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### Lesson 21: Curve Sketching (slides)

• 1. Sec on 4.4 Curve Sketching V63.0121.001: Calculus I Professor Ma hew Leingang New York University April 13, 2011 .
• 2. Announcements Quiz 4 on Sec ons 3.3, 3.4, 3.5, and 3.7 this week (April 14/15) Quiz 5 on Sec ons 4.1–4.4 April 28/29 Final Exam Thursday May 12, 2:00–3:50pm I am teaching Calc II MW 2:00pm and Calc III TR 2:00pm both Fall ’11 and Spring ’12
• 3. Objectives given a func on, graph it completely, indica ng zeroes (if easy) asymptotes if applicable cri cal points local/global max/min inﬂec on points
• 4. Why? Graphing func ons is like dissec on
• 5. Why? Graphing func ons is like dissec on … or diagramming sentences
• 6. Why? Graphing func ons is like dissec on … or diagramming sentences You can really know a lot about a func on when you know all of its anatomy.
• 7. 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). Example f(x) f(x) = x3 + x2 .
• 8. 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). Example f(x) f′ (x) f(x) = x3 + x2 f′ (x) = 3x2 + 2x .
• 9. Testing for Concavity Theorem (Concavity Test) If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave downward on (a, b). Example f(x) f(x) = x3 + x2 .
• 10. Testing for Concavity Theorem (Concavity Test) If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave downward on (a, b). Example f′ (x) f(x) f(x) = x3 + x2 f′ (x) = 3x2 + 2x .
• 11. Testing for Concavity Theorem (Concavity Test) If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave downward on (a, b). Example f′′ (x) f′ (x) f(x) f(x) = x3 + x2 f′ (x) = 3x2 + 2x . f′′ (x) = 6x + 2
• 12. Graphing Checklist To graph a func on f, follow this plan: 0. Find when f is posi ve, nega ve, zero, not deﬁned.
• 13. Graphing Checklist To graph a func on f, follow this plan: 0. Find when f is posi ve, nega ve, zero, not deﬁned. 1. Find f′ and form its sign chart. Conclude informa on about increasing/decreasing and local max/min.
• 14. Graphing Checklist To graph a func on f, follow this plan: 0. Find when f is posi ve, nega ve, zero, not deﬁned. 1. Find f′ and form its sign chart. Conclude informa on about increasing/decreasing and local max/min. 2. Find f′′ and form its sign chart. Conclude concave up/concave down and inﬂec on.
• 15. Graphing Checklist To graph a func on f, follow this plan: 3. Put together a big chart to assemble monotonicity and concavity data
• 16. Graphing Checklist To graph a func on f, follow this plan: 3. Put together a big chart to assemble monotonicity and concavity data 4. Graph!
• 17. Outline Simple examples A cubic func on A quar c func on More Examples Points of nondiﬀeren ability Horizontal asymptotes Ver cal asymptotes Trigonometric and polynomial together Logarithmic
• 18. Graphing a cubic Example Graph f(x) = 2x3 − 3x2 − 12x.
• 19. 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 quadra c, 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.
• 20. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: .
• 21. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: . x−2 2 x+1 −1 f′ (x) −1 2 f(x)
• 22. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 x+1 −1 f′ (x) −1 2 f(x)
• 23. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) −1 2 f(x)
• 24. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + −1 2 f(x)
• 25. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − −1 2 f(x)
• 26. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − + −1 2 f(x)
• 27. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − + ↗−1 2 f(x)
• 28. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − + ↗−1 ↘ 2 f(x)
• 29. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − + ↗−1 ↘ 2 ↗ f(x)
• 30. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − + ↗−1 ↘ 2 ↗ f(x) max
• 31. Step 1: Monotonicity f(x) = 2x3 − 3x2 − 12x =⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2) We can form a sign chart from this: − .− + x−2 2 − + + x+1 −1 f′ (x) + − + ↗−1 ↘ 2 ↗ f(x) max min
• 32. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: .
