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### Lesson18 -maximum_and_minimum_values_slides

1. 1. Section 4.1 Maximum and Minimum Values V63.0121.002.2010Su, Calculus I New York University June 8, 2010Announcements Exams not graded yet Assignment 4 is on the website Quiz 3 on Thursday covering 3.3, 3.4, 3.5, 3.7
2. 2. Announcements Exams not graded yet Assignment 4 is on the website Quiz 3 on Thursday covering 3.3, 3.4, 3.5, 3.7V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 2 / 32
3. 3. Objectives Understand and be able to explain the statement of the Extreme Value Theorem. Understand and be able to explain the statement of Fermat’s Theorem. Use the Closed Interval Method to ﬁnd the extreme values of a function deﬁned on a closed interval.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 3 / 32
4. 4. Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method ExamplesV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 4 / 32
5. 5. Optimize
6. 6. Why go to the extremes? Rationally speaking, it is advantageous to ﬁnd the extreme values of a function (maximize proﬁt, minimize costs, etc.) Pierre-Louis MaupertuisV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values (1698–1759) 8, 2010 June 5 / 32
7. 7. DesignImage credit: Jason TrommV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 6 / 32
8. 8. Why go to the extremes? Rationally speaking, it is advantageous to ﬁnd the extreme values of a function (maximize proﬁt, minimize costs, etc.) Many laws of science are derived from minimizing principles. Pierre-Louis MaupertuisV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values (1698–1759) 8, 2010 June 7 / 32
9. 9. OpticsImage credit: jacreativeV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 8 / 32
10. 10. Why go to the extremes? Rationally speaking, it is advantageous to ﬁnd the extreme values of a function (maximize proﬁt, minimize costs, etc.) Many laws of science are derived from minimizing principles. Maupertuis’ principle: “Action is minimized through the wisdom of God.” Pierre-Louis MaupertuisV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values (1698–1759) 8, 2010 June 9 / 32
11. 11. Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method ExamplesV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 10 / 32
12. 12. Extreme points and valuesDeﬁnitionLet f have domain D.Image credit: Patrick QV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 11 / 32
13. 13. Extreme points and valuesDeﬁnitionLet f have domain D. The function f has an absolute maximum (or global maximum) (respectively, absolute minimum) at c if f (c) ≥ f (x) (respectively, f (c) ≤ f (x)) for all x in DImage credit: Patrick QV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 11 / 32
14. 14. Extreme points and valuesDeﬁnitionLet f have domain D. The function f has an absolute maximum (or global maximum) (respectively, absolute minimum) at c if f (c) ≥ f (x) (respectively, f (c) ≤ f (x)) for all x in D The number f (c) is called the maximum value (respectively, minimum value) of f on D.Image credit: Patrick QV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 11 / 32
15. 15. Extreme points and valuesDeﬁnitionLet f have domain D. The function f has an absolute maximum (or global maximum) (respectively, absolute minimum) at c if f (c) ≥ f (x) (respectively, f (c) ≤ f (x)) for all x in D The number f (c) is called the maximum value (respectively, minimum value) of f on D. An extremum is either a maximum or a minimum. An extreme value is either a maximum value or minimum value.Image credit: Patrick QV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 11 / 32
16. 16. The Extreme Value Theorem Theorem (The Extreme Value Theorem) Let f be a function which is continuous on the closed interval [a, b]. Then f attains an absolute maximum value f (c) and an absolute minimum value f (d) at numbers c and d in [a, b].V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 12 / 32
17. 17. The Extreme Value Theorem Theorem (The Extreme Value Theorem) Let f be a function which is continuous on the closed interval [a, b]. Then f attains an absolute maximum value f (c) and an absolute minimum value f (d) at numbers c and d in [a, b]. a bV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 12 / 32
18. 18. The Extreme Value Theorem Theorem (The Extreme Value Theorem) Let f be a function which is continuous on the closed interval [a, b]. Then f attains an absolute maximum value f (c) and an absolute minimum value f (d) at numbers c and d in [a, b]. maximum f (c) value minimum f (d) value a d c b minimum maximumV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 12 / 32
19. 19. No proof of EVT forthcoming This theorem is very hard to prove without using technical facts about continuous functions and closed intervals. But we can show the importance of each of the hypotheses.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 13 / 32
20. 20. Bad Example #1 Example Consider the function x 0≤x <1 f (x) = x − 2 1 ≤ x ≤ 2.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 14 / 32
