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Homework Solution Chapter 2

# Homework Solution Chapter 2 - C H A P T E R 2...

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Unformatted text preview: C H A P T E R 2 Differentiation Section 2.1 Section 2.2 Section 2.3 Section 2.4 Section 2.5 Section 2.6 The Derivative and the Tangent Line Problem . . . . . 95 Basic Differentiation Rules and Rates of Change . . 109 Product and Quotient Rules and Higher-Order Derivatives . . . . . . . . . . . . . . . 120 The Chain Rule . . . . . . . . . . . . . . . . . . . . 134 Implicit Differentiation . . . . . . . . . . . . . . . . 147 Related Rates . . . . . . . . . . . . . . . . . . . . . 161 . . . . . . . . . . . . . . . . . . . . . . . . . . 173 Review Exercises Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 C H A P T E R Differentiation Section 2.1 2 The Derivative and the Tangent Line Problem 0. 5 2. 5 2. 1. (a) At x1, y1 , slope At x2, y2 , slope (b) At x1, y1 , slope At x2, y2 , slope 2. (a) At x1, y1 , slope At x2, y2 , slope (b) At x1, y1 , slope At x2, y2 , slope f 4 4 f 4 4 f 1 1 f 3 3 f 4 4 5 3 5 2 3. 2 5. 4 3. 5 4. 2. 3. (a), (b) y y 6 5 4 3 2 1 f )4) 4 f )1) )x 1 4. (a) 1) 2 4.75 1 1 0.25 f 3 . 3 f )1) x 1 f )4) 5 )4, 5) f )4) f )1) )1, 2) x 1 2 3 4 5 6 f )1) 3 Thus, f 1 f 4 > 1 4 2 (b) The slope of the tangent line at 1, 2 equals f 1 . This slope is steeper than the slope of the line through 1, 2 and 4, 5 . Thus, f 1) f 4 4 f 1 < f 1. 1 (c) y f 4 4 3 x 3 1x x 1 f 1 x 1 1 1 2 2 1 5. f x 3 2x is a line. Slope g1 1 1 2 x x x 2 2 g1 4 x x x x t t t t f 0 2 6. g x 3 2x 1 is a line. Slope g2 5 5 4 h 2 4 4 2 2 x x 3 2 7. Slope at 1, 3 lim lim lim x0 8. Slope at 2, 1 3 lim lim g2 x 2 x0 1 x 2 x0 x0 x 4 4 x t t t 4 t t t 4 2 x 2 2 1 x0 lim x x 4 h 3 t 2 1 x0 lim 2 x0 lim x0 9. Slope at 0, 0 lim lim f 0 3 t t t0 10. Slope at 0 2, 7 lim lim 2 7 4 t0 t0 t0 t lim 3 t0 3 lim lim t0 t0 95 96 Chapter 2 3 lim lim f x 3 x Differentiation 12. g x x x 3 0 f x g x 5 lim gx 5 x 0 x 0 x x 5 gx 13. f x f x 5x lim f x 5x 5 5 x0 11. f x f x x0 x0 x x x x f x 5x x0 lim x0 lim lim x0 lim 0 x0 lim x0 x0 14. f x f x 3x lim 2 f x 3x x x x f x 15. h s h s 2 x 3x 2 3 lim 2 s 3 hs 3 2 3 2 3 s0 x0 s s s 2 3 hs s s 3 2 3s lim x0 lim s0 3 x lim x0 x lim 3 x0 lim 3 s0 s s 16. f x f x 9 lim 1 x 2 f x 9 1 2 x f x 2x 2x 2 4x x 1 x x x 2 x0 x x 1 2 x 1 2 f x x x 9 1 2x lim x0 lim x0 17. f x f x 2x2 lim f x x 2 2 x0 lim x x x x 2 1 x x 2x 2 x 1 x 1 2x 2 x 1 x0 lim lim 4x x 2 x x x0 x0 lim 4x x0 2 x 1 4x 