• 33. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . f′′ (x) 1/2 f(x)
• 34. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . −− f′′ (x) 1/2 f(x)
• 35. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . −− ++ f′′ (x) 1/2 f(x)
• 36. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . −− ++ f′′ (x) ⌢ 1/2 f(x)
• 37. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . −− ++ f′′ (x) ⌢ 1/2 ⌣ f(x)
• 38. Step 2: Concavity f′ (x) = 6x2 − 6x − 12 =⇒ f′′ (x) = 12x − 6 = 6(2x − 1) Another sign chart: . −− ++ f′′ (x) ⌢ 1/2 ⌣ f(x) IP
• 39. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. .
• 40. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity
• 41. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity
• 42. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
• 43. monotonicity and concavity II I . III IV
• 44. monotonicity and concavity decreasing, concave down II I . III IV
• 45. monotonicity and concavity increasing, decreasing, concave concave down down II I . III IV
• 46. monotonicity and concavity increasing, decreasing, concave concave down down II I . III IV decreasing, concave up
• 47. monotonicity and concavity increasing, decreasing, concave concave down down II I . III IV decreasing, increasing, concave concave up up
• 48. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
• 49. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
• 50. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
• 51. Step 3: One sign chart to rule them all Remember, f(x) = 2x3 − 3x2 − 12x. + −. − + f′ (x) ↗−1 ↘ ↘ 2 ↗ monotonicity −− −− ++ ++ f′′ (x) ⌢ ⌢ 1/2 ⌣ ⌣ concavity 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
• 52. f(x) Step 4: Graph f(x) = 2x3 − 3x2 − 12x ( √ ) (−1, 7) 3− 105 4 ,0 (0, 0) . ( x√ ) (1/2, −61/2) 3+ 105 4 ,0 (2, −20) 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
• 53. f(x) Step 4: Graph f(x) = 2x3 − 3x2 − 12x ( √ ) (−1, 7) 3− 105 4 ,0 (0, 0) . ( x√ ) (1/2, −61/2) 3+ 105 4 ,0 (2, −20) 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
• 54. f(x) Step 4: Graph f(x) = 2x3 − 3x2 − 12x ( √ ) (−1, 7) 3− 105 4 ,0 (0, 0) . ( x√ ) (1/2, −61/2) 3+ 105 4 ,0 (2, −20) 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
• 55. f(x) Step 4: Graph f(x) = 2x3 − 3x2 − 12x ( √ ) (−1, 7) 3− 105 4 ,0 (0, 0) . ( x√ ) (1/2, −61/2) 3+ 105 4 ,0 (2, −20) 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
• 56. f(x) Step 4: Graph f(x) = 2x3 − 3x2 − 12x ( √ ) (−1, 7) 3− 105 4 ,0 (0, 0) . ( x√ ) (1/2, −61/2) 3+ 105 4 ,0 (2, −20) 7 −61/2 −20 f(x) −1 1/2 2 shape of f max IP min
• 57. Graphing a quartic Example Graph f(x) = x4 − 4x3 + 10
• 58. 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.
• 59. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
• 60. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. .
• 61. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. 0. 4x2 0
• 62. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0. 4x2 0
• 63. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0. + 4x2 0
• 64. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0. + + 4x2 0
• 65. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0. + + 4x2 0 0 (x − 3) 3
• 66. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − 0 (x − 3) 3
• 67. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0 (x − 3) 3
• 68. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3
• 69. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 f′ (x) 0 0 0 3 f(x)
• 70. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 f′ (x) −0 0 0 3 f(x)
• 71. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 f′ (x) −0 − 0 0 3 f(x)
• 72. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 ′ −0 − 0 + f (x) 0 3 f(x)
• 73. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 ′ −0 − 0 + f (x) ↘0 3 f(x)
• 74. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 ′ −0 − 0 + f (x) ↘0 ↘ 3 f(x)
• 75. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 ′ −0 − 0 + f (x) ↘0 ↘ 3 ↗ f(x)
• 76. Step 1: Monotonicity f(x) = x4 − 4x3 + 10 =⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3) We make its sign chart. +0 . + + 4x2 0 − − 0+ (x − 3) 3 ′ −0 − 0 + f (x) ↘0 ↘ 3 ↗ f(x) min
• 77. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) .
