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Lesson 22: Graphing
The document provides steps for graphing the cubic function f(x) = 2x^3 - 3x^2 - 12x. Step 1 analyzes the monotonicity of the function by examining the sign chart of its derivative f'(x). Step 2 analyzes concavity by examining the sign chart of the second derivative f''(x). Step 3 combines these analyses into a single sign chart summarizing the function's properties over its domain. The goal is to completely graph the function, indicating zeros, asymptotes, critical points, maxima/minima, and inflection points.
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Lesson 22: Graphing
- 1. Section 4.4 Curve Sketching V63.0121.027, Calculus I November 17, 2009 Announcements Next written assignment will be due Wednesday, Nov 25 next and last quiz will be the week after Thanksgiving Final Exam: Friday, December 18, 2:00–3:50pm . . . . . .
- 2. Outline The Procedure Simple examples A cubic function A quartic function More Examples Points of nondifferentiability Horizontal asymptotes Vertical asymptotes Trigonometric and polynomial together Logarithmic . . . . . .
- 3. Objective Given a function, graph it completely, indicating zeroes asymptotes if applicable critical points local/global max/min inflection points . . Image credit: Image Of Surgery . . . . . .
- 4. The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b), then f is decreasing on (a, b). Proof. Pick two points x and y in (a, b) with x < y. We must show f(x) < f(y). By MVT there exists a point c in (x, y) such that f(y) − f(x) = f′ (c) > 0. y−x So f(y) − f(x) = f′ (c)(y − x) > 0. . . . . . .
- 5. Theorem (Concavity Test) If f′′ (x) > 0 for all x in I, then the graph of f is concave upward on I If f′′ (x) < 0 for all x in I, then the graph of f is concave downward on I Proof. Suppose f′′ (x) > 0 on I. This means f′ is increasing on I. Let a and x be in I. The tangent line through (a, f(a)) is the graph of L(x) = f(a) + f′ (a)(x − a) f(x) − f(a) By MVT, there exists a b between a and x with = f′ (b). x−a So f(x) = f(a) + f′ (b)(x − a) ≥ f(a) + f′ (a)(x − a) = L(x) . . . . . .
- 6. Graphing Checklist To graph a function f, follow this plan: 0. Find when f is positive, negative, zero, not defined. 1. Find f′ and form its sign chart. Conclude information about increasing/decreasing and local max/min. 2. Find f′′ and form its sign chart. Conclude concave up/concave down and inflection. 3. Put together a big chart to assemble monotonicity and concavity data 4. Graph! . . . . . .
- 7. Outline The Procedure Simple examples A cubic function A quartic function More Examples Points of nondifferentiability Horizontal asymptotes Vertical asymptotes Trigonometric and polynomial together Logarithmic . . . . . .
- 8. Graphing a cubic Example Graph f(x) = 2x3 − 3x2 − 12x. . . . . . .
- 9. Graphing a cubic Example Graph f(x) = 2x3 − 3x2 − 12x. (Step 0) First, let’s find the zeros. We can at least factor out one power of x: f(x) = x(2x2 − 3x − 12) so f(0) = 0. The other factor is a quadratic, so we the other two roots are √ √ 3 ± 32 − 4(2)(−12) 3 ± 105 x= = 4 4 It’s OK to skip this step for now since the roots are so complicated. . . . . . .
- 10. Step 1: Monotonicity f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: . . . . . . .
- 11. Step 1: Monotonicity f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: . . . −2 x 2 . . x . +1 − . 1 .′ (x) f . . − . 1 2 . f .(x) . . . . . .
- 12. Step 1: Monotonicity f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − . − . . . . + . −2 x 2 . . x . +1 − . 1 .′ (x) f . . − . 1 2 . f .(x) . . . . . .
- 13. Step 1: Monotonicity f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 .′ (x) f . . − . 1 2 . f .(x) . . . . . .
- 14. Step 1: Monotonicity f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + .′ (x) f . − . 1 2 . f .(x) . . . . . .
- 15. Step 1: Monotonicity f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . .′ (x) f . − . 1 2 . f .(x) . . . . . .
- 16. Step 1: Monotonicity f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . . + .′ (x) f . − . 1 2 . f .(x) . . . . . .
- 17. Step 1: Monotonicity f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . . + .′ (x) f . ↗− . . 1 2 . f .(x) . . . . . .
- 18. Step 1: Monotonicity f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . . + .′ (x) f . ↗− . . 1 ↘ . 2 . f .(x) . . . . . .
- 19. Step 1: Monotonicity f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . . + .′ (x) f . ↗− . . 1 ↘ . 2 . ↗ . f .(x) . . . . . .
- 20. Step 1: Monotonicity f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . . + .′ (x) f . ↗− . . 1 ↘ . 2 . ↗ . f .(x) m . ax . . . . . .
- 21. Step 1: Monotonicity f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − . − . . . . + . −2 x 2 . − . . . + . + x . +1 − . 1 . . + − . . + .′ (x) f . ↗− . . 1 ↘ . 2 . ↗ . f .(x) m . ax m . in . . . . . .
- 22. Step 2: Concavity f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . . . . . . .
- 23. Step 2: Concavity f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . .′′ (x) f . . 1/2 f .(x) . . . . . .
- 24. Step 2: Concavity f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . − . − .′′ (x) f . . 1/2 f .(x) . . . . . .
- 25. Step 2: Concavity f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . − . − . + + .′′ (x) f . . 1/2 f .(x) . . . . . .
- 26. Step 2: Concavity f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . − . − . + + .′′ (x) f . . ⌢ . 1/2 f .(x) . . . . . .
