Lesson 23: Antiderivatives (slides)

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At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.

At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.

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  • 1. Sec on 4.7 An deriva ves V63.0121.001: Calculus I Professor Ma hew Leingang New York University April 19, 2011.
  • 2. Announcements 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 ”simple“ elementary func on, find a func on whose deriva ve is that func on. Remember that a func on whose deriva ve is zero along an interval must be zero along that interval. Solve problems involving rec linear mo on.
  • 4. Outline What is an an deriva ve? Tabula ng An deriva ves Power func ons Combina ons Exponen al func ons Trigonometric func ons An deriva ves of piecewise func ons Finding An deriva ves Graphically Rec linear mo on
  • 5. What is an antiderivative? Defini on Let f be a func on. An an deriva ve for f is a func on F such that F′ = f.
  • 6. Who cares? Ques on Why would we want the an deriva ve of a func on? Answers For the challenge of it For applica ons when the deriva ve of a func on is known but the original func on is not Biggest applica on will be a er the Fundamental Theorem of Calculus (Chapter 5)
  • 7. Hard problem, easy check Example Find an an deriva ve for f(x) = ln x.
  • 8. Hard problem, easy check Example Find an an deriva ve for f(x) = ln x. Solu on ???
  • 9. Hard problem, easy check Example is F(x) = x ln x − x an an deriva ve for f(x) = ln x?
  • 10. Hard problem, easy check Example is F(x) = x ln x − x an an deriva ve for f(x) = ln x? Solu on d dx 1 (x ln x − x) = 1 · ln x + x · − 1 = ln x x 
  • 11. Why the MVT is the MITCMost Important Theorem In Calculus! Theorem Let f′ = 0 on an interval (a, b). Then f is constant on (a, b). Proof. Pick any points x and y in (a, b) with x < y. By MVT there exists a point z in (x, y) such that f(y) = f(x) + f′ (z)(y − x) But f′ (z) = 0, so f(y) = f(x). Since this is true for all x and y in (a, b), then f is constant.
  • 12. Functions with the same derivative Theorem Suppose f and g are two differen able func ons on (a, b) with f′ = g′ . Then f and g differ by a constant. That is, there exists a constant C such that f(x) = g(x) + C. Proof. Let h(x) = f(x) − g(x) Then h′ (x) = f′ (x) − g′ (x) = 0 on (a, b) So h(x) = C, a constant This means f(x) − g(x) = C on (a, b)
  • 13. Outline What is an an deriva ve? Tabula ng An deriva ves Power func ons Combina ons Exponen al func ons Trigonometric func ons An deriva ves of piecewise func ons Finding An deriva ves Graphically Rec linear mo on
  • 14. Antiderivatives of power functions Recall that the deriva ve of a y power func on is a power f(x) = x2 func on. Fact (The Power Rule) If f(x) = xr , then f′ (x) = rxr−1 . . x
  • 15. Antiderivatives of power functions ′ Recall that the deriva ve of a yf (x) = 2x power func on is a power f(x) = x2 func on. Fact (The Power Rule) If f(x) = xr , then f′ (x) = rxr−1 . . x
  • 16. Antiderivatives of power functions ′ Recall that the deriva ve of a yf (x) = 2x power func on is a power f(x) = x2 func on. Fact (The Power Rule) F(x) = ? If f(x) = xr , then f′ (x) = rxr−1 . . x
  • 17. Antiderivatives of power functions ′ Recall that the deriva ve of a yf (x) = 2x power func on is a power f(x) = x2 func on. Fact (The Power Rule) F(x) = ? If f(x) = xr , then f′ (x) = rxr−1 . So in looking for an deriva ves of power . x func ons, try power func ons!
  • 18. Antiderivatives of power functions Example Find an an deriva ve for the func on f(x) = x3 . Solu on
  • 19. Antiderivatives of power functions Example Find an an deriva ve for the func on f(x) = x3 . Solu on Try a power func on F(x) = axr
  • 20. Antiderivatives of power functions Example Find an an deriva ve for the func on f(x) = x3 . Solu on Try a power func on F(x) = axr Then F′ (x) = arxr−1 , so we want arxr−1 = x3 .
