) = 2 ln x + 3 = 0 ⇒ x = e−1.5 . f 000 (x) = Thus, f 0 (x) has a minimum at x = e−1.5 . Math 1174 Minh Phuong Bui & Pichmony Anhaouy 2 ⇒ f 000 (e−1.5 ) > 0. x Solutions 1.0 36 Chapter 4 Applications of the Derivatives 4.1 Related Rates Problems 3. Here, we want to find dr . dt 10 mm r V = πr2 h ⇒ dV dV = 2πrh ⇒ dr = dr 2πrh 5 = 7.96 (mm) 2π(1)0.1 5 b) dr = 1 = 0.796 (mm) 2π(10)0.1 5 c) dr = 1 = 0.0796 (mm) 2π(100)0.1 a) dr = 5. Diagram is similar to the one in question 4, but instead of East we have West. 1 1 1 A = ;B = ⇒ C = √ . 2 2 2 dA dC = −50; = −80. dt dt A2 + B 2 = C 2 ⇒ 2A dB dC dA + 2B = 2C . dt dt dt C dC − A dA dB dt dt ⇒ = = −63.14 (mph). dt B 37 7. Here, we want to find dθ . dt A 10, 000 f t G ✓ x V = πr2 h ⇒ dV dV = 2πrh ⇒ dr = dr 2πrh 5 = 7.96 (mm) 2π(1)0.1 5 b) dr = 1 = 0.796 (mm) 2π(10)0.1 5 1 = 0.0796 (mm) c) dr = 2π(100)0.1 a) dr = 9. Here, we want to find dy . dt y 24 f t 1 f t/s x x2 + y 2 = 242 ⇒ 2x √ dy dy x dx dx + 2y =0⇒ =− . dt dt dt y dt dy 1 =− · 1 = −0.0417 (ft/s). dt 23.98 √ dy 10 =− · 1 = −0.458 (ft/s). b) y = 242 − 102 = 21.82 ⇒ dt 21.82 √ dy 23 c) y = 242 − 232 = 6.86 ⇒ =− · 1 = −3.353 (ft/s). dt 6.86 dy x = lim − √ · 1 = −∞. So it does not exist. d) dt x→24 242 − x2 a) y = Math 1174 242 − 12 = 23.98 ⇒ Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 38 11. x 1 = ⇒ y = 4x. y 4 1 1 y2 1 dV 1 dy V = πx2 y = πy = πy 3 ⇒ = πy 2 . 3 3 16 48 dt 16 dt 10 f t x 20 f t y dy 16 · 10 = = 50.93 (ft/min). dt π(1)2 16 · 10 dy = = 0.5093 (ft/min). b) dt π(10)2 dy 16 · 10 c) = = 0.141 (ft/min). dt π(19)2 a) 523.6 1 = 52.36 (mins). V = π52 · 20 = 523.6 ⇒ t = 3 10 13. Solution here z y x a) z 2 = x2 + y 2 = 302 + 402 = 2500 ⇒ z = 50 ⇒ L = 30 + 50 = 80 (ft). √ √ b) x = 50 ⇒ z = 502 + 302 = 10 34. dx dz dz 50 · 2 302 + x2 = z 2 ⇒ 0 + x =z ⇒ = √ = 1.715. dt dt dt 10 34 √ √ c) x = 80 ⇒ z = 802 + 302 = 10 73. dx dz dz 80 · 2 302 + x2 = z 2 ⇒ 0 + x =z ⇒ = √ = 1.873. dt dt dt 10 73 √ d) z = 80 ⇒ x = 802 − 302 = 74.16. The man needs to walk 74.16 − 40 = 34.16 (ft). Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 39 2 4 3 15. h = d = r ⇒ r = h. 3 3 4 x 2 3D y D 1 9 3 1 V = πr2 h = π h3 = πh3 . 3 3 16 16 dh dV 9 dh = πh2 ⇒ = 0.00314 (ft/s). dt 16 dt dt Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 40 4.2 Applied Optimization Problems 3. Let P be the product of two numbers x and y, such that x + y = 100. Here, we want to maximize P and find it value. y = 100 − x ⇒ P = xy = x(100 − x) = −x2 + 100x. Then, P 0 (x) = −2x + 100 = 0 ⇒ x = 50. P 00 (x) = −2 < 0 ⇒ P (x) has a maximum at x = 50 ⇒ y = 50. 