1 18. f x f x 1 lim x2 f x x x x x2 2x x x x f x 2 x0 lim 1 1 1 x x 2 x2 1 x2 x0 x 2x x x 2 lim lim x0 x0 lim x0 2x x 2x Section 2.1 19. f x f x x3 lim x0 The Derivative and the Tangent Line Problem 97 12x f x x x3 x x x 3 f x 12 x x 3x x 2 lim lim lim x x 2 x3 x 3 12x 12x 12 x x3 12x x0 3x 2 x 3x 3x x x0 x 3x 2 x x x x 2 3 12 x 3x 2 12 x0 lim 3x 2 x0 12 20. f x f x x3 lim x0 x2 f x x x3 x x x 3 f x x x 3x x 2 lim lim x x 2 x3 x 3 x2 x2 x 2x x x 3x 2 2 x0 3x 2 x 3x 3x x 2 x 2 x3 x2 x0 lim 3x 2 x x x x 2 3 2x x x x0 lim 3x 2 x0 2x 2x 21. f x f x 1 x lim 1 f x 1 x 1 xx xx x 1 x 1 2 22. f x x x 1 x x x x x x 1 x 1 1 1 1 x f x 1 x x 1 x 1 1 1 f x 1 x2 lim f x 1 lim lim lim lim x x0 x0 x0 x x x2 x f x 1 x2 lim lim lim lim x x x0 x0 x2 x xx 2x x xx 2x x x2 x 2x2 x2 x 2x2 x0 x0 x0 x0 x x 2x2 x0 2x x4 2 x3 23. f x f x x lim lim lim lim x0 1 f x x x x x x x x x x x x 1 1 f x 1 x 1 1 x x x x 1 1 1 1 1 x 1 x 1 1 2 x 1 x x x x 1 1 x x 1 1 x0 x0 x0 98 Chapter 2 4 x lim f x x0 Differentiation 24. f x f x 25. (a) f x x x 4 x x f x 4 x x x 4x x x x x 4 x 2 x x (b) x x x x x x x x x x x f x x2 lim lim lim 1 f x x 2x x x x 2x 22 4. x x x 2 f x 1 x x2 1 x0 lim lim lim lim x x0 x0 x 2 x0 4 x 4 x x0 x x x x0 lim 2x x0 4x x x x x x 4 x At 2, 5 , the slope of the tangent line is m The equation of the tangent line is y y 5 5 y 4x 4x 4x 8 2 8 3. x0 x x (2, 5) -5 -2 5 26. (a) f x f x x2 lim x0 2x f x x 2x x 1 x x x 2 (b) f x 2x x x x 2 5 (-3, 4) lim x x 2 x 2x 2 1 x2 2x 1 -6 -1 3 x0 lim x0 lim 2x x0 2 3, 4 , the slope of the tangent line is m At The equation of the tangent line is y 4 y 4x 4x 3 8. 2 3 2 4. 27. (a) f x f x x3 lim f x x 3x 2 x 3x 2 x0 (b) x x x3 x 3x x 3x x x 2 10 (2, 8) f x -5 5 -4 lim lim x3 x 2 x0 x 3 x0 lim x0 3x 2 32 2 At 2, 8 , the slope of the tangent is m The equation of the tangent line is y 8 y 12 x 12x 2 16. 12. Section 2.1 28. (a) f x f x x3 lim x0 The Derivative and the Tangent Line Problem (b) 4 99 1 f x x x3 x x x3 3x 2 3x x x f x (1, 2) -6 6 lim lim x0 1 x 3x x3 x x 2 2 1 -4 x x 3x 2 3 1 x3 1 x0 x0 lim 3x 2 At 1, 2 , the slope of the tangent line is m The equation of the tangent line is y 2 y 3x 3x 1 1. 31 2 3. 29. (a) f x f x x lim lim lim f x x x x x x 1 x x0 (b) x x x x x x x f x -1 3 (1, 1) 5 -1 x x x x0 x x x x x x x0 lim 1 2 x x0 At 1, 1 , the slope of the tangent line is m 1 2 1 1 x 2 1 x 2 x lim lim lim x0 1 . 2 The equation of the tangent line is y 1 y 1 1 . 