• 78. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: .
• 79. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: 0 . 12x 0
• 80. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0 . 12x 0
• 81. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0 . + 12x 0
• 82. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0 . + + 12x 0
• 83. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0 . + + 12x 0 0 x−2 2
• 84. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − 0 x−2 2
• 85. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 x−2 2
• 86. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2
• 87. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) 0 0 0 2 f(x)
• 88. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 0 0 2 f(x)
• 89. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 −− 0 0 2 f(x)
• 90. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 −− 0 ++ 0 2 f(x)
• 91. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 −− 0 ++ ⌣0 2 f(x)
• 92. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 −− 0 ++ ⌣0 ⌢ 2 f(x)
• 93. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 −− 0 ++ ⌣0 ⌢ 2 ⌣ f(x)
• 94. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − − 0 + x−2 2 f′′ (x) ++0 −− 0 ++ ⌣0 ⌢ 2 ⌣ f(x) IP
• 95. Step 2: Concavity f′ (x) = 4x3 − 12x2 =⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2) Here is its sign chart: −0. + + 12x 0 − −0 + x−2 2 f′′ (x) ++0 −− 0 ++ ⌣0 ⌢ 2 ⌣ f(x) IP IP
• 96. Step 3: Grand Uniﬁed Sign Chart . Remember, f(x) = x − 4x3 + 10. 4 −0 − −0+ f′ (x) ↘0 ↘ ↘3↗ monotonicity f′′ (x) ++0 −− 0++ ++ ⌣0 ⌢ 2⌣ ⌣ concavity 10 −6 −17 f(x) 0 2 3 shape IP IP min
• 97. Step 3: Grand Uniﬁed Sign Chart . Remember, f(x) = x − 4x3 + 10. 4 −0 − −0+ f′ (x) ↘0 ↘ ↘3↗ monotonicity f′′ (x) ++0 −− 0++ ++ ⌣0 ⌢ 2⌣ ⌣ concavity 10 −6 −17 f(x) 0 2 3 shape IP IP min
• 98. Step 3: Grand Uniﬁed Sign Chart . Remember, f(x) = x − 4x3 + 10. 4 −0 − −0+ f′ (x) ↘0 ↘ ↘3↗ monotonicity f′′ (x) ++0 −− 0++ ++ ⌣0 ⌢ 2⌣ ⌣ concavity 10 −6 −17 f(x) 0 2 3 shape IP IP min
• 99. Step 3: Grand Uniﬁed Sign Chart . Remember, f(x) = x − 4x3 + 10. 4 −0 − −0+ f′ (x) ↘0 ↘ ↘3↗ monotonicity f′′ (x) ++0 −− 0++ ++ ⌣0 ⌢ 2⌣ ⌣ concavity 10 −6 −17 f(x) 0 2 3 shape IP IP min
• 100. Step 3: Grand Uniﬁed Sign Chart . Remember, f(x) = x − 4x3 + 10. 4 −0 − −0+ f′ (x) ↘0 ↘ ↘3↗ monotonicity f′′ (x) ++0 −− 0++ ++ ⌣0 ⌢ 2⌣ ⌣ concavity 10 −6 −17 f(x) 0 2 3 shape IP IP min
• 101. y Step 4: Graph f(x) = x4 − 4x3 + 10 (0, 10) . x (2, −6) (3, −17) 10 −6 −17 f(x) 0 2 3 shape IP IP min
• 102. y Step 4: Graph f(x) = x4 − 4x3 + 10 (0, 10) . x (2, −6) (3, −17) 10 −6 −17 f(x) 0 2 3 shape IP IP min
• 103. y Step 4: Graph f(x) = x4 − 4x3 + 10 (0, 10) . x (2, −6) (3, −17) 10 −6 −17 f(x) 0 2 3 shape IP IP min
• 104. y Step 4: Graph f(x) = x4 − 4x3 + 10 (0, 10) . x (2, −6) (3, −17) 10 −6 −17 f(x) 0 2 3 shape IP IP min
• 105. y Step 4: Graph f(x) = x4 − 4x3 + 10 (0, 10) . x (2, −6) (3, −17) 10 −6 −17 f(x) 0 2 3 shape IP IP min
• 106. Outline Simple examples A cubic func on A quar c func on More Examples Points of nondiﬀeren ability Horizontal asymptotes Ver cal asymptotes Trigonometric and polynomial together Logarithmic
• 107. Graphing a function with a cusp Example √ Graph f(x) = x + |x|
• 108. Graphing a function with a cusp Example √ Graph f(x) = x + |x| This func on looks strange because of the absolute value. But whenever we become nervous, we can just take cases.