5. Let S be the sum of two numbers x and y, such that x · y = 500. Here, we want to maximize S and find it value. √ √ 500 500 500 . Then, S 0 (x) = 1− 2 = 0 ⇒ x2 = 500 ⇒ x = −10 5, 10 5. ⇒ S = x+y = x+ x x x √ √ √ 1000 00 00 00 S (x) = 3 ⇒ S (−10 5) < 0 and S (10 5) > 0. Thus, S(x) has maximum at x = −10 5. x y= 7. Let A be the area of the right triangle below. Here, we want to maximize A and find it value. 1 y x √ x2 + y 2 = 1 ⇒ y = 1 − x2 . 1√ x2 1 1 1 √ 0 2 2 √ 1−x − =0⇒x= √ . A = xy = x 1 − x . Then, A (x) = 2 2 2 2 2 1−x 2 1 A0 (x) changes the sign from positive to negative at x = √ . Thus, the maximum area happens 2 1 if x = y = √ . 2 9. Here, we want to find r, h. r h Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 41 V = πr2 h ⇒ h = V V 710 + 2πr2 . . Also, A = 2πrh + 2πr2 = 2πr 2 + 2πr2 = 2 πr πr r a) Minimize the amount of material M needed to produce the can. 710 A0 (r) = − 2 + 4πr = 0 ⇒ r = 3.837 ⇒ h = 7.596. A0 (r) changes the sign from negative r to positive at r = 3.837, h = 7.596. Thus, the minimum happens if r = 3.837, A = M = 277.545. b) Minimize the costs C needed to produce the can. Let P be price unit for each cm2 of the material. C = A · P . Thus, the standard can is produced to minimize the cost. 11. Let V be the volume of the box below. Here, we want to maximize V when w = h. h w ` ` + 2w + 2h = 108 ⇒ ` + 4w = 108 ⇒ ` = 108 − 4w. We have V = ` · w · h = (108 − 4w)ww = 108w2 − 4w3 . Then, V 0 (w) = 216w − 12w2 = 0 ⇒ w = 18 ⇒ h = 18 ⇒ ` = 36. The sign of V 0 (w) changes from positive to negative at w = 18. Thus, V has a maximum at w = 18. 13. Let C be the cost to lay to power line underground. Here, we want find x to minimize C. 2 x C = 50x + 80 5 x p √ 4 + (5 − x)2 = 50x + 80 x2 − 10x + 29. 40(2x − 10) = 0 ⇒ x = 6.6 (invalid because we need x < 5, and x = 3.4. So x2 − 10x + 29 the cost is minimum at x = 3.4. C 0 (x) = 50 + √ Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 42 15. Let T be the time the dog takes to get the stick. Here, we want to find x to minimize T . p 152 + (20 − x)2 x y x T = + = + . Then, we have 22 1.5 22 1.5 1 2x − 40 + √ = 0 ⇒ x = 19.4. So, the dog needs to run 19.4 feet along the T 0 (x) = 2 22 3 x − 40x + 625 shore to minimize the time. y x 20 15 x 17. Let A be the area of the rectangle inscribed inside the unit circle (r = 1), depicted in the following diagram. Here, we want to find w and ` to maximize A. 1 w ` √ w2 + `2 = (2r)2 = 4 ⇒ ` = 4 − w2 . √ 1 1 √ 1 √ w2  A = ` · w = w( 4 − w2 ). Then, A0 (w) = 4 − w2 − √ =0⇒w= 2⇒`= 2 2 4 − w2 √ 2 2 ⇒ A = 2. Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 43 4.3 Economics and Business Applications 1. x = 500 − 10p ⇒ x0 = −10. a) E(p) = − 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 3. x2 + 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. c) Now, we can solve for E in term of x from (a). Here, we get E = 0 < x < 1.42 and inelastic on 1.42 < x < 2. x2 − 4 . So, elastic on x2 5 x 2 √ 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 ⇒ x0 2 = − 0.02. x p 100 − p . Currently, at p = 200, |E| = 2 > 1, the demand is elastic. a) Here, we have E = 50 Therefore, we should increase the revenue. b) When |E| = 1 ⇒ 100 − p = 50 ⇒ p = 150. 9. V (t) = 100(60 + t2 ) ⇒ V 0 (t) = 100(0 + 2t) = 200t. V0 2t 2(60 − t2 ) 0 ⇒ 0.0625 = ⇒ t = 2, 30. Now, we check r (t) = . Since r0 (2) > 0 2 2 2 V 60 + t (60 + t ) and r0 (30) < 0, the present value is a minimum at t = 2 (good to buy), and is maximum at t = 30, when the asset should be sold. r= Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 44 11. V (t) = 250(5 + 0.2t)3/2 ⇒ V 0 (t) = 75(5 + 0.2t)1/2 . V0 75(5 + 0.2t)1/2 3 ⇒ 0.04 = = ⇒ t = 12.5. Sell in about 12.5 years. V 250(5 + 0.2t)3/2 10(5 + 0.2t) V (12.5) = $5134.90. r = 13. V (t) = 50te−0.10t ⇒ V 0 (t) = 50e−0.10t (1 − 0.10t). 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. 50e−0.10t (1 − 0.10t) V 0 (t) ⇒ 0.025 = b) r = = 1 − 0.10t ⇒ t = 8 years. V (t) 50te−0.10t 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 x x ·s= · (4.50) = . Cs (x) = · r = · 5.76 = 2.88x. The total x x x 2 2 45, 000 45, 000 0 cost function is then C(x) = + 2.88x. C (x) = − + 2.88 = 0 ⇒ x = 125. Since x x2 90, 000 C 00 (x) = > 0 for all x > 0, C(x) has a min. at x = 125. The number of times that the x3 10, 000 merchant should order is = 80 in order minimize the total cost. The minimum total 125 45, 000 cost is C(125) = + 2.88(125) = 720. 