2 (b) x x x x x x x x x x 1 1 1 x 1 1 x f x -2 10 4 30. (a) f x f x 1 f x (5, 2) x x x 1 1 x 1 1 x0 x x x x 1 1 x x 1 1 -4 x0 lim x0 1 2 x 1 At 5, 2 , the slope of the tangent line is 1 1 2 5 1 4 The equation of the tangent line is m y 2 y 1 x 4 1 x 4 5 3 4 100 Chapter 2 4 x lim Differentiation 31. (a) f x f x x (b) f x x x x f x 4 x x xx x3 x2 x2 xx 4 1 x x 2x 2 x x x x x x x x x x 4 x2 4 2 10 (4, 5) -12 12 x0 x lim lim lim lim lim x2 x2 x0 x x x x 4 x x x 4x 4 x x -6 x0 4x x 2 x x x x x3 x x2 x0 x x2 x x x 4 x x x0 x0 At 4, 5 , the slope of the tangent line is m 1 4 16 3 . 4 The equation of the tangent line is y 5 y 3 x 4 3 x 4 1 x lim 1 f x 1 x 1 xx x 1 x 1 2 4 2. 32. (a) f x f x 33. From Exercise 27 we know that f x slope of the given line is 3, we have x x 1 x x x 1 x 1 x 1 f x 1 x x 1 x 1 1 1 3x 2 x 3 1. 3x2. Since the x0 lim lim lim x x x0 1, 1 the tangent Therefore, at the points 1, 1 and lines are parallel to 3x y 1 0. These lines have equations y 1 y 3x 3x 1 2 and y 1 y 3x 3x 1 2. x0 x0 At 0, 1 , the slope of the tangent line is m 0 1 1 2 1. x 1. The equation of the tangent line is y (b) (0, 1) -6 3 3 -3 Section 2.1 34. Using the limit definition of derivative, f x 3x 2 x 2 The Derivative and the Tangent Line Problem 101 3x 2. Since the slope of the given line is 3, we have 3 1x 1. Therefore, at the points 1, 3 and y 3 y 3x 3x 1 and y 1, 1 the tangent lines are parallel to 3x 1 y 3x 3x 1 4. y 4 0. These lines have equations 35. Using the limit definition of derivative, f x 1 2x x . 1 2, 36. Using the limit definition of derivative, f x we have 2x 1 1 3 2. Since the slope of the given line is 1 2x x x 1. 1 2 Since the slope of the given line is 2x 1 1 32 1 2, we have 1 2 x x 1 32 1 1 Therefore, at the point 1, 1 the tangent line is parallel to x 2y 6 0. The equation of this line is y y 1 1 y 1 x 2 1 x 2 1 x 2 1 1 2 3 . 2 1 Matches (b). 1 x 2. At the point 2, 1 , the tangent line is parallel to x 2y 7 0. The equation of the tangent line is y 1 y 1 x 2 1 x 2 2 2. 37. f x 39. f x x f x x f x 38. f x x2 f x 2x Matches (d). Matches (a). 40. f does not exist at x 0. Matches (c). decreasing slope as x 41. g 5 g 5 2 because the tangent line passes through 5, 2 . 2 5 0 9 2 4 1 2 44. The slope of the graph of f is 0 f x 0. y y 4 3 2 1 2 4 42. h 1 h 4 because the tangent line passes through 1 6 3 4 1 2 4 1 2 1, 4 . 