{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework Solution Chapter 2

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}