125 17. Here C0 (x) = Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 45 4.4 Linear Approximation and Differentials 7. f (x) = x2 ⇒ f 0 (x) = 2x. Using linear approximation (or tangent line approximation), we take a = 6 ⇒ f 0 (a) = 12 ⇒ f (a) = 36 ⇒ y − 36 = 12(x − 6). ⇒ y = 12x − 36 ⇒ 5.932 ≈ 12(5.93) − 36 = 35.16. Or we can use differential form of linear approximation as follows. Here, dx = 5.93 − 6 = −0.07. dy = 12 dx = 12 · (−0.07) = −0.84 ⇒ 5.932 ≈ 62 − 0.84 = 35.16. 9. f (x) = x3 ⇒ f 0 (x) = 3x2 . Here, dx = 6.8 − 7 = −0.2. Take a = 7 ⇒ dy = 3(7)2 dx = 147 · (−0.2) = −29.4 ⇒ 6.83 = 73 − 29.4 ≈ 313.6. 11. f (x) = √ 1 x ⇒ f 0 (x) = √ . Here, dx = 24 − 25 = −1. 2 x Take a = 25 ⇒ dy = √ √ 1 dx = −0.1 ⇒ 24 ≈ 25 − 0.1 = 4.9. 2·5 √ 1 1 2 13. f (x) = 3 x = x 3 ⇒ f 0 (x) = x− 3 . Here, dx = 8.5 − 8 = 0.5. 3 √ 1 1 −2 1 1 3 Take a = 8 ⇒ dy = 8 3 dx = ⇒ 8.5 3 ≈ 8 + ≈ 2.083. 3 12 12 15. f (x) = cos x ⇒ f 0 (x) = − sin x. π π π ≈ 1.571, we take a = ⇒ dx = 1.5 − = −0.071 2 2 2 π ⇒ dy = − sin dx = (−1) · (−0.071) = 0.071 2 π ⇒ cos 1.5 ≈ cos + 0.071 = 0.071. 2 Since Note: These problems above can be done using the tangent line approximation as done in #7 as well. The answers will be the same. Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 46 17. y = x2 + 3x − 5 ⇒ dy = (2x + 3) dx. 19. y = 1 1 −3 x dx. ⇒ dy = − 4x2 2 21. y = x2 e3x ⇒ dy = (2x · e3x + x2 · e3x · 3)dx = x(2 + 3x)e3x dx. 23. y = 2x 2(tan x + 1) + 2x sec2 x ⇒ dy = dx. tan x + 1 (tan x + 1)2 25. y = ex sin x ⇒ dy = (ex sin x + ex cos x) dx = (sin x + cos x)ex dx. 27. y = 1 · (x + 2) − (x + 1) · 1 1 x+1 ⇒ dy = dx = dx. x+2 (x + 2)2 (x + 2)2 29. y = x ln x − x ⇒ dy = (1 · ln x + x · Math 1174 1 − 1) dx = ln x dx. x Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 47 4.5 L’Hopital’s Rule 1 sin x − cos x cos x + sin x = limπ = −√ . x→ 4 x→ 4 cos 2x −2 sin 2x 2 11. limπ sin 2x sin 0 = = 0. x→0 x + 2 0+2 13. lim a cos ax a sin ax = lim = . x→0 b cos bx x→0 sin bx b 15 lim 17. lim+ x→0 ex − x − 1 ex 1 ex − 1 = lim = . = lim 2 + + x→0 2 x→0 x 2x 2 √ ex 21. lim √ = lim 2ex x = +∞. x→+∞ x x→+∞ 3x2 + 8x + 4 6x + 8 x3 + 4x2 + 4x = lim = lim = −2. 25. lim 3 x→−2 3x2 + 14x + 16 x→−2 6x + 14 x→−2 x + 7x2 + 16x + 12 ln x2 2 = lim = 0. x→∞ x x→∞ x 27. lim 29. lim+ x · ln x = lim+ x→0 x→0 ln x 1 x = lim+ x→0 1 x − x12 = lim+ (−x) = 0. x→0 1 −2x−2 1 −1 −2x−3 2 x2 x = lim+ 35. lim+ 2 e = lim+ 1 = lim+ 1 1 −2 = lim 1 = 0. x + −2 x→0 −e x x→0 x x→0 −e x x x→0 e x x→0 e x 37. Let y = lim+ (2x)x ⇒ ln y = lim+ ln(2x)x = lim+ x ln 2x = lim+ x→0 = lim+ x→0 2 x − x12 x→0 x→0 ln 2x x→0 1 x = lim+ (−2x) = 