- 27. Step 2: Concavity f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . − . − . + + .′′ (x) f . . ⌢ . 1/2 . ⌣ f .(x) . . . . . .
- 28. Step 2: Concavity f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . − . − . + + .′′ (x) f . . ⌢ . 1/2 . ⌣ f .(x) I .P . . . . . .
- 29. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . . . . . .
- 30. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. − . . . . + − . . + .′ (x) f . ↗− ↘ . . 1 . ↘ . 2 . ↗ . m . onotonicity . . . . . .
- 31. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . + − . . − . . + .′ (x) f . ↗− . . 1 ↘ . ↘ . . 2 ↗ . m .′′ onotonicity − . − − . − . . + + . + + f . (x) . ⌢ . ⌢ 1/2 . . ⌣ . ⌣ c . oncavity . . . . . .
- 32. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . + − . . − . . + .′ (x) f . ↗− . . 1 ↘ . ↘ . . 2 ↗ . m .′′ onotonicity − . − − . − . . + + . + + f . (x) . ⌢ ⌢ ./2 . . 1 ⌣ . ⌣ c . oncavity 7 .. − . 6 1/2 −. . 20 f .(x) . − . 1 . 1/2 2 . . hape of f s m . ax I .P m . in . . . . . .
- 33. Combinations of monotonicity and concavity . . increasing, decreasing, concave concave down down I .I I . . I .II I .V . . decreasing, increasing, concave up concave up . . . . . .
- 34. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . + − . . − . . + .′ (x) f . ↗− . . 1 ↘ . ↘ . . 2 ↗ . m .′′ onotonicity − . − − . − . . + + . + + f . (x) . ⌢ ⌢ ./2 . . 1 ⌣ . ⌣ c . oncavity 7 .. − . 6 1/2 −. . 20 f .(x) . . . 1 − . 1/2 2 . . hape of f s m . ax I .P m . in . . . . . .
- 35. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . + − . . − . . + .′ (x) f . ↗− . . 1 ↘ . ↘ . . 2 ↗ . m .′′ onotonicity − . − − . − . . + + . + + f . (x) . ⌢ ⌢ ./2 . . 1 ⌣ . ⌣ c . oncavity 7 .. − . 6 1/2 −. . 20 f .(x) . . . 1 . ./2 − 1 2 . . hape of f s m . ax I .P m . in . . . . . .
- 36. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . + − . . − . . + .′ (x) f . ↗− . . 1 ↘ . ↘ . . 2 ↗ . m .′′ onotonicity − . − − . − . . + + . + + f . (x) . ⌢ ⌢ ./2 . . 1 ⌣ . ⌣ c . oncavity 7 .. − . 6 1/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . . hape of f s m . ax I .P m . in . . . . . .
- 37. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. . . + − . . − . . + .′ (x) f . ↗− . . 1 ↘ . ↘ . . 2 ↗ . m .′′ onotonicity − . − − . − . . + + . + + f . (x) . ⌢ ⌢ ./2 . . 1 ⌣ . ⌣ c . oncavity 7 .. − . 6 1/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . . . hape of f s m . ax I .P m . in . . . . . .
- 38. Step 4: Graph f .(x) .(x) = 2x3 − 3x2 − 12x f ( √ ) . −1, 7) ( . . 3− 4105 , 0 . 0, 0 ) ( . . . x . 1/2, −61/2) ( ( . √ ) . . 3+ 4105 , 0 . 2, −20) ( . 7 .. − . 61/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . . . hape of f s m . ax I .P m . in . . . . . .
- 39. Step 4: Graph f .(x) .(x) = 2x3 − 3x2 − 12x f ( √ ) . −1, 7) ( . . 3− 4105 , 0 . 0, 0 ) ( . . . x . 1/2, −61/2) ( ( . √ ) . . 3+ 4105 , 0 . 2, −20) ( . 7 .. − . 61/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . . . hape of f s m . ax I .P m . in . . . . . .
- 40. Step 4: Graph f .(x) .(x) = 2x3 − 3x2 − 12x f ( √ ) . −1, 7) ( . . 3− 4105 , 0 . 0, 0 ) ( . . . x . 1/2, −61/2) ( ( . √ ) . . 3+ 4105 , 0 . 2, −20) ( . 7 .. − . 61/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . . . hape of f s m . ax I .P m . in . . . . . .
- 41. Step 4: Graph f .(x) .(x) = 2x3 − 3x2 − 12x f ( √ ) . −1, 7) ( . . 3− 4105 , 0 . 0, 0 ) ( . . . x . 1/2, −61/2) ( ( . √ ) . . 3+ 4105 , 0 . 2, −20) ( . 7 .. − . 61/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . . . hape of f s m . ax I .P m . in . . . . . .
- 42. Step 4: Graph f .(x) .(x) = 2x3 − 3x2 − 12x f ( √ ) . −1, 7) ( . . 3− 4105 , 0 . 0, 0 ) ( . . . x . 1/2, −61/2) ( ( . √ ) . . 3+ 4105 , 0 . 2, −20) ( . 7 .. − . 61/2 −. . 20 f .(x) . . . 1 . ./2 . − 1 2 . . . hape of f s m . ax I .P m . in . . . . . .
- 43. Graphing a quartic Example Graph f(x) = x4 − 4x3 + 10 . . . . . .
- 44. Graphing a quartic Example Graph f(x) = x4 − 4x3 + 10 (Step 0) We know f(0) = 10 and lim f(x) = +∞. Not too many x→±∞ other points on the graph are evident. . . . . . .