21. 21. Bad Example #1 Example Consider the function x 0≤x <1 f (x) = | x − 2 1 ≤ x ≤ 2. 1V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 14 / 32
22. 22. Bad Example #1 Example Consider the function x 0≤x <1 f (x) = | x − 2 1 ≤ x ≤ 2. 1 Then although values of f (x) get arbitrarily close to 1 and never bigger than 1, 1 is not the maximum value of f on [0, 1] because it is never achieved.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 14 / 32
23. 23. Bad Example #1 Example Consider the function x 0≤x <1 f (x) = | x − 2 1 ≤ x ≤ 2. 1 Then although values of f (x) get arbitrarily close to 1 and never bigger than 1, 1 is not the maximum value of f on [0, 1] because it is never achieved. This does not violate EVT because f is not continuous.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 14 / 32
24. 24. Bad Example #2 Example Consider the function f (x) = x restricted to the interval [0, 1).V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 15 / 32
25. 25. Bad Example #2 Example Consider the function f (x) = x restricted to the interval [0, 1). | 1V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 15 / 32
26. 26. Bad Example #2 Example Consider the function f (x) = x restricted to the interval [0, 1). | 1 There is still no maximum value (values get arbitrarily close to 1 but do not achieve it).V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 15 / 32
27. 27. Bad Example #2 Example Consider the function f (x) = x restricted to the interval [0, 1). | 1 There is still no maximum value (values get arbitrarily close to 1 but do not achieve it). This does not violate EVT because the domain is not closed.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 15 / 32
28. 28. Final Bad Example Example 1 Consider the function f (x) = is continuous on the closed interval [1, ∞). xV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 16 / 32
29. 29. Final Bad Example Example 1 Consider the function f (x) = is continuous on the closed interval [1, ∞). x 1V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 16 / 32
30. 30. Final Bad Example Example 1 Consider the function f (x) = is continuous on the closed interval [1, ∞). x 1 There is no minimum value (values get arbitrarily close to 0 but do not achieve it).V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 16 / 32
31. 31. Final Bad Example Example 1 Consider the function f (x) = is continuous on the closed interval [1, ∞). x 1 There is no minimum value (values get arbitrarily close to 0 but do not achieve it). This does not violate EVT because the domain is not bounded.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 16 / 32
32. 32. Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method ExamplesV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 17 / 32
33. 33. Local extrema Deﬁnition A function f has a local maximum or relative maximum at c if f (c) ≥ f (x) when x is near c. This means that f (c) ≥ f (x) for all x in some open interval containing c. Similarly, f has a local minimum at c if f (c) ≤ f (x) when x is near c.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 18 / 32
34. 34. Local extrema Deﬁnition A function f has a local maximum or relative maximum at c if f (c) ≥ f (x) when x is near c. This means that f (c) ≥ f (x) for all x in some open interval containing c. Similarly, f has a local minimum at c if f (c) ≤ f (x) when x is near c. | | a local local b maximum minimumV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 18 / 32
35. 35. Local extrema So a local extremum must be inside the domain of f (not on the end). A global extremum that is inside the domain is a local extremum. | | a local local and global b global maximum min maxV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 18 / 32
36. 36. Fermat’s Theorem Theorem (Fermat’s Theorem) Suppose f has a local extremum at c and f is diﬀerentiable at c. Then f (c) = 0. | | a local local b maximum minimumV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 19 / 32
37. 37. Fermat’s Theorem Theorem (Fermat’s Theorem) Suppose f has a local extremum at c and f is diﬀerentiable at c. Then f (c) = 0. | | a local local b maximum minimumV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 19 / 32
38. 38. Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 20 / 32
39. 39. Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f (x) ≤ f (c). This means f (x) − f (c) ≤0 x −cV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 20 / 32
40. 40. Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f (x) ≤ f (c). This means f (x) − f (c) f (x) − f (c) ≤ 0 =⇒ lim+ ≤0 x −c x→c x −cV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 20 / 32
41. 41. Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f (x) ≤ f (c). This means f (x) − f (c) f (x) − f (c) ≤ 0 =⇒ lim+ ≤0 x −c x→c x −c The same will be true on the other end: if x is slightly less than c, f (x) ≤ f (c). This means f (x) − f (c) ≥0 x −cV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 20 / 32