43. The slope of the graph of f is 1 f x 1. y 45. The slope of the graph of f is negative for x < 4, positive for x > 4, and 0 at x 4. f x f x -2 -1 -1 -2 1 1 2 3 2 f 2 x -6 -4 -2 -3 -2 -1 -2 -4 -6 -8 2 4 6 -1 -2 102 Chapter 2 Differentiation 1 47. Answers will vary. Sample answer: y y 4 3 2 4 3 2 1 x 1 2 3 4 -4 -3 -2 -2 -3 -4 x 1 2 3 4 46. The slope of the graph of f is for x < 4, 1 for x > 4, and undefined at x 4. y 3 2 1 x 1 2 3 4 5 6 x 48. Answers will vary. Sample answer: y y x f 1 -4 -3 -2 -1 -1 -2 -3 -4 49. f x 53. f 0 f x 5 3x and c 1 3, 50. f x x 3 and c 2 51. f x x 2 and c 6 52. f x 2 x and c 0; f x > 0 if 9 2 and f x < x < 54. f 0 4, f 0 0; f x < 0 for x < 0, f x > 0 for x > 0 f x x2 y 12 55. f 0 0; f 0 x 0 f x x3 y 3x y 2 4 f 3 2 1 -3 -2 f 2 1 -3 -2 -1 x -1 -2 -3 2 3 10 8 6 4 x -1 -2 1 2 3 f -6 -4 -2 2 x 2 4 6 -3 56. (a) If f c 3 and f is odd, then f c f c 3. (b) If f c 3 and f is even, then f c f c 3. 57. Let x 0, y0 be a point of tangency on the graph of f. By the limit definition for the derivative, f x 4 2x. The slope of the line through 2, 5 and x 0, y0 equals the derivative of f at x 0: 5 2 5 5 4x 0 y0 x0 y0 x0 2 58. Let x 0, y0 be a point of tangency on the graph of f. By the limit definition for the derivative, f x 2x. The slope of the line through 1, 3 and x 0, y0 equals the derivative of f at x 0: 3 y0 1 x0 2x 0 1 2x 0 0 0 x0 3, 1 x 0 2x 0 2x 02 4 2 8 x0 2 2x 0 x0 4 8x 0 4x 0 1 x0 2x 0 2x 0 3 3 x0 1, 3 2 3 3 x 02 x0 2x 0 3 x0 y0 x 02 3 1 0 0 x0 Therefore, the points of tangency are 1, 3 and 3, 3 , and the corresponding slopes are 2 and 2. The equations of the tangent lines are: y 5 y y 7 6 5 4 3 2 1 -2 x 1 2 3 6 Therefore, the points of tangency are 3, 9 and ( 1, 1 , and the corresponding slopes are 6 and 2. The equations of the tangent lines are: y 3 y 6x 6x y 10 2x 2x 2 1 y 5 y 2x 2x 2 9 1 9 y 3 y 2x 2x 1 1 (2, 5) (3, 3) (1, 3) (-1, 1) 8 6 4 (3, 9) -8 -6 -4 -2 -2 -4 x 2 4 6 (1, -3) Section 2.1 59. (a) g 0 (b) g 3 0 8 3, The Derivative and the Tangent Line Problem 103 3 (c) Because g 1 (d) Because g 4 g is decreasing (falling) at x g is increasing (rising) at x 1. 4. g 4 > 0. 2. g x x x x3 x 3x 2 x g x x3 3x x x 3x 2 3x x 3x x x 2 7 3, (e) Because g 4 and g 6 are both positive, g 6 is greater than g 4 , and g 6 (f) No, it is not possible. All you can say is that g is decreasing (falling) at x 60. (a) f x f x x2 lim lim f x x x2 x0 (b) g x x x x2 x 2x x x x 2x x x 1 2x 1 2x At x 2 and the tangent line is or y 2x 1. 0. 2x 1. y 2 lim x0 f x lim x2 lim x 2 x0 x x3 x0 x 2 x 3 x3 lim lim x2 x0 x0 lim x x 2 x x0 x0 lim 3x 2 x0 3x 2 lim 2x x0 1, g 1 3x 0, g 0 1, g 1 1 3x 2 1 1 3 and the tangent line is or y 3x 2. 0. At x y At x At x 1 1, f 0, f 0 1, f 1 y At x At x 0 and the tangent line is y 3 and the tangent line is 1 or y 3x 2. 