- 45. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) . . . . . .
- 46. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . . . . . . .
- 47. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. 0 .. . x2 4 0 . . . . . . .
- 48. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . x2 4 0 . . . . . . .
- 49. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . x2 4 0 . . . . . . .
- 50. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . . . . . . .
- 51. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . 0 .. . x − 3) ( 3 . . . . . . .
- 52. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . 0 .. . x − 3) ( 3 . . . . . . .
- 53. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . 0 .. . x − 3) ( 3 . . . . . . .
- 54. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . . . . . . .
- 55. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . 0 .. 0 .. .′ (x) f 0 . 3 . f .(x) . . . . . .
- 56. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. 0 .. .′ (x) f 0 . 3 . f .(x) . . . . . .
- 57. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. − . 0 .. .′ (x) f 0 . 3 . f .(x) . . . . . .
- 58. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. − . .. . 0 + .′ (x) f 0 . 3 . f .(x) . . . . . .
- 59. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. − . .. . 0 + .′ (x) f ↘ 0 . . 3 . f .(x) . . . . . .
- 60. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. − . .. . 0 + .′ (x) f ↘ 0 . . ↘ . 3 . f .(x) . . . . . .
- 61. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. − . .. . 0 + .′ (x) f ↘ 0 . . ↘ . 3 ↗ . . f .(x) . . . . . .
- 62. Step 1: Monotonicity f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. . .. + 0 . + . + . x2 4 0 . − . − . .. . 0 + . x − 3) ( 3 . − 0 . .. − . .. . 0 + .′ (x) f ↘ 0 . . ↘ . 3 ↗ . . f .(x) m . in . . . . . .
- 63. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) . . . . . .
- 64. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: . . . . . . .
- 65. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: 0 .. 1 . 2x 0 . . . . . . .
- 66. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. 1 . 2x 0 . . . . . . .
- 67. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + 1 . 2x 0 . . . . . . .
- 68. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . . . . . . .
- 69. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . 0 .. . −2 x 2 . . . . . . .
- 70. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . 0 .. . −2 x 2 . . . . . . .
- 71. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . −2 x 2 . . . . . . .
- 72. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . . . . . .
- 73. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . 0 .. 0 .. .′′ (x) f 0 . 2 . f . (x ) . . . . . .
- 74. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 0 .. .′′ (x) f 0 . 2 . f . (x ) . . . . . .
- 75. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. .′′ (x) f 0 . 2 . f . (x ) . . . . . .
- 76. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. . + + .′′ (x) f 0 . 2 . f . (x ) . . . . . .
- 77. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. . + + .′′ (x) f . . ⌣ 0 2 . f . (x ) . . . . . .
- 78. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. . + + .′′ (x) f . . ⌣ 0 . ⌢ 2 . f . (x ) . . . . . .
- 79. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. . + + .′′ (x) f . . ⌣ 0 . ⌢ 2 . . ⌣ f . (x ) . . . . . .
- 80. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. . + + .′′ (x) f . . ⌣ 0 . ⌢ 2 . . ⌣ f . (x ) I .P . . . . . .
- 81. Step 2: Concavity f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: − 0 . .. . + . + 1 . 2x 0 . − . − . 0 .. . + . −2 x 2 . . + .. + 0 − . − 0 .. . + + .′′ (x) f . . ⌣ 0 . ⌢ 2 . . ⌣ f . (x ) I .P I .P . . . . . .
- 82. Step 3: Grand Unified Sign Chart . Remember, f(x) = x4 − 4x3 + 10. − 0 . .. − . − 0 + . .. . .′ (x) f ↘ 0 . . ↘ . ↘ 3 ↗ . . . m .′′ onotonicity . + .. + 0 − . − .. . + . + 0+ + f . (x) . . ⌣ 0 . ⌢ 2 . . ⌣ . ⌣ c . oncavity 1. .0 − −. . .6 . 17 f .(x) 0 . 2 . 3 . s . hape I .P I .P . inm . . . . . .
- 83. Step 3: Grand Unified Sign Chart . Remember, f(x) = x4 − 4x3 + 10. − 0 . .. − . − 0 + . .. . .′ (x) f ↘ 0 . . ↘ . ↘ 3 ↗ . . . m .′′ onotonicity . + .. + 0 − . − .. . + . + 0+ + f . (x) . . ⌣ 0 . ⌢ 2 . . ⌣ . ⌣ c . oncavity 1. .0 − −. . .6 . 17 f .(x) . .0 2 . 3 . s . hape I .P I .P . inm . . . . . .
- 84. Step 3: Grand Unified Sign Chart . Remember, f(x) = x4 − 4x3 + 10. − 0 . .. − . − 0 + . .. . .′ (x) f ↘ 0 . . ↘ . ↘ 3 ↗ . . . m .′′ onotonicity . + .. + 0 − . − .. . + . + 0+ + f . (x) . . ⌣ 0 . ⌢ 2 . . ⌣ . ⌣ c . oncavity 1. .0 − −. . .6 . 17 f .(x) . .0 . 2 . 3 . s . hape I .P I .P . inm . . . . . .
- 85. Step 3: Grand Unified Sign Chart . Remember, f(x) = x4 − 4x3 + 10. − 0 . .. − . − 0 + . .. . .′ (x) f ↘ 0 . . ↘ . ↘ 3 ↗ . . . m .′′ onotonicity . + .. + 0 − . − .. . + . + 0+ + f . (x) . . ⌣ 0 . ⌢ 2 . . ⌣ . ⌣ c . oncavity 1. .0 − −. . .6 . 17 f .(x) . .0 . 2 . . . 3 s . hape I .P I .P . inm . . . . . .