42. 42. Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f (x) ≤ f (c). This means f (x) − f (c) f (x) − f (c) ≤ 0 =⇒ lim+ ≤0 x −c x→c x −c The same will be true on the other end: if x is slightly less than c, f (x) ≤ f (c). This means f (x) − f (c) f (x) − f (c) ≥ 0 =⇒ lim ≥0 x −c x→c − x −cV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 20 / 32
43. 43. Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If x is slightly greater than c, f (x) ≤ f (c). This means f (x) − f (c) f (x) − f (c) ≤ 0 =⇒ lim+ ≤0 x −c x→c x −c The same will be true on the other end: if x is slightly less than c, f (x) ≤ f (c). This means f (x) − f (c) f (x) − f (c) ≥ 0 =⇒ lim ≥0 x −c x→c − x −c f (x) − f (c) Since the limit f (c) = lim exists, it must be 0. x→c x −cV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 20 / 32
44. 44. Meet the Mathematician: Pierre de Fermat 1601–1665 Lawyer and number theorist Proved many theorems, didn’t quite prove his last oneV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 21 / 32
45. 45. Tangent: Fermat’s Last Theorem Plenty of solutions to x 2 + y 2 = z 2 among positive whole numbers (e.g., x = 3, y = 4, z = 5)V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 22 / 32
46. 46. Tangent: Fermat’s Last Theorem Plenty of solutions to x 2 + y 2 = z 2 among positive whole numbers (e.g., x = 3, y = 4, z = 5) No solutions to x 3 + y 3 = z 3 among positive whole numbersV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 22 / 32
47. 47. Tangent: Fermat’s Last Theorem Plenty of solutions to x 2 + y 2 = z 2 among positive whole numbers (e.g., x = 3, y = 4, z = 5) No solutions to x 3 + y 3 = z 3 among positive whole numbers Fermat claimed no solutions to x n + y n = z n but didn’t write down his proofV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 22 / 32
48. 48. Tangent: Fermat’s Last Theorem Plenty of solutions to x 2 + y 2 = z 2 among positive whole numbers (e.g., x = 3, y = 4, z = 5) No solutions to x 3 + y 3 = z 3 among positive whole numbers Fermat claimed no solutions to x n + y n = z n but didn’t write down his proof Not solved until 1998! (Taylor–Wiles)V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 22 / 32
49. 49. Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method ExamplesV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 23 / 32
50. 50. Flowchart for placing extremaThanks to Fermat Suppose f is a continuous function on the closed, bounded interval [a, b], and c is a global maximum point. c is a start local max Is f Is c an f is not no diﬀ’ble at no endpoint? diﬀ at c c? yes yes c = a or f (c) = 0 c = bV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 24 / 32
51. 51. The Closed Interval Method This means to ﬁnd the maximum value of f on [a, b], we need to: Evaluate f at the endpoints a and b Evaluate f at the critical points or critical numbers x where either f (x) = 0 or f is not diﬀerentiable at x. The points with the largest function value are the global maximum points The points with the smallest or most negative function value are the global minimum points.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 25 / 32
52. 52. Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method ExamplesV63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 26 / 32
53. 53. Extreme values of a linear function Example Find the extreme values of f (x) = 2x − 5 on [−1, 2].V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 27 / 32
54. 54. Extreme values of a linear function Example Find the extreme values of f (x) = 2x − 5 on [−1, 2]. Solution Since f (x) = 2, which is never zero, we have no critical points and we need only investigate the endpoints: f (−1) = 2(−1) − 5 = −7 f (2) = 2(2) − 5 = −1V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 27 / 32
55. 55. Extreme values of a linear function Example Find the extreme values of f (x) = 2x − 5 on [−1, 2]. Solution Since f (x) = 2, which is never zero, we have no critical points and we need only investigate the endpoints: f (−1) = 2(−1) − 5 = −7 f (2) = 2(2) − 5 = −1 So The absolute minimum (point) is at −1; the minimum value is −7. The absolute maximum (point) is at 2; the maximum value is −1.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 27 / 32
56. 56. Extreme values of a quadratic function Example Find the extreme values of f (x) = x 2 − 1 on [−1, 2].V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
57. 57. Extreme values of a quadratic function Example Find the extreme values of f (x) = x 2 − 1 on [−1, 2]. Solution We have f (x) = 2x, which is zero when x = 0.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
58. 58. Extreme values of a quadratic function Example Find the extreme values of f (x) = x 2 − 1 on [−1, 2]. Solution We have f (x) = 2x, which is zero when x = 0. So our points to check are: f (−1) = f (0) = f (2) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