0 and the tangent line is y 2 and the tangent line is y -3 3 -3 3 -3 For this function, the slopes of the tangent lines are always distinct for different values of x. -2 For this function, the slopes of the tangent lines are sometimes the same. 61. f x 1 3 4x 3 2 4x . -2 2 By the limit definition of the derivative we have f x x f x f x 2 2 3 1.5 27 32 27 16 3 4 2 1 1 4 0.5 1 32 3 16 0 0 0 0.5 1 32 3 16 1 1 4 3 4 1.5 27 32 27 16 2 2 3 -2 62. f x 1 2 2x 3 By the limit definition of the derivative we have f x x f x f x 2 2 2 1.5 1.125 1.5 1 0.5 1 0.5 0.125 0.5 0 0 0 0.5 x. 1 0.5 1 1.5 1.125 1.5 2 2 2 -2 -1 2 0.125 0.5 104 Chapter 2 f x 2x 2 3 Differentiation 0.01 0.01 0.01 f x x 0.01 2 63. g x 64. g x 2x x 2 100 8 f x 3 x 0.01 0.01 0.01 f x 3 x 100 2x 0.01 f g f -2 -1 4 The graph of g x is approximately the graph of f x 2 2x. g -1 -1 8 The graph of g x is approximately 3 the graph of f x . 2 x 65. f 2 f 2 24 3.99 2.1 2 4 2 4, f 2.1 0.1 2.1 4 Exact: f 2 2.1 0 3.99 66. f 2 f 2 1 3 2 4 2, f 2.1 2.31525 3.1525 Exact: f 2 3 2.31525 2 2.1 2 x3 4 6 67. f x 5 1 x and f x . 2x 3 2 As x 5 1 68. f x 3x and f x 3 2 x 4 3 f f -2 , f is nearly horizontal 0. -9 9 f -5 and thus f f -6 69. f x S x 4 x x f 2 4 2 1: S 0.5: S 0.1: S 3 2 x x x x x f 2 3 2 x 3 3 2 2 2 x x 3 3 f 2 2 1 3 x 2 19 x 10 4 5 -2 3 1 x x 1 2 x 2 3 x 5 2 x S 0.1 2 3 (a) x x x x 2 x 3 x 2 19 x 10 S1 -1 f 7 x S 0.5 (b) As x 0, the line approaches the tangent line to f at 2, 3 . 1 x f 2 22 x x x2 22 x 70. f x x S x x f 2 2 x 5 x 6 4 x 5 16 x 21 2 x 52 x 2 5 2 5 2 2 5 2 2 x f 2 x 5 x 6 4 x 5 16 x 21 2 5 6 9 10 41 42 5 2 x 1 2 x 2 x 22 x 3 x x 5 2 x 2 5 2 5 2 2 (a) x 1: S 4 x x (b) As 0.5: S 0.1: S x 2 -6 S0.1 S0.5 S1 f -4 6 x x 0, the line approaches the tangent line to f at 2, 5 . 2 Section 2.1 71. f x f 2 x2 x2 The Derivative and the Tangent Line Problem 105 1, c f x x 1 gx x 2x 2 lim f x 2 f 2 2 x2 g1 1 1, c x 1 f 1 1 0 g0 0 x0 x1 x2 lim lim x2 x 1 x2 x x 1 2 3 x 2 lim x 2 x x 2 2 x2 lim x 2 4 72. g x g 1 xx x1 x, c x1 lim lim 0 1 x1 lim xx x 1 1 x1 lim x 1 73. f x f 2 x3 2 f 2 2 lim x3 2x 2 1 x 2 1 x x 2 x 2 lim x2 x 2 x 2 2 x 2 lim x 2 4 74. f x f 1 x3 x1 2x, c f x x lim lim x3 x 2x 1 3 x 1 lim x 1 x2 x x 1 3 x1 lim x 2 x 3 5 75. g x g 0 x ,c x0 lim gx x x x x x lim x x . . . Does not exist. As x 0 , As x 0 , 1 x 1 x 76. f x f 3 1 ,c x lim 3 f x x 6 2 3, x3 f 3 3 c f 6 6 6 x3 lim 1 x x 1 3 3 x3 lim 3 3x x x 1 3 x3 lim 1 3x 1 9 77. f x f 6 x x6 lim f x x x6 lim x 623 x 6 0 x6 lim x 1 6 1 3 Does not exist. 