- 86. Step 3: Grand Unified Sign Chart . Remember, f(x) = x4 − 4x3 + 10. − 0 . .. − . − 0 + . .. . .′ (x) f ↘ 0 . . ↘ . ↘ 3 ↗ . . . m .′′ onotonicity . + .. + 0 − . − .. . + . + 0+ + f . (x) . . ⌣ 0 . ⌢ 2 . . ⌣ . ⌣ c . oncavity 1. .0 − −. . .6 . 17 f .(x) . .0 . 2 . . . . 3 s . hape I .P I .P . inm . . . . . .
- 87. Step 4: Graph y . .(x) = x4 − 4x3 + 10 f . 0, 10) ( . . . x . . . 2, −6) ( . 3, −17) ( 1. .0 − −. . .6 . 17 f . (x ) . .0 . 2 . . . . 3 s . hape I .P I .P . in m . . . . . .
- 88. Step 4: Graph y . .(x) = x4 − 4x3 + 10 f . 0, 10) ( . . . x . . . 2, −6) ( . 3, −17) ( 1. .0 − −. . .6 . 17 f . (x ) . .0 . 2 . . . . 3 s . hape I .P I .P . in m . . . . . .
- 89. Step 4: Graph y . .(x) = x4 − 4x3 + 10 f . 0, 10) ( . . . x . . . 2, −6) ( . 3, −17) ( 1. .0 − −. . .6 . 17 f . (x ) . .0 . 2 . . . . 3 s . hape I .P I .P . in m . . . . . .
- 90. Step 4: Graph y . .(x) = x4 − 4x3 + 10 f . 0, 10) ( . . . x . . . 2, −6) ( . 3, −17) ( 1. .0 − −. . .6 . 17 f . (x ) . .0 . 2 . . . . 3 s . hape I .P I .P . in m . . . . . .
- 91. Step 4: Graph y . .(x) = x4 − 4x3 + 10 f . 0, 10) ( . . . x . . . 2, −6) ( . 3, −17) ( 1. .0 − −. . .6 . 17 f . (x ) . .0 . 2 . . . . 3 s . hape I .P I .P . in m . . . . . .
- 92. Outline The Procedure Simple examples A cubic function A quartic function More Examples Points of nondifferentiability Horizontal asymptotes Vertical asymptotes Trigonometric and polynomial together Logarithmic . . . . . .
- 93. Example √ Graph f(x) = x + |x| . . . . . .
- 94. Example √ Graph f(x) = x + |x| This function looks strange because of the absolute value. But whenever we become nervous, we can just take cases. . . . . . .
- 95. Step 0: Finding Zeroes √ f(x) = x + |x| First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if x is positive. . . . . . .
- 96. Step 0: Finding Zeroes √ f(x) = x + |x| First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if x is positive. Are there negative numbers which are zeroes for f? . . . . . .
- 97. Step 0: Finding Zeroes √ f(x) = x + |x| First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if x is positive. Are there negative numbers which are zeroes for f? √ x + −x = 0 √ −x = −x −x = x2 x2 + x = 0 The only solutions are x = 0 and x = −1 . . . . . .
- 98. Step 0: Asymptotic behavior √ f(x) = x + |x| lim f(x) = ∞, because both terms tend to ∞. x→∞ . . . . . .
- 99. Step 0: Asymptotic behavior √ f(x) = x + |x| lim f(x) = ∞, because both terms tend to ∞. x→∞ lim f(x) is indeterminate of the form −∞ + ∞. It’s the x→−∞ √ same as lim (−y + y) y→+∞ . . . . . .
- 100. Step 0: Asymptotic behavior √ f(x) = x + |x| lim f(x) = ∞, because both terms tend to ∞. x→∞ lim f(x) is indeterminate of the form −∞ + ∞. It’s the x→−∞ √ same as lim (−y + y) y→+∞ √ √ √ y+y lim (−y + y) = lim ( y − y) · √ y→+∞ y→∞ y+y y − y2 = lim √ = −∞ y→∞ y+y . . . . . .
- 101. Step 1: The derivative √ Remember, f(x) = x + |x|. To find f′ , first assume x > 0. Then d ( √ ) 1 f ′ (x ) = x+ x =1+ √ dx 2 x . . . . . .
- 102. Step 1: The derivative √ Remember, f(x) = x + |x|. To find f′ , first assume x > 0. Then d ( √ ) 1 f ′ (x ) = x+ x =1+ √ dx 2 x Notice f′ (x) > 0 when x > 0 . . . . . .
- 103. Step 1: The derivative √ Remember, f(x) = x + |x|. To find f′ , first assume x > 0. Then d ( √ ) 1 f ′ (x ) = x+ x =1+ √ dx 2 x Notice f′ (x) > 0 when x > 0 lim f′ (x) = ∞ x→0+ . . . . . .
- 104. Step 1: The derivative √ Remember, f(x) = x + |x|. To find f′ , first assume x > 0. Then d ( √ ) 1 f ′ (x ) = x+ x =1+ √ dx 2 x Notice f′ (x) > 0 when x > 0 lim f′ (x) = ∞ x→0+ lim f′ (x) = 1 x→∞ . . . . . .
- 105. Step 1: The derivative √ Remember, f(x) = x + |x|. If x is negative, we have d ( √ ) 1 f′ (x) = x + −x = 1 − √ dx 2 −x √ Again, this looks weird because −x appears to be a negative number. But since x < 0, −x > 0. . . . . . .