• 109. 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 posi ve.
• 110. 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 posi ve. Are there nega ve numbers which are zeroes for f?
• 111. 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 posi ve. Are there nega ve numbers which are zeroes for f? √ √ x + −x = 0 =⇒ −x = −x −x = x2 =⇒ x2 + x = 0 The only solu ons are x = 0 and x = −1.
• 112. Step 0: Asymptotic behavior √ f(x) = x + |x| lim f(x) = ∞, because both terms tend to ∞. x→∞
• 113. 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 same x→−∞ √ as lim (−y + y) y→+∞
• 114. 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 same x→−∞ √ as lim (−y + y) y→+∞ √ √ √ y+y lim (−y + y) = lim ( y − y) · √ y→+∞ y→∞ y+y y − y2 = lim √ = −∞ y→∞ y+y
• 115. 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
• 116. 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 No ce f′ (x) > 0 when x > 0 (so no cri cal points here)
• 117. 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 No ce f′ (x) > 0 when x > 0 (so no cri cal points here) lim+ f′ (x) = ∞ (so 0 is a cri cal point) x→0
• 118. 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 No ce f′ (x) > 0 when x > 0 (so no cri cal points here) lim+ f′ (x) = ∞ (so 0 is a cri cal point) x→0 lim f′ (x) = 1 (so the graph is asympto c to a line of slope 1) x→∞
• 119. Step 1: The derivative √ Remember, f(x) = x + |x|. If x is nega ve, we have d ( √ ) 1 f′ (x) = x + −x = 1 − √ dx 2 −x No ce lim− f′ (x) = −∞ (other side of the cri cal point) x→0
• 120. Step 1: The derivative √ Remember, f(x) = x + |x|. If x is nega ve, we have d ( √ ) 1 f′ (x) = x + −x = 1 − √ dx 2 −x No ce lim− f′ (x) = −∞ (other side of the cri cal point) x→0 lim f′ (x) = 1 (asympto c to a line of slope 1) x→−∞
• 121. Step 1: The derivative √ Remember, f(x) = x + |x|. If x is nega ve, we have d ( √ ) 1 f′ (x) = x + −x = 1 − √ dx 2 −x No ce lim− f′ (x) = −∞ (other side of the cri cal point) x→0 lim f′ (x) = 1 (asympto c to a line of slope 1) x→−∞ ′ f (x) = 0 when 1 √ 1 1 1 1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = − 2 −x 2 4 4
• 122. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. f′ (x) . f(x)
• 123. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. 0 f′ (x) . −1 4 f(x)
• 124. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. 0 ∞ f′ (x) . −1 4 0 f(x)
• 125. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0 ∞ f′ (x) . −1 4 0 f(x)
• 126. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ f′ (x) . −4 0 1 f(x)
• 127. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ + f′ (x) . −4 0 1 f(x)
• 128. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ + f′ (x) . ↗ −4 0 1 f(x)
• 129. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ + f′ (x) . ↗ −4 1↘ 0 f(x)
• 130. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ + f′ (x) . ↗ −4 1↘ 0 ↗ f(x)
• 131. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ + f′ (x) . ↗ −41↘ 0 ↗ f(x) max
• 132. Step 1: Monotonicity  1 1 + √  if x > 0 ′ f (x) = 2 x 1 − √ 1  if x < 0 2 −x We can’t make a mul -factor sign chart because of the absolute value, but we can test points in between cri cal points. + 0− ∞ + f′ (x) . ↗ −41↘ 0 ↗ f(x) max min
• 133. Step 2: Concavity ( ) d 1 1 If x > 0, then f′′ (x) = 1 + x−1/2 = − x−3/2 This is dx 2 4 nega ve whenever x > 0.