78. g x g 3 x 3 lim 1 3 ,c g 3 3 3 x 3 x 3 gx x lim x 313 x 3 0 x 3 lim x 1 3 2 3 Does not exist. 79. h x h 5 x 5 ,c x 5 5 h 5 5 x 5 lim hx x lim x x 5 5 0 x 5 lim x x 5 5 Does not exist. 80. f x f 4 x x4 4 ,c f x x 4 f 4 4 x4 lim lim x x 4 4 0 x4 lim x x 4 4 81. f x is differentiable everywhere except at x (Discontinuity) 1. Does not exist. 106 Chapter 2 Differentiation 3. 82. f x is differentiable everywhere except at x (Sharp turns in the graph) 84. f x is differentiable everywhere except at x (Discontinuities) 86. f x is differentiable everywhere except at x 83. f x is differentiable everywhere except at x (Sharp turn in the graph) 85. f x is differentiable on the interval 1, (At x 1 the tangent line is vertical.) . 3. 2. 0. (Discontinuity) 2x is differentiable x 1 for all x 1. f is not defined at x 1. (Vertical asymptote) 6 87. f x x 3 is differentiable for all x 3. There is a sharp corner at x 3. 5 88. f x 89. f x all x at x x 2 5 is differentiable for 0. There is a sharp corner 0. 5 -6 -7 -1 2 -6 -2 6 -3 6 90. f is differentiable for all x 1. f is not continuous at x 1. 3 91. f x x 1 The derivative from the left is x1 lim f x x f 1 1 x1 lim x x 1 1 0 1. -4 5 The derivative from the right is x1 -3 lim f x x f 1 1 x1 lim x x 1 1 0 1. The one-sided limits are not equal. Therefore, f is not differentiable at x 1. 92. f x 1 x2 The derivative from the left does not exist because x1 lim f x x f 1 1 x1 lim 1 x x2 1 0 x1 lim 1 x x2 1 1 1 x2 x2 x1 lim 1 1 x x2 . (Vertical tangent) 1. The limit from the right does not exist since f is undefined for x > 1. Therefore, f is not differentiable at x x x 1 3, x 1 1 2, x > 1 93. f x The derivative from the left is lim f x x f 1 1 lim x x 1 1 3 0 0. x1 x1 1 2 x1 lim x The derivative from the right is lim f x x f 1 1 lim x x 1 1 2 0 0. 1. f 1 0 x1 x1 1 x1 lim x These one-sided limits are equal. Therefore, f is differentiable at x Section 2.1 x, x 1 x 2, x > 1 The Derivative and the Tangent Line Problem 107 94. f x The derivative from the left is lim f x x f 1 1 lim x x 1 1 lim 1 1. x1 x1 x1 The derivative from the right is x1 lim f x x f 1 1 x1 lim x2 x 1 1 x1 lim x 1 2. 1. These one-sided limits are not equal. Therefore, f is not differentiable at x 95. Note that f is continuous at x 2. f x f x x f x x x2 4x f 2 2 f 2 2 1, x 2 3, x > 2 lim lim x2 x 4x x 1 2 3 2 2. f 2 5 5 lim x lim 4 4 2 4. 4. The derivative from the left is lim x2 x2 x2 The derivative from the right is lim x2 x2 x2 The one-sided limits are equal. Therefore, f is differentiable at x 1 2x 96. Note that f is continuous at x The derivative from the left is 2. f x 2x, 1, x < 2 x 2 1 1 f x f 2 2x lim x2 x2 x 2 x 2 The derivative from the right is lim 2 x2 lim 1 2 x x 2 2 1 . 