  • 21. Antiderivatives of power functions Example Find an an deriva ve for the func on f(x) = x3 . Solu on Try a power func on F(x) = axr Then F′ (x) = arxr−1 , so we want arxr−1 = x3 . 1 r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4
  • 22. Antiderivatives of power functions Example Find an an deriva ve for the func on f(x) = x3 . Solu on Try a power func on F(x) = axr Then F′ (x) = arxr−1 , so we want arxr−1 = x3 . 1 r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4
  • 23. Antiderivatives of power functions Example Find an an deriva ve for the func on f(x) = x3 . Solu on 1 So F(x) = x4 is an an deriva ve. 4
  • 24. Antiderivatives of power functions Example Find an an deriva ve for the func on f(x) = x3 . Solu on 1 So F(x) = x4 is an an deriva ve. 4 ( ) Check: d 1 4 dx 4 1 x = 4 · x4−1 = x3 4 
  • 25. Antiderivatives of power functions Example Find an an deriva ve for the func on f(x) = x3 . Solu on 1 So F(x) = x4 is an an deriva ve. 4 ( ) Check: d 1 4 dx 4 1 x = 4 · x4−1 = x3 4  Any others?
  • 26. Antiderivatives of power functions Example Find an an deriva ve for the func on f(x) = x3 . Solu on 1 So F(x) = x4 is an an deriva ve. 4 ( ) Check: d 1 4 dx 4 1 x = 4 · x4−1 = x3 4  1 Any others? Yes, F(x) = x4 + C is the most general form. 4
  • 27. General power functions Fact (The Power Rule for an deriva ves) If f(x) = xr , then 1 r+1 F(x) = x r+1 is an an deriva ve for f…
  • 28. General power functions Fact (The Power Rule for an deriva ves) If f(x) = xr , then 1 r+1 F(x) = x r+1 is an an deriva ve for f as long as r ̸= −1.
  • 29. General power functions Fact (The Power Rule for an deriva ves) If f(x) = xr , then 1 r+1 F(x) = x r+1 is an an deriva ve for f as long as r ̸= −1. Fact 1 If f(x) = x−1 = , then F(x) = ln |x| + C is an an deriva ve for f. x
  • 30. What’s with the absolute value? { ln(x) if x > 0; F(x) = ln |x| = ln(−x) if x < 0. The domain of F is all nonzero numbers, while ln x is only defined on posi ve numbers.
  • 31. What’s with the absolute value? { ln(x) if x > 0; F(x) = ln |x| = ln(−x) if x < 0. The domain of F is all nonzero numbers, while ln x is only defined on posi ve numbers. d If x > 0, ln |x| dx
  • 32. What’s with the absolute value? { ln(x) if x > 0; F(x) = ln |x| = ln(−x) if x < 0. The domain of F is all nonzero numbers, while ln x is only defined on posi ve numbers. d d If x > 0, ln |x| = ln(x) dx dx
  • 33. What’s with the absolute value? { ln(x) if x > 0; F(x) = ln |x| = ln(−x) if x < 0. The domain of F is all nonzero numbers, while ln x is only defined on posi ve numbers. d d 1 If x > 0, ln |x| = ln(x) = dx dx x
  • 34. What’s with the absolute value? { ln(x) if x > 0; F(x) = ln |x| = ln(−x) if x < 0. The domain of F is all nonzero numbers, while ln x is only defined on posi ve numbers. If x > 0, d dx ln |x| = d dx ln(x) = 1 x 
  • 35. What’s with the absolute value? { ln(x) if x > 0; F(x) = ln |x| = ln(−x) if x < 0. The domain of F is all nonzero numbers, while ln x is only defined on posi ve numbers. If x > 0, d dx ln |x| = d dx ln(x) = 1 x  d If x < 0, ln |x| dx
  • 36. What’s with the absolute value? { ln(x) if x > 0; F(x) = ln |x| = ln(−x) if x < 0. The domain of F is all nonzero numbers, while ln x is only defined on posi ve numbers. If x > 0, d dx ln |x| = d dx ln(x) = 1 x  d d If x < 0, ln |x| = ln(−x) dx dx