- 106. Step 1: The derivative √ Remember, f(x) = x + |x|. If x is negative, we have d ( √ ) 1 f′ (x) = x + −x = 1 − √ dx 2 −x √ Again, this looks weird because −x appears to be a negative number. But since x < 0, −x > 0. Notice lim f′ (x) = −∞ x→0− . . . . . .
- 107. Step 1: The derivative √ Remember, f(x) = x + |x|. If x is negative, we have d ( √ ) 1 f′ (x) = x + −x = 1 − √ dx 2 −x √ Again, this looks weird because −x appears to be a negative number. But since x < 0, −x > 0. Notice lim f′ (x) = −∞ x→0− lim f′ (x) = 1 x→−∞ . . . . . .
- 108. Step 1: Monotonicity Where are the critical points? We see that f′ (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4 . . . . . .
- 109. Step 1: Monotonicity Where are the critical points? We see that f′ (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4 We know f is not differentiable at 0 as well. . . . . . .
- 110. Step 1: Monotonicity Where are the critical points? We see that f′ (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4 We know f is not differentiable at 0 as well. We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. 0 .. ∓. . ∞ . ′ (x ) f . 1 −4 0 . f .(x) . . . . . .
- 111. Step 1: Monotonicity Where are the critical points? We see that f′ (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4 We know f is not differentiable at 0 as well. We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 .. ∓. . ∞ . ′ (x ) f . 1 −4 0 . f .(x) . . . . . .
- 112. Step 1: Monotonicity Where are the critical points? We see that f′ (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4 We know f is not differentiable at 0 as well. We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . ′ (x ) f . 1 −4 0 . f .(x) . . . . . .
- 113. Step 1: Monotonicity Where are the critical points? We see that f′ (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4 We know f is not differentiable at 0 as well. We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + . ′ (x ) f . 1 −4 0 . f .(x) . . . . . .
- 114. Step 1: Monotonicity Where are the critical points? We see that f′ (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4 We know f is not differentiable at 0 as well. We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + . ′ (x ) f ↗ . . 1 −4 0 . f .(x) . . . . . .
- 115. Step 1: Monotonicity Where are the critical points? We see that f′ (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4 We know f is not differentiable at 0 as well. We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + . ′ (x ) f ↗ . −4 ↘ 0 . 1. . f .(x) . . . . . .
- 116. Step 1: Monotonicity Where are the critical points? We see that f′ (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4 We know f is not differentiable at 0 as well. We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + . ′ (x ) f ↗ . −4 ↘ 0 . 1. . ↗ . f .(x) . . . . . .
- 117. Step 1: Monotonicity Where are the critical points? We see that f′ (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4 We know f is not differentiable at 0 as well. We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + . ′ (x ) f ↗ . −4 ↘ 0 . 1. . ↗ . f .(x) . max . . . . . .
- 118. Step 1: Monotonicity Where are the critical points? We see that f′ (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4 We know f is not differentiable at 0 as well. We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + . ′ (x ) f ↗ . −4 ↘ 0 . .1 . . ↗ . f .(x) . max min . . . . . .
- 119. Step 1: Monotonicity Where are the critical points? We see that f′ (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4 We know f is not differentiable at 0 as well. We can’t make a multi-factor sign chart because of the absolute value, but we can test points in between critical points. . + 0 −∓ . .. . . ∞ . + . ′ (x ) f ↗ . −4 ↘ 0 . .1 . . ↗ . f .(x) . max min . . . . . .
- 120. Step 2: Concavity If x > 0, then ( ) ′′ d 1 1 f (x) = 1 + x−1/2 = − x−3/2 dx 2 4 This is negative whenever x > 0. . . . . . .
- 121. Step 2: Concavity If x > 0, then ( ) ′′ d 1 1 f (x) = 1 + x−1/2 = − x−3/2 dx 2 4 This is negative whenever x > 0. If x < 0, then ( ) ′′ d 1 −1/2 1 f (x) = 1 − (−x) = − (−x)−3/2 dx 2 4 which is also always negative for negative x. . . . . . .
- 122. Step 2: Concavity If x > 0, then ( ) ′′ d 1 1 f (x) = 1 + x−1/2 = − x−3/2 dx 2 4 This is negative whenever x > 0. If x < 0, then ( ) ′′ d 1 −1/2 1 f (x) = 1 − (−x) = − (−x)−3/2 dx 2 4 which is also always negative for negative x. 1 In other words, f′′ (x) = − |x|−3/2 . 4 . . . . . .
- 123. Step 2: Concavity If x > 0, then ( ) ′′ d 1 1 f (x) = 1 + x−1/2 = − x−3/2 dx 2 4 This is negative whenever x > 0. If x < 0, then ( ) ′′ d 1 −1/2 1 f (x) = 1 − (−x) = − (−x)−3/2 dx 2 4 which is also always negative for negative x. 1 In other words, f′′ (x) = − |x|−3/2 . 4 Here is the sign chart: − . − −. . ∞ − . − f′′ . . (x ) . ⌢ . ⌢ . 0 . f .(x) . . . . . .
- 124. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| ′ . 1 + . + 0 −∓ . .. . . ∞ . + . 1 (x) +. f ↗ . ↗ . −4 ↘ 0 . 1. . ↗ . ↗m . .′′ onotonicity − . ∞ − . − − . . . − ∞ − − . − . ∞ (x) − . f . ⌢ . ⌢ . . ⌢ 0 . ⌢ . . oncavity ⌢c − . ∞ 0 .. .1 4 0 .. . ∞ x) + .(f . − . 1 − . .41 0 . s . hape . zero . max min . . . . . .