2 x2 lim f x x f 2 2 x2 lim 2x 2 x 2 x 2x 2 4 2x 2x 2x 2 2 2 lim x 2x 2 2. 2 2x f 2 2 1 2 x2 lim x2 x2 lim 2 2x 2 1 . 2 The one-sided limits are equal. Therefore, f is differentiable at x 97. (a) The distance from 3, 1 to the line mx d Ax 1 m3 By1 A2 m2 (b) 5 y 4 0 is y C B2 3 2 11 1 4 3m m2 3 1 . 1 x 1 2 3 4 -4 -1 4 The function d is not differentiable at m 1. This corresponds to the line y x 4, which passes through the point 3, 1 . 108 Chapter 2 Differentiation 2x (b) g x x 3 and g x y 3 98. (a) f x x 2 and f x y 5 3x 2 f 4 3 2 1 1 x 1 2 3 4 -2 -1 -1 1 2 x (c) The derivative is a polynomial of degree 1 less than the original function. If h x x n, then n 1. h x nx g 2 g -4 -3 -2 -1 f' -3 (d) If f x f x x4, then lim lim lim lim f x x x4 x x0 x x x4 x 4x 3 x 6x 2 6x 2 f x x4 6x 2 x x 4x x 4x x 2 2 x0 4x x x 2 x 3 x 3 4 x4 x0 4x 3 x x 3 x0 lim x0 4x 3 x 4x 3. x 4, then f x 4x 3 which is consistent with the conjecture. However, this is not Hence, if f x a proof since you must verify the conjecture for all integer values of n, n 2. f 2 x x f 2 99. False. The slope is lim x0 . x 2 is continuous at x 2, but is not 100. False. y differentiable at x 2. (Sharp turn in the graph) 101. False. If the derivative from the left of a point does not equal the derivative from the right of a point, x , then the derivative from the then the derivative does not exist at that point. For example, if f x left at x 0 is 1 and the derivative from the right at x 0 is 1. At x 0, the derivative does not exist. 102. True--see Theorem 2.1. x sin 1 x , 0, x x 0 0 0 f 0 and 103. f x x x sin 1 x x , x 0. Thus, lim x sin 1 x Using the Squeeze Theorem, we have x0 f is continuous at x 0. Using the alternative form of the derivative, we have f x x sin 1 x 1 f 0 0 lim lim sin . x0 x0 x 0 x 0 x Since this limit does not exist (sin 1 x oscillates between x0 lim 1 and 1), the function is not differentiable at x 0. gx x 2 sin 1 x , 0, x x 0 0 x 2 x 2 sin 1 x x 2, x 0. Thus, lim x 2 sin 1 x x0 Using the Squeeze Theorem again, we have and g is continuous at x x0 0 g0 0. Using the alternative form of the derivative again, we have 2 lim gx x g0 0 x0 lim x sin 1 x x 0 0 0, g 0 x0 lim x sin 0. 1 x 0. Therefore, g is differentiable at x Section 2.2 Basic Differentiation Rules and Rates of Change 109 104. 