• 134. Step 2: Concavity ( ) d 1 −1/2 1 If x > 0, then f′′ (x) = 1+ x = − x−3/2 This is dx 2 4 nega ve whenever x > 0. ( ) ′′ d 1 −1/2 1 If x < 0, then f (x) = 1 − (−x) = − (−x)−3/2 dx 2 4 which is also always nega ve for nega ve x.
• 135. Step 2: Concavity ( ) d 1 −1/2 1 If x > 0, then f′′ (x) = 1+ x = − x−3/2 This is dx 2 4 nega ve whenever x > 0. ( ) ′′ d 1 −1/2 1 If x < 0, then f (x) = 1 − (−x) = − (−x)−3/2 dx 2 4 which is also always nega ve for nega ve x. 1 In other words, f′′ (x) = − |x|−3/2 . 4
• 136. Step 2: Concavity ( ) d 1 −1/2 1 If x > 0, then f′′ (x) = 1+ x = − x−3/2 This is dx 2 4 nega ve whenever x > 0. ( ) ′′ d 1 −1/2 1 If x < 0, then f (x) = 1 − (−x) = − (−x)−3/2 dx 2 4 which is also always nega ve for nega ve x. 1 In other words, f′′ (x) = − |x|−3/2 . 4 Here is the sign chart: −− −∞ −− f′′ (x) . ⌢ 0 ⌢ f(x)
• 137. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| +1 + 0− ∞ + f′ +1 (x) . ↗ ↗ 1↘ 0 ↗ ↗monotonicity −∞ −− − −− 4 −∞ −− f′′ −∞ (x) ⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity −∞ 0 4 0 +∞f(x) −1 −4 0 1 shape zero max min
• 138. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| +1 + 0− ∞ + f′ +1 (x) . ↗ ↗ 1↘ 0 ↗ ↗monotonicity −∞ −− − −− 4 −∞ −− f′′ −∞ (x) ⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity −∞ 0 4 0 +∞f(x) −1 −4 0 1 shape zero max min
• 139. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| +1 + 0− ∞ + f′ +1 (x) . ↗ ↗ 1↘ 0 ↗ ↗monotonicity −∞ −− − −− 4 −∞ −− f′′ −∞ (x) ⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity −∞ 0 4 0 +∞f(x) −1 −4 0 1 shape zero max min
• 140. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| +1 + 0− ∞ + f′ +1 (x) . ↗ ↗ 1↘ 0 ↗ ↗monotonicity −∞ −− − −− 4 −∞ −− f′′ −∞ (x) ⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity −∞ 0 4 0 +∞f(x) −1 −4 0 1 shape zero max min
• 141. Step 3: Synthesis Now we can put these things together. √ f(x) = x + |x| +1 + 0− ∞ + f′ +1 (x) . ↗ ↗ 1↘ 0 ↗ ↗monotonicity −∞ −− − −− 4 −∞ −− f′′ −∞ (x) ⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity −∞ 0 4 0 +∞f(x) −1 −4 0 1 shape zero max min
• 142. Graph √ f(x) = x + |x| . x 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
• 143. Graph √ f(x) = x + |x| (−1, 0) . x 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
• 144. Graph √ f(x) = x + |x| (− 1 , 1 ) 4 4 (−1, 0) . x 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
• 145. Graph √ f(x) = x + |x| (− 1 , 1 ) 4 4 (−1, 0) . x (0, 0) 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
• 146. Graph √ f(x) = x + |x| (− 1 , 1 ) 4 4 (−1, 0) . x (0, 0) 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
• 147. Graph √ f(x) = x + |x| (− 1 , 1 ) 4 4 (−1, 0) . x (0, 0) 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
• 148. Graph √ f(x) = x + |x| (− 1 , 1 ) 4 4 (−1, 0) . x (0, 0) 1 −∞0 4 0 +∞ x −1 −1 0 shape 4 zero max min
• 149. Example with Horizontal Asymptotes Example Graph f(x) = xe−x 2
• 150. Example with Horizontal Asymptotes Example Graph f(x) = xe−x 2 Before taking deriva ves, we no ce that f is odd, that f(0) = 0, and lim f(x) = 0 x→∞
• 151. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e . √0 √ 1− 2x 0 1/2 √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
• 152. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + . √0 √ 1− 2x 0 1/2 √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