- 125. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| ′ . 1 + . + 0 −∓ . .. . . ∞ . + . 1 (x) +. f ↗ . ↗ . −4 ↘ 0 . 1. . ↗ . ↗m . .′′ onotonicity − . ∞ − . − − . . . − ∞ − − . − . ∞ (x) − . f . ⌢ . ⌢ . . ⌢ 0 . ⌢ . . oncavity ⌢c − . ∞ 0 .. .1 4 0 .. . ∞ x) + .(f . . . 1 − − . .41 0 . s . hape . zero . max min . . . . . .
- 126. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| ′ . 1 + . + 0 −∓ . .. . . ∞ . + . 1 (x) +. f ↗ . ↗ . −4 ↘ 0 . 1. . ↗ . ↗m . .′′ onotonicity − . ∞ − . − − . . . − ∞ − − . − . ∞ (x) − . f . ⌢ . ⌢ . . ⌢ 0 . ⌢ . . oncavity ⌢c − . ∞ 0 .. .1 4 0 .. . ∞ x) + .(f . . . 1 − . − . .41 0 . s . hape . zero . max min . . . . . .
- 127. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| ′ . 1 + . + 0 −∓ . .. . . ∞ . + . 1 (x) +. f ↗ . ↗ . −4 ↘ 0 . 1. . ↗ . ↗m . .′′ onotonicity − . ∞ − . − − . . . − ∞ − − . − . ∞ (x) − . f . ⌢ . ⌢ . . ⌢ 0 . ⌢ . . oncavity ⌢c − . ∞ 0 .. .1 4 0 .. . ∞ x) + .(f . . . 1 − . − . .41 . .0 s . hape . zero . max min . . . . . .
- 128. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| ′ . 1 + . + 0 −∓ . .. . . ∞ . + . 1 (x) +. f ↗ . ↗ . −4 ↘ 0 . 1. . ↗ . ↗m . .′′ onotonicity − . ∞ − . − − . . . − ∞ − − . − . ∞ (x) − . f . ⌢ . ⌢ . . ⌢ 0 . ⌢ . . oncavity ⌢c − . ∞ 0 .. .1 4 0 .. . ∞ x) + .(f . . . 1 − . − . .41 . .0 . s . hape . zero . max min . . . . . .
- 129. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| ′ . 1 + . + 0 −∓ . .. . . ∞ . + . 1 (x) +. f ↗ . ↗ . −4 ↘ 0 . 1. . ↗ . ↗m . .′′ onotonicity − . ∞ − . − − . . . − ∞ − − . − . ∞ (x) − . f . ⌢ . ⌢ . . ⌢ 0 . ⌢ . . oncavity ⌢c − . ∞ 0 .. .1 4 0 .. . ∞ x) + .(f . . . 1 − . − . .41 . .0 . s . hape . zero . max min . . . . . .
- 130. Graph √ f(x) = x + |x| f .(x) .−1, 1) ( 4 4 . −1, 0) ( . . . x . . 0, 0 ) ( − 0 . ∞ .. .1 4 0 .. . ∞ .(x) + f . . − . 1 . − . .41 . . 0 . s . hape . zero . max min . . . . . .
- 131. Graph √ f(x) = x + |x| f .(x) .−1, 1) ( 4 4 . −1, 0) ( . . . x . . 0, 0 ) ( − 0 . ∞ .. .1 4 0 .. . ∞ .(x) + f . . − . 1 . − . .41 . . 0 . s . hape . zero . max min . . . . . .
- 132. Graph √ f(x) = x + |x| f .(x) .−1, 1) ( 4 4 . −1, 0) ( . . . x . . 0, 0 ) ( − 0 . ∞ .. .1 4 0 .. . ∞ .(x) + f . . − . 1 . − . .41 . . 0 . s . hape . zero . max min . . . . . .
- 133. Graph √ f(x) = x + |x| f .(x) .−1, 1) ( 4 4 . −1, 0) ( . . . x . . 0, 0 ) ( − 0 . ∞ .. .1 4 0 .. . ∞ .(x) + f . . − . 1 . − . .41 . . 0 . s . hape . zero . max min . . . . . .
- 134. Graph √ f(x) = x + |x| f .(x) .−1, 1) ( 4 4 . −1, 0) ( . . . x . . 0, 0 ) ( − 0 . ∞ .. .1 4 0 .. . ∞ .(x) + f . . − . 1 . − . .41 . . 0 . s . hape . zero . max min . . . . . .
- 135. Example 2 Graph f(x) = xe−x . . . . . .
- 136. Example 2 Graph f(x) = xe−x Before taking derivatives, we notice that f is odd, that f(0) = 0, and lim f(x) = 0 x→∞ . . . . . .
- 137. Step 1: Monotonicity 2 If f(x) = xe−x , then 2 2 ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x ( √ )( √ ) 2 = 1 − 2x 1 + 2x e−x 2 The factor e−x is always positive so it doesn’t figure into the sign of f′ (x). So our sign chart looks like this: . + .. + 0 . − . √ √. . − 1 2x . 1/2 − . 0 .. . + . + √ √ 1 . + 2x − . 1/2 ′ − . 0 .. . + 0 . − . f . (x) √ √. ↘ . − 1/2 ↗ . ↘ . f . (x ) . . . . 1/2 max min . . . . . .