3 -3 -1 3 As you zoom in, the graph of y1 x 2 1 appears to be locally the graph of a horizontal line, whereas the graph of y2 x 1 always has a sharp corner at 0, 1 . y2 is not differentiable at 0, 1 . Section 2.2 1. (a) y y y 1 3. y y 8 0 1 x7 7x x1 1 2x 1 2 2 Basic Differentiation Rules and Rates of Change (b) 1 2 y y y 1 x3 3x 2 3 2. (a) y y y 1 x 1 2 (b) y y y 1 x 1 1 3 2 2x 1 2 x 1 x8 8x 7 2 4. f x f x 0 2 5. y y x6 6x 5 6. y y 7. y y x 8 7 8. y 7 x8 y 1 x8 8x x 9 8 9. y 8 x9 y 5 x x1 4 5 5 10. y 5 4 x x1 3 4 4 1 x 5 1 5x 4 y 1 x 4 1 4x 3 4 11. f x f x x 1 x2 2x 1 12. g x g x 4x 3 12x 2 3x 3 1 13. f t f t t3 2t 2 4t 3 3t 6 14. y y t2 2t 2t 2 2x 3 6x 2 3 15. g x g x 16. y y 8 x3 3x 2 17. s t s t 2t 2 1 cos x 2 1 sin x 2 4 18. f x f x x2 2x 3x 3 3t 2 19. y y 2 2 sin cos cos sin 20. g t g t cos t sin t 21. y y x2 2x 22. y y 5 sin x cos x 23. y y 1 x 1 x2 3 sin x 3 cos x 24. y y 5 2x 5 8 3 2 cos x 4 5 x 8 3 2 cos x 15 8x 4 2 sin x 3x 2 sin x Function 25. y 5 2x 2 2 3x 2 3 2x Rewrite y 5 x 2 2 x 3 3 x 8 2 Differentiate y 5x 4 x 3 9 x 8 3 Simplify y 5 x3 4 3x 3 9 8x 4 26. y y 2 y 3 y 27. y 3 y 3 y 4 y 110 Chapter 2 Function Differentiation Rewrite y 9 x x 2 Differentiate y 2 x 9 1 x 2 12x 2 3 Simplify y 2 9x 3 1 2x 3 12x 2 28. y 3x x x 4 x 3 2 29. y y y 1 2 y 3 2 y 2 30. y 4x 3 y y 31. f x 3 x2 6x 6 3x 3 2 , 1, 3 6 x3 32. f t f t f 3 5 y 3 3 5t 2 5 3 2x 4x 2 3 , 5t 3 ,2 5 33. f x f x f 0 1 2 21 2 x 5 0 7 3 x , 5 0, 1 2 f x f 1 34. y y y 2 3x 3 9x 2 36 6, 2, 18 35. 1 2, 0, 1 4x 4 1 36. f x 35 3x 2 x 2, 5, 0 30x 30 75 y y 0 8x 4 2 f x f 5 6x 0 x2 2x 37. f f f 0 4 sin 4 cos 41 1 , 0, 0 1 3 38. gt g t g 3 cos t, 3 sin t , 1 39. f x f x 5 6x 3x 3 2 2x 0 6 x3 40. f x f x x2 2x 2x 3x 3 3 3x 6x 6 x3 4 x3 x3 x3 2 3 41. g t g t t2 2t 4 t3 12t t2 4 4t 2t 3 42. f x 12 t4 f x x 1 1 x 2x 2 x3 2 3 43. f x f x x3 3x 2 x2 8 x3 1 1 63 x 1 2 x 8 3 4x 2 44. h x h x 2x 2 3x x 1 x2 5x 2 45x 2 1 2x 3 x 1 1 2 2x 2 1 x2 18x 2 15x 3 45. y y x x2 3x 2 x 46. y y 3x 6x 36x 3 47. f x f x x 1 x 2 s4 4 s 5 x1 2 3 2 6x 1 1 2 x 3 48. f x 2 x2 3 f x x 5 x x1 1 x 5 3 x1 5 2x 1 x 3 2 3 4 5 1 3x 2 3 1 5x 4 5 49. h s h (s 5 s2 1 5 3 50. f t 1 3 t2 2 t 3 3 t1 1 3 3 4 2 3 2 s 3 4 5s 1 5 2 3s 1 3 f t 1 t 3 2 3t 1 3 1 3t 2 3 Section 2.2 Basic Differentiation Rules and Rates of Change 2 3 111 51. f x f x 6 x 3x 1 2 5 cos x 5 sin x 6x 1 3 2 5 cos x 5 sin x 52. f x f x x 3 cos x 4 3 2x 1 3 3 cos x 3 sin x x 2 x 3 x3 3x 2 1, 3 sin x 2 3x 4 3 53. (a) y y x4 4x 3 3x 2 6x 2 54. (a) y y x 1 2: y 3 y 4x y 1 2 2 2 At 1, 0 : y Tangent ...
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