  • 37. What’s with the absolute value? { ln(x) if x > 0; F(x) = ln |x| = ln(−x) if x < 0. The domain of F is all nonzero numbers, while ln x is only defined on posi ve numbers. If x > 0, d dx ln |x| = d dx ln(x) = 1 x  d d 1 If x < 0, ln |x| = ln(−x) = · (−1) dx dx −x
  • 38. What’s with the absolute value? { ln(x) if x > 0; F(x) = ln |x| = ln(−x) if x < 0. The domain of F is all nonzero numbers, while ln x is only defined on posi ve numbers. If x > 0, d dx ln |x| = d dx ln(x) = 1 x  d d 1 1 If x < 0, ln |x| = ln(−x) = · (−1) = dx dx −x x
  • 39. What’s with the absolute value? { ln(x) if x > 0; F(x) = ln |x| = ln(−x) if x < 0. The domain of F is all nonzero numbers, while ln x is only defined on posi ve numbers. If x > 0, d dx ln |x| = d dx ln(x) = 1 x  If x < 0, d dx ln |x| = d dx ln(−x) = 1 −x · (−1) = 1 x 
  • 40. What’s with the absolute value? { ln(x) if x > 0; F(x) = ln |x| = ln(−x) if x < 0. The domain of F is all nonzero numbers, while ln x is only defined on posi ve numbers. If x > 0, d dx ln |x| = d dx ln(x) = 1 x  If x < 0, d dx ln |x| = d dx ln(−x) = 1 −x · (−1) = 1 x  We prefer the an deriva ve with the larger domain.
  • 41. Graph of ln |x| y . x = 1/x f(x)
  • 42. Graph of ln |x| y F(x) = ln(x) . x = 1/x f(x)
  • 43. Graph of ln |x| y F(x) = ln |x| . x = 1/x f(x)
  • 44. Combinations of antiderivatives Fact (Sum and Constant Mul ple Rule for An deriva ves) If F is an an deriva ve of f and G is an an deriva ve of g, then F + G is an an deriva ve of f + g. If F is an an deriva ve of f and c is a constant, then cF is an an deriva ve of cf.
  • 45. Combinations of antiderivatives Proof. These follow from the sum and constant mul ple rule for deriva ves: If F′ = f and G′ = g, then (F + G)′ = F′ + G′ = f + g Or, if F′ = f, (cF)′ = cF′ = cf
  • 46. Antiderivatives of Polynomials Example Find an an deriva ve for f(x) = 16x + 5.
  • 47. Antiderivatives of Polynomials Example Find an an deriva ve for f(x) = 16x + 5. Solu on 1 The expression x2 is an an deriva ve for x, and x is an 2 an deriva ve for 1. So ( ) 1 2 F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C 2 is the an deriva ve of f.
  • 48. Antiderivatives of Polynomials Ques on Do we need two C’s or just one?
  • 49. Antiderivatives of Polynomials Ques on Do we need two C’s or just one? Answer Just one. A combina on of two arbitrary constants is s ll an arbitrary constant.
  • 50. Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax .
  • 51. Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . Accordingly, Fact 1 x If f(x) = ax , then F(x) = a + C is the an deriva ve of f. ln a
  • 52. Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . Accordingly, Fact 1 x If f(x) = ax , then F(x) = a + C is the an deriva ve of f. ln a Proof. Check it yourself.
  • 53. Exponential Functions In par cular,
  • 54. Exponential Functions In par cular, Fact If f(x) = ex , then F(x) = ex + C is the an deriva ve of f.
  • 55. Logarithmic functions? Remember we found F(x) = x ln x − x is an an deriva ve of f(x) = ln x.
  • 56. Logarithmic functions? Remember we found F(x) = x ln x − x is an an deriva ve of f(x) = ln x. This is not obvious. See Calc II for the full story.
  • 57. Logarithmic functions? Remember we found F(x) = x ln x − x is an an deriva ve of f(x) = ln x. This is not obvious. See Calc II for the full story. ln x However, using the fact that loga x = , we get: ln a Fact If f(x) = loga (x) 1 1 F(x) = (x ln x − x) + C = x loga x − x+C ln a ln a is the an deriva ve of f(x).