0 ⇒ lim+ (2x)x = e0 = 1. x→0 x→0 x lim+ ln(2x)x Or: We write lim+ (2x) = e x→0 . Now, we compute x→0 lim x ln 2x = lim+ x→0+ Math 1174 x→0 ln 2x 1 x = lim+ x→0 2 x − x12 = lim+ (−2x) = 0 ⇒ lim+ (2x)x = e0 = 1. x→0 x→0 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 48 1 1 1 1 ln x = lim x = 0 x→∞ 1 x→∞ x 41. Let y = lim x x ⇒ ln y = lim ln x x = lim x→∞ x→∞ 1 ⇒ lim x x = e0 = 1. x→∞ 43. Let y = lim+ (ln x1−x ) ⇒ ln y = lim+ ln(ln x1−x ) = lim+ (1 − x) ln(ln x) x→1 = lim+ x→1 x→1 ln(ln x) 1 1−x 1 = lim+ x ln1 x = lim+ x→1 (1−x)2 x→1 x→1 −2(1 − x) (1 − x)2 = lim+ =0 x→1 x ln x ln x + 1 ⇒ lim+ (ln x1−x ) = e0 = 1. x→1 1 2 45. We write lim (1 + x ) x→∞ 1 x lim ln[(1 + x2 ) x ] = e x→∞ . Now, we compute 2x 1 2 2 lim ln[(1 + x ) ] = lim ln(1 + x2 ) = lim 1+x = lim = 0. x→∞ x→∞ x x→∞ 1 x→∞ 2x 2 1 x 1 So, we get lim (1 + x2 ) x = e0 = 1. x→∞ 49. lim+ x→3 5 x 5 − x(x + 3) −x2 − 3x + 5 − = lim = lim = −∞. x2 − 9 x − 3 x→3+ (x − 3)(x + 3) x→3+ (x − 3)(x + 3) 6 (ln x)3 3(ln x)2 6 ln x = lim = lim = lim x = 0. x→∞ x→∞ x→∞ x→∞ 1 x x x 51. lim Math 1174 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 49 4.6 Antiderivatives and Indefinite Integration Z 9. 1 x8 dx = x9 + C. 9 Z 11. dt = t + C. Z 1 1 dt = − + C. 2 3t 3t Z √ 1 √ dx = 2 x + C. x 13. 15. Z 17. Z 19. Z 21. Z 23. Z 25. sin θ dθ = − cos θ + C. 5eθ dθ = 5eθ + C. 5t 5t dt = + C. 2 2 ln 5 2 3 (t + 3)(t − 2t) dt = Z 1 1 (t5 + t3 − 6t) dt = t6 + t4 − 3t2 + C. 6 4 eπ dx = eπ x + C. Z 29. f (x) = 5ex dx = 5ex + C. f (0) = 10 ⇒ 5 + C = 10 ⇒ C = 5 ⇒ f (x) = 5ex + 5. Z 31. f (x) = f π  4 Math 1174 sec2 x dx = tan x + C. = 5 ⇒ 1 + C = 5 ⇒ C = 4 ⇒ f (x) = tan x + 4. Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 50 0 Z 33. f (x) = 5 dx = 5x + C. f 0 (0) = 7 ⇒ C = 7 ⇒ f 0 (x) = 5x + 7. Z f (x) = 5 (5x + 7) dx = x2 + 7x + D. 2 5 f (0) = 3 ⇒ D = 3 ⇒ f (x) = x2 + 7x + 3. 2 0 Z 35. f (x) = 5ex dx = 5ex + C. f 0 (0) = 3 ⇒ C = −2 ⇒ f 0 (x) = 5ex − 2. Z f (x) = (5ex − 2) dx = 5ex − 2x + D. f (0) = 5 ⇒ D = 0 ⇒ f (x) = 5ex − 2x. 0 37. f (x) = Z 2x (24x + 2 − cos x) dx = 8x + − sin x + C. ln 2 2 x f 0 (0) = 5 ⇒ C = − Z  f (x) = 3 1 2x 1 ⇒ f 0 (x) = 8x3 + − sin x − . ln 2 ln 2 ln 2 1 2x − sin x − 8x + ln 2 ln 2 3 f (0) = 0 ⇒ D = −  dx = 2x4 + 1 − 1. (ln 2)2 ⇒ f (x) = 2x4 + 2x (ln 2)2 + cos x − x ln 2 − Math 1174 x 2x + cos x − + D. 2 (ln 2) ln 2 1 − 1. (ln 2)2 Minh Phuong Bui & Pichmony Anhaouy Solutions 1.0 51