- 138. Step 2: Concavity 2 If f′ (x) = (1 − 2x2 )e−x , we know 2 2 ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 = 2x(2x2 − 3)e−x − . − . 0 .. . + . + .x 2 0 . − . − . − . 0 . . + √ √ √. . 2x − 3 . 3/2 − . 0 .. . + . + . + √ √ √ . 2x + 3 − 3/2 . − . − .. . + 0 + 0 .. − . − .. 0 . + + .′′ (x) f . ⌢ √ . ⌣ ⌢ √3 . . ⌣ − 3/2 . 0 . f .(x) . . . . /2 IP IP IP . . . . . .
- 139. Step 3: Synthesis 2 f(x) = xe−x − . − 0 + . .. . + . − − .′ (x) f √ . . √. . 0 . ↘ . ↘ 1/2 . .. − ↗ ↗ ↘ . . 1/2. ↘ . m . onotonic − . − .. . + 0+ + 0 − . + .. . − − 0 . − .. . + + .′′ (x) f . ⌢ √. ⌣ ⌣ . . . . √3 . − 3/2 . 0 ⌢ ⌢ . /2 ⌣ c . oncavity √ √ − 2e3 . √1 . 3 − 2e .√1 . 2e3 3 f .(x) . . 0 .. 2e . √. . √ . √ √ . − . . . . 3/2− 1/2 . . . . 0 . . 1/2 . 3/2 . . s . hape IP min IP max IP . . . . . .
- 140. Step 4: Graph f .(x) (√ )( . 1/2, √1 √ √ ) 2 2e . 3/2, 3 .(x) = xe−x f . 2e3 . . x . . 0, 0 ) ( ( . √ √ ) . . − 3/2, − 2e3 ( √ 3 ) . − 1/2, − √1 2e √ √ − 2e3 √1 . 3 − . 2e .√1 . 2e33 f .(x) . . 0 .. 2e . √ √ √. √. . − .. . . 3/2 1/2 . . . . . .1/2.. 3/2 − . 0 . s . hape IP min IP max IP . . . . . .
- 141. Example 1 1 Graph f(x) = + 2 x x . . . . . .
- 142. Step 0 Find when f is positive, negative, zero, not defined. . . . . . .
- 143. Step 0 Find when f is positive, negative, zero, not defined. We need to factor f: 1 1 x+1 f(x) = + 2 = . x x x2 This means f is 0 at −1 and has trouble at 0. In fact, x+1 lim = ∞, x→0 x2 so x = 0 is a vertical asymptote of the graph. . . . . . .
- 144. Step 0 Find when f is positive, negative, zero, not defined. We need to factor f: 1 1 x+1 f(x) = + 2 = . x x x2 This means f is 0 at −1 and has trouble at 0. In fact, x+1 lim = ∞, x→0 x2 so x = 0 is a vertical asymptote of the graph. We can make a sign chart as follows: − . 0 .. . . + x . +1 − . 1 . + 0 .. . + .2 x 0 . − . .. . 0 + ∞ .. . + f . (x ) − . 1 0 . . . . . . .
- 145. For horizontal asymptotes, notice that x+1 lim = 0, x→∞ x2 so y = 0 is a horizontal asymptote of the graph. The same is true at −∞. . . . . . .
- 146. Step 1: Monotonicity . . . . . .
- 147. Step 1: Monotonicity We have 1 2 x+2 f ′ (x ) = − 2 − 3 =− 3 . x x x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . 0 .. . + .3 x 0 . . . . . . .
- 148. Step 1: Monotonicity We have 1 2 x+2 f ′ (x ) = − 2 − 3 =− 3 . x x x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . 0 .. . + .3 x 0 . − . 0 .. . + ∞ .. − . . ′ (x ) f − . 2 0 . f .(x) . . . . . .
- 149. Step 1: Monotonicity We have 1 2 x+2 f ′ (x ) = − 2 − 3 =− 3 . x x x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . 0 .. . + .3 x 0 . − .. . 0 . + ∞ .. − . . ′ (x ) f . − ↘ . 2 0 . f .(x) . . . . . .
- 150. Step 1: Monotonicity We have 1 2 x+2 f ′ (x ) = − 2 − 3 =− 3 . x x x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . 0 .. . + .3 x 0 . − .. . 0 . + ∞ .. − . . ′ (x ) f . − ↘ . 2 ↗ . 0 . f .(x) . . . . . .
- 151. Step 1: Monotonicity We have 1 2 x+2 f ′ (x ) = − 2 − 3 =− 3 . x x x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . .. . 0 + .3 x 0 . − .. . 0 . + ∞ − .. . . ′ (x ) f . − ↘ . 2 ↗ . 0 ↘ . . f .(x) . . . . . .
- 152. Step 1: Monotonicity We have 1 2 x+2 f ′ (x ) = − 2 − 3 =− 3 . x x x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . .. . 0 + .3 x 0 . − .. . 0 . + ∞ − .. . . ′ (x ) f . − ↘ . 2 ↗ . 0 ↘ . . f .(x) m . in . . . . . .
- 153. Step 1: Monotonicity We have 1 2 x+2 f ′ (x ) = − 2 − 3 =− 3 . x x x The critical points are x = −2 and x = 0. We have the following sign chart: . + 0 .. . − . − . (x + 2) − . 2 − . .. . 0 + .3 x 0 . − .. . 0 . + ∞ − .. . . ′ (x ) f . − ↘ . 2 ↗ . 0 ↘ . . f .(x) m . in V .A . . . . . .
- 154. Step 2: Concavity . . . . . .