  • 58. Trigonometric functions Fact d d sin x = cos x cos x = − sin x dx dx
  • 59. Trigonometric functions Fact d d sin x = cos x cos x = − sin x dx dx So to turn these around, Fact The func on F(x) = − cos x + C is the an deriva ve of f(x) = sin x.
  • 60. Trigonometric functions Fact d d sin x = cos x cos x = − sin x dx dx So to turn these around, Fact The func on F(x) = − cos x + C is the an deriva ve of f(x) = sin x. The func on F(x) = sin x + C is the an deriva ve of f(x) = cos x.
  • 61. More Trig Example Find an an deriva ve of f(x) = tan x.
  • 62. More Trig Example Find an an deriva ve of f(x) = tan x. Solu on ???
  • 63. More Trig Example Find an an deriva ve of f(x) = tan x. Answer F(x) = ln | sec x|.
  • 64. More Trig Example Find an an deriva ve of f(x) = tan x. Answer F(x) = ln | sec x|. Check d 1 d = · sec x dx sec x dx
  • 65. More Trig Example Find an an deriva ve of f(x) = tan x. Answer F(x) = ln | sec x|. Check d 1 d = · sec x dx sec x dx
  • 66. More Trig Example Find an an deriva ve of f(x) = tan x. Answer F(x) = ln | sec x|. Check d 1 d 1 = · sec x = · sec x tan x dx sec x dx sec x
  • 67. More Trig Example Find an an deriva ve of f(x) = tan x. Answer F(x) = ln | sec x|. Check d 1 d 1 = · sec x = · sec x tan x = tan x dx sec x dx sec x
  • 68. More Trig Example Find an an deriva ve of f(x) = tan x. Answer F(x) = ln | sec x|. Check d = 1 · d dx sec x dx sec x = 1 sec x · sec x tan x = tan x 
  • 69. More Trig Example Find an an deriva ve of f(x) = tan x. Answer F(x) = ln | sec x|. Check d = 1 · d dx sec x dx sec x = 1 sec x · sec x tan x = tan x  More about this later.
  • 70. Antiderivatives of piecewise functions Example Let { x if 0 ≤ x ≤ 1; f(x) = 1 − x2 if 1 < x. Find the an deriva ve of f with F(0) = 1.
  • 71. Antiderivatives of piecewise functions Solu on We can an differen ate each piece:  1 2  x + C1 if 0 ≤ x ≤ 1; F(x) = 2 1 x − x3 + C  if 1 < x. 2 3 The constants need to be chosen so that F(0) = 1 and F is con nuous (at 1).
  • 72.  1 2  x + C1 if 0 ≤ x ≤ 1; F(x) = 2 1 x − x3 + C  if 1 < x. 2 3 1Note F(0) = 02 + C1 = C1 , so if F(0) is to be 1, C1 = 1. 2
  • 73.  1 2  x + C1 if 0 ≤ x ≤ 1; F(x) = 2 1 x − x3 + C  if 1 < x. 2 3 1Note F(0) = 02 + C1 = C1 , so if F(0) is to be 1, C1 = 1. 2 1 3This means lim− F(x) = 12 + 1 = . x→1 2 2
  • 74.  1 2  x + C1 if 0 ≤ x ≤ 1; F(x) = 2 1 x − x3 + C  if 1 < x. 2 3 1Note F(0) = 02 + C1 = C1 , so if F(0) is to be 1, C1 = 1. 2 1 3This means lim− F(x) = 12 + 1 = . x→1 2 2On the other hand, 1 2 lim+ F(x) = 1 − + C 2 = + C2 x→1 3 3 3 2 5So for F to be con nuous we need = + C2 . Solving, C2 = . 2 3 6
  • 75. Outline What is an an deriva ve? Tabula ng An deriva ves Power func ons Combina ons Exponen al func ons Trigonometric func ons An deriva ves of piecewise func ons Finding An deriva ves Graphically Rec linear mo on
  • 76. Finding Antiderivatives Graphically y Problem Pictured is the graph of a y = f(x) func on f. Draw the graph of . an an deriva ve for f. x 1 2 3 4 5 6
  • 77. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: y f = F′ . 1 2 3 4 5 6F f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 78. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: y + f = F′ . 1 2 3 4 5 6F f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 79. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: y + + f = F′ . 1 2 3 4 5 6F f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 80. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: y + + − f = F′ . 1 2 3 4 5 6F f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 81. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: y + + − − f = F′ . 1 2 3 4 5 6F f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 82. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1 2 3 4 5 6F f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 83. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2 3 4 5 6 F f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 84. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3 4 5 6 F f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 85. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4 5 6 F f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 86. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5 6 F f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 87. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 88. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 89. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 90. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ++ f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 91. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ++ −− f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 92. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ++ −−−− f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 93. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ++ −−−− ++ f′ = F′′ . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 94. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 95. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 96. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ ⌢ 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 97. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ ⌢ ⌢ 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 98. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ ⌢ ⌢ ⌣ 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 99. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ ⌢ ⌢ ⌣ ⌣ 1 2 3 4 5 6 x 1 2 3 4 5 6F F 1 2 3 4 5 6 shape
  • 100. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ ⌢ ⌢ ⌣ ⌣ 1 2 3 4 5 6 x 1 2 3 4 5 6F IP F 1 2 3 4 5 6 shape
  • 101. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ ⌢ ⌢ ⌣ ⌣ 1 2 3 4 5 6 x 1 2 3 4 5 6F IP IP F 1 2 3 4 5 6 shape
  • 102. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ ⌢ ⌢ ⌣ ⌣ 1 2 3 4 5 6 x 1 2 3 4 5 6F IP IP F 1 2 3 4 5 6 shape
  • 103. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ ⌢ ⌢ ⌣ ⌣ 1 2 3 4 5 6 x 1 2 3 4 5 6F IP IP F 1 2 3 4 5 6 shape
  • 104. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ ⌢ ⌢ ⌣ ⌣ 1 2 3 4 5 6 x 1 2 3 4 5 6F IP IP F 1 2 3 4 5 6 shape
  • 105. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ ⌢ ⌢ ⌣ ⌣ 1 2 3 4 5 6 x 1 2 3 4 5 6F IP IP F 1 2 3 4 5 6 shape
  • 106. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ ⌢ ⌢ ⌣ ⌣ 1 2 3 4 5 6 x 1 2 3 4 5 6F IP IP F 1 2 3 4 5 6 shape
  • 107. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for F: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ ⌢ ⌢ ⌣ ⌣ 1 2 3 4 5 6 x 1 2 3 4 5 6F IP IP ? ? ? ? ? ?F 1 2 3 4 5 6 shape The only ques on le is: What are the func on values?
  • 108. Could you repeat the question? Problem Below is the graph of a func on f. Draw the graph of the an deriva ve for f with F(1) = 0. Solu on y f . x 1 2 3 4 5 6 F 1 2 3 4 5 6 shape IP max IP min
  • 109. Could you repeat the question? Problem Below is the graph of a func on f. Draw the graph of the an deriva ve for f with F(1) = 0. Solu on y We start with F(1) = 0. f . x 1 2 3 4 5 6 F 1 2 3 4 5 6 shape IP max IP min
  • 110. Could you repeat the question? Problem Below is the graph of a func on f. Draw the graph of the an deriva ve for f with F(1) = 0. Solu on y We start with F(1) = 0. f . Using the sign chart, we draw arcs x 1 2 3 4 5 6 with the specified monotonicity and concavity F 1 2 3 4 5 6 shape IP max IP min
  • 111. Could you repeat the question? Problem Below is the graph of a func on f. Draw the graph of the an deriva ve for f with F(1) = 0. Solu on y We start with F(1) = 0. f . Using the sign chart, we draw arcs x 1 2 3 4 5 6 with the specified monotonicity and concavity F 1 2 3 4 5 6 shape IP max IP min
  • 112. Could you repeat the question? Problem Below is the graph of a func on f. Draw the graph of the an deriva ve for f with F(1) = 0. Solu on y We start with F(1) = 0. f . Using the sign chart, we draw arcs x 1 2 3 4 5 6 with the specified monotonicity and concavity F 1 2 3 4 5 6 shape IP max IP min