• 153. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ √ 1− 2x 0 1/2 √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
• 154. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x 0 1/2 √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
• 155. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 1/2 √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
• 156. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
• 157. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
• 158. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 0 f′ (x) √ √ − 1/2 1/2 f(x)
• 159. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 f′ (x) √ √ − 1/2 1/2 f(x)
• 160. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 − f′ (x) √ √ ↘ − 1/2 1/2 f(x)
• 161. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 − f′ (x) √ √ ↘ − 1/2 ↗ 1/2 f(x)
• 162. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 − f′ (x) √ √ ↘ − 1/2 ↗ 1/2 ↘ f(x)
• 163. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 − f′ (x) √ √ ↘ − 1/2 ↗ 1/2 ↘ f(x)
• 164. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 − f′ (x) √ √ ↘ − 1/2 ↗ 1/2 ↘ f(x) min
• 165. Step 1: −xMonotonicity 2 If f(x) = xe , then ( ) 2 f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x 2 2 ( √ )( √ ) −x2 = 1 − 2x 1 + 2x e + + 0 . √ − √ 1− 2x − 0 + 1/2 + √ √ 1 + 2x − − 0 1/2 + 0 − f′ (x) √ √ ↘ − 1/2 ↗ 1/2 ↘ f(x) min max
• 166. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 0. 2x 0 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 167. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − 0. 2x 0 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 168. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. 2x 0 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 169. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + 2x 0 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 170. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 171. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 172. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 173. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 174. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 175. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 176. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 177. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 178. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 179. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 180. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 181. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 f′′ (x) √ √ − 3/2 0 3/2 f(x)
• 182. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ √ − 3/2 0 3/2 f(x)
• 183. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 3/2 f(x)
• 184. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 ⌢ 3/2 f(x)
• 185. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 ⌢ 3/2 ⌣ f(x)
• 186. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 ⌢ 3/2 ⌣ f(x)
• 187. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 ⌢ 3/2 ⌣ f(x) IP
• 188. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 ⌢ 3/2 ⌣ f(x) IP IP
• 189. Step 2: Concavity ′ 2 −x 2 If f (x) = (1 − 2x )e , we know ( ) 2 f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x 2 2 = 2x(2x2 − 3)e−x 2 − − 0. + + 2x 0 − − − 0 + √ √ √ 2x − 3 − 0 + + 3/2 + √ √ √ 2x + 3 −− − 0 3/2 ++ 0 −− 0 ++ f′′ (x) ⌢ √ ⌣ √ − 3/2 0 ⌢ 3/2 ⌣ f(x) IP IP IP
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