- 155. Step 2: Concavity We have 2 6 2 (x + 3 ) f′′ (x) = + = . x 3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + 0 .. . + .4 x 0 . 0 .. ∞ .. .′ (x) f − . 3 0 . f .(x) . . . . . .
- 156. Step 2: Concavity We have 2 6 2 (x + 3 ) f′′ (x) = + = . x 3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + 0 .. . + .4 x 0 . − . − .. 0 ∞ .. .′ (x) f − . 3 0 . f .(x) . . . . . .
- 157. Step 2: Concavity We have 2 6 2 (x + 3 ) f′′ (x) = + = . x 3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + 0 .. . + .4 x 0 . − . − .. 0 . + + ∞ .. .′ (x) f − . 3 0 . f .(x) . . . . . .
- 158. Step 2: Concavity We have 2 6 2 (x + 3 ) f′′ (x) = + = . x 3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + .. . 0 + .4 x 0 . − . − .. 0 . + + .. . + ∞ + .′ (x) f − . 3 0 . f .(x) . . . . . .
- 159. Step 2: Concavity We have 2 6 2 (x + 3 ) f′′ (x) = + = . x 3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + .. . 0 + .4 x 0 . − . − .. 0 . + + .. . + ∞ + .′ (x) f . ⌢ . 3 − 0 . f .(x) . . . . . .
- 160. Step 2: Concavity We have 2 6 2 (x + 3 ) f′′ (x) = + = . x 3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + .. . 0 + .4 x 0 . − . − .. 0 . + + .. . + ∞ + .′ (x) f . ⌢ . 3 − . ⌣ 0 . f .(x) . . . . . .
- 161. Step 2: Concavity We have 2 6 2 (x + 3 ) f′′ (x) = + = . x 3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + .. . 0 + .4 x 0 . − . − .. 0 . + + .. . + ∞ + .′ (x) f . ⌢ . 3 − . ⌣ . . 0 ⌣ f .(x) . . . . . .
- 162. Step 2: Concavity We have 2 6 2 (x + 3 ) f′′ (x) = + = . x 3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + .. . 0 + .4 x 0 . − . − .. 0 . + + .. . + ∞ + .′ (x) f . ⌢ . 3 − . ⌣ . . 0 ⌣ f .(x) I .P . . . . . .
- 163. Step 2: Concavity We have 2 6 2 (x + 3 ) f′′ (x) = + = . x 3 x4 x4 The critical points of f′ are −3 and 0. Sign chart: − . 0 .. . . + . x + 3) ( − . 3 . + .. . 0 + .4 x 0 . − . − .. 0 . + + .. . + ∞ + .′ (x) f . ⌢ . 3 − . ⌣ . . 0 ⌣ f .(x) I .P V .A . . . . . .
- 164. Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f . − ↘ . 2 ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ − .. . − .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 . 1 . − + 0 . . + ∞s . . hape of f . . . . . .
- 165. Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f . − ↘ . 2 ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ − .. . − .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 . 1 . − + 0 . . + ∞s . . hape of f H . A . . . . . .
- 166. Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f . − ↘ . 2 ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ − .. . − .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 . 1 . − + 0 . . + ∞s . . hape of f . A . H . . . . . .
- 167. Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f . − ↘ . 2 ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ − .. . − .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 . 1 . − + 0 . . + ∞s . . hape of f . A . H I .P . . . . . .
- 168. Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f . − ↘ . 2 ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ − .. . − .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 . 1 . − + 0 . . + ∞s . . hape of f . A . H .P . I . . . . . .
- 169. Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f . − ↘ . 2 ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ − .. . − .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 . 1 . − + 0 . . + ∞s . . hape of f . A . H .P . . in I m . . . . . .
- 170. Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f . − ↘ . 2 ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ − .. . − .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 . 1 . − + 0 . . + ∞s . . hape of f . A . H .P . . in . I m . . . . . .
- 171. Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f . − ↘ . 2 ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ − .. . − .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 . 1 . − + 0 . . + ∞s . . hape of f . A . H .P . . in . I m 0 . . . . . . .
- 172. Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f . − ↘ . 2 ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ − .. . − .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 . 1 . − + 0 . . + ∞s . . hape of f . A . H .P . . in . I m 0 . . . . . . . .
- 173. Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f . − ↘ . 2 ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ − .. . − .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 . 1 . − + . 0 . + ∞s . . hape of f . A . H .P . . in . I m 0 . .A . V . . . . . .
- 174. Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f . − ↘ . 2 ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ − .. . − .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 . 1 . − + . 0 . + ∞s . . hape of f . A . H .P . . in . I m 0 . .A . . V . . . . . .
- 175. Step 3: Synthesis . − .. . 0 . + ∞ − .. . .′ f . − ↘ . 2 ↗ . 0 ↘ . . m . onotonicity − . − .. 0 . + + ∞ − .. . − .′′ f . ⌢ . 3 − . ⌣ . . 0 ⌣ c . oncavity 0 . − . 2/9 − . 1/4 0 .. ∞ .. 0f .. . . − . ∞ . . 3 − − − . 2 . 1 . − + . 0 . + ∞s . . hape of f . A . H .P . . in . I m 0 . .A . . A . V H . . . . . .
- 176. Step 4: Graph y . . x . . . . −3, −2/9) . −2, −1/4) ( ( . . . . . .
- 177. Problem Graph f(x) =cos x − x . . . . . .
- 178. Problem Graph f(x) =cos x − x y . . . . . . . .
- 179. Problem Graph f(x) =x ln x2 . . . . . .
- 180. Problem Graph f(x) =x ln x2 y . . x . . . . . . .