  • 113. Could you repeat the question? Problem Below is the graph of a func on f. Draw the graph of the an deriva ve for f with F(1) = 0. Solu on y We start with F(1) = 0. f . Using the sign chart, we draw arcs x 1 2 3 4 5 6 with the specified monotonicity and concavity F 1 2 3 4 5 6 shape IP max IP min
  • 114. Could you repeat the question? Problem Below is the graph of a func on f. Draw the graph of the an deriva ve for f with F(1) = 0. Solu on y We start with F(1) = 0. f . Using the sign chart, we draw arcs x 1 2 3 4 5 6 with the specified monotonicity and concavity F 1 2 3 4 5 6 shape IP max IP min
  • 115. Could you repeat the question? Problem Below is the graph of a func on f. Draw the graph of the an deriva ve for f with F(1) = 0. Solu on y We start with F(1) = 0. f . Using the sign chart, we draw arcs x 1 2 3 4 5 6 with the specified monotonicity and concavity F 1 2 3 4 5 6 shape IP max IP min
  • 116. Could you repeat the question? Problem Below is the graph of a func on f. Draw the graph of the an deriva ve for f with F(1) = 0. Solu on y We start with F(1) = 0. f . Using the sign chart, we draw arcs x 1 2 3 4 5 6 with the specified monotonicity and concavity F 1 2 3 4 5 6 shape IP max IP min
  • 117. Could you repeat the question? Problem Below is the graph of a func on f. Draw the graph of the an deriva ve for f with F(1) = 0. Solu on y We start with F(1) = 0. f . Using the sign chart, we draw arcs x 1 2 3 4 5 6 with the specified monotonicity and concavity F 1 2 3 4 5 6 shape IP max IP min
  • 118. Could you repeat the question? Problem Below is the graph of a func on f. Draw the graph of the an deriva ve for f with F(1) = 0. Solu on y We start with F(1) = 0. f . Using the sign chart, we draw arcs x 1 2 3 4 5 6 with the specified monotonicity and concavity F 1 2 3 4 5 6 shape IP max IP min
  • 119. Could you repeat the question? Problem Below is the graph of a func on f. Draw the graph of the an deriva ve for f with F(1) = 0. Solu on y We start with F(1) = 0. f . Using the sign chart, we draw arcs x 1 2 3 4 5 6 with the specified monotonicity and concavity F It’s harder to tell if/when F crosses 1 2 3 4 5 6 shape IP max IP min the axis; more about that later.
  • 120. Outline What is an an deriva ve? Tabula ng An deriva ves Power func ons Combina ons Exponen al func ons Trigonometric func ons An deriva ves of piecewise func ons Finding An deriva ves Graphically Rec linear mo on
  • 121. Say what? “Rec linear mo on” just means mo on along a line. O en we are given informa on about the velocity or accelera on of a moving par cle and we want to know the equa ons of mo on.
  • 122. Application: Dead Reckoning
  • 123. Application: Dead Reckoning
  • 124. ProblemSuppose a par cle of mass m is acted upon by a constant force F.Find the posi on func on s(t), the velocity func on v(t), and theaccelera on func on a(t).
  • 125. ProblemSuppose a par cle of mass m is acted upon by a constant force F.Find the posi on func on s(t), the velocity func on v(t), and theaccelera on func on a(t).Solu on By Newton’s Second Law (F = ma) a constant force induces a F constant accelera on. So a(t) = a = . m
  • 126. ProblemSuppose a par cle of mass m is acted upon by a constant force F.Find the posi on func on s(t), the velocity func on v(t), and theaccelera on func on a(t).Solu on By Newton’s Second Law (F = ma) a constant force induces a F constant accelera on. So a(t) = a = . m ′ Since v (t) = a(t), v(t) must be an an deriva ve of the constant func on a. So v(t) = at + C = at + v0
  • 127. ProblemSuppose a par cle of mass m is acted upon by a constant force F.Find the posi on func on s(t), the velocity func on v(t), and theaccelera on func on a(t).Solu on By Newton’s Second Law (F = ma) a constant force induces a F constant accelera on. So a(t) = a = . m ′ Since v (t) = a(t), v(t) must be an an deriva ve of the constant func on a. So v(t) = at + C = at + v0
  • 128. An earlier Hatsumon Example Drop a ball off the roof of the Silver Center. What is its velocity when it hits the ground?
  • 129. An earlier Hatsumon Example Drop a ball off the roof of the Silver Center. What is its velocity when it hits the ground? Solu on Assume s0 = 100 m, and v0 = 0. Approximate a = g ≈ −10. Then s(t) = 100 − 5t2 √ √ So s(t) = 0 when t = 20 = 2 5. Then v(t) = −10t,
  • 130. Finding initial velocity fromstopping distance Example The skid marks made by an automobile indicate that its brakes were fully applied for a distance of 160 before it came to a stop. Suppose that the car in ques on has a constant decelera on of 20 ft/s2 under the condi ons of the skid. How fast was the car traveling when its brakes were first applied?
  • 131. Finding initial velocity fromstopping distance Example The skid marks made by an automobile indicate that its brakes were fully applied for a distance of 160 before it came to a stop. Suppose that the car in ques on has a constant decelera on of 20 ft/s2 under the condi ons of the skid. How fast was the car traveling when its brakes were first applied? Solu on (Setup) While braking, the car has accelera on a(t) = −20
  • 132. Finding initial velocity fromstopping distance Example The skid marks made by an automobile indicate that its brakes were fully applied for a distance of 160 before it came to a stop. Suppose that the car in ques on has a constant decelera on of 20 ft/s2 under the condi ons of the skid. How fast was the car traveling when its brakes were first applied? Solu on (Setup) While braking, the car has accelera on a(t) = −20 Measure me 0 and posi on 0 when the car starts braking. So
  • 133. Finding initial velocity fromstopping distance Example The skid marks made by an automobile indicate that its brakes were fully applied for a distance of 160 before it came to a stop. Suppose that the car in ques on has a constant decelera on of 20 ft/s2 under the condi ons of the skid. How fast was the car traveling when its brakes were first applied? Solu on (Setup) While braking, the car has accelera on a(t) = −20 Measure me 0 and posi on 0 when the car starts braking. So
  • 134. Implementing the Solution In general, 1 s(t) = s0 + v0 t + at2 2 Since s0 = 0 and a = −20, we have s(t) = v0 t − 10t2 v(t) = v0 − 20t for all t.
  • 135. Implementing the Solution In general, 1 s(t) = s0 + v0 t + at2 2 Since s0 = 0 and a = −20, we have s(t) = v0 t − 10t2 v(t) = v0 − 20t for all t. Plugging in t = t1 , 160 = v0 t1 − 10t2 1 0 = v0 − 20t1
  • 136. Solving We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0
  • 137. Solving We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 The second gives t1 = v0 /20, so subs tute into the first: v0 ( v )2 0 v0 · − 10 = 160 20 20 Solve: v2 0 10v2 − 0 = 160 20 400 2v2 − v2 = 160 · 40 = 6400 0 0
  • 138. Solving We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 The second gives t1 = v0 /20, so subs tute into the first: v0 ( v )2 0 v0 · − 10 = 160 20 20 Solve: v2 0 10v2 − 0 = 160 20 400 2v2 − v2 = 160 · 40 = 6400 0 0
  • 139. Summary of Antiderivatives so far f(x) F(x) 1 r+1 xr , r ̸= 1 x +C r+1 1 = x−1 ln |x| + C x x e ex + C 1 x ax a +C ln a ln x x ln x − x + C x ln x − x loga x +C ln a sin x − cos x + C cos x sin x + C
  • 140. Final Thoughts An deriva ves are a useful concept, especially in mo on y We can graph an f an deriva ve from the . x graph of a func on 1 2 3 4 5 6F We can compute an deriva ves, but not f(x) = e−x 2 always