59. 59. Extreme values of a quadratic function Example Find the extreme values of f (x) = x 2 − 1 on [−1, 2]. Solution We have f (x) = 2x, which is zero when x = 0. So our points to check are: f (−1) = 0 f (0) = f (2) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
60. 60. Extreme values of a quadratic function Example Find the extreme values of f (x) = x 2 − 1 on [−1, 2]. Solution We have f (x) = 2x, which is zero when x = 0. So our points to check are: f (−1) = 0 f (0) = − 1 f (2) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
61. 61. Extreme values of a quadratic function Example Find the extreme values of f (x) = x 2 − 1 on [−1, 2]. Solution We have f (x) = 2x, which is zero when x = 0. So our points to check are: f (−1) = 0 f (0) = − 1 f (2) = 3V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
62. 62. Extreme values of a quadratic function Example Find the extreme values of f (x) = x 2 − 1 on [−1, 2]. Solution We have f (x) = 2x, which is zero when x = 0. So our points to check are: f (−1) = 0 f (0) = − 1 (absolute min) f (2) = 3V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
63. 63. Extreme values of a quadratic function Example Find the extreme values of f (x) = x 2 − 1 on [−1, 2]. Solution We have f (x) = 2x, which is zero when x = 0. So our points to check are: f (−1) = 0 f (0) = − 1 (absolute min) f (2) = 3 (absolute max)V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
64. 64. Extreme values of a cubic function Example Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2].V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
65. 65. Extreme values of a cubic function Example Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2]. Solution Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
66. 66. Extreme values of a cubic function Example Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2]. Solution Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f (−1) = f (0) = f (1) = f (2) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
67. 67. Extreme values of a cubic function Example Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2]. Solution Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f (−1) = − 4 f (0) = f (1) = f (2) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
68. 68. Extreme values of a cubic function Example Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2]. Solution Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f (−1) = − 4 f (0) = 1 f (1) = f (2) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
69. 69. Extreme values of a cubic function Example Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2]. Solution Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f (−1) = − 4 f (0) = 1 f (1) = 0 f (2) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
70. 70. Extreme values of a cubic function Example Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2]. Solution Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f (−1) = − 4 f (0) = 1 f (1) = 0 f (2) = 5V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
71. 71. Extreme values of a cubic function Example Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2]. Solution Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f (−1) = − 4 (global min) f (0) = 1 f (1) = 0 f (2) = 5V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
72. 72. Extreme values of a cubic function Example Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2]. Solution Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f (−1) = − 4 (global min) f (0) = 1 f (1) = 0 f (2) = 5 (global max)V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
73. 73. Extreme values of a cubic function Example Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2]. Solution Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f (−1) = − 4 (global min) f (0) = 1 (local max) f (1) = 0 f (2) = 5 (global max)V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
74. 74. Extreme values of a cubic function Example Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2]. Solution Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f (−1) = − 4 (global min) f (0) = 1 (local max) f (1) = 0 (local min) f (2) = 5 (global max)V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
75. 75. Extreme values of an algebraic function Example Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2].V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
76. 76. Extreme values of an algebraic function Example Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2]. Solution Write f (x) = x 5/3 + 2x 2/3 , then 5 4 1 f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4) 3 3 3 Thus f (−4/5) = 0 and f is not diﬀerentiable at 0.V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
77. 77. Extreme values of an algebraic function Example Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2]. Solution Write f (x) = x 5/3 + 2x 2/3 , then 5 4 1 f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4) 3 3 3 Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check are: f (−1) = f (−4/5) = f (0) = f (2) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
78. 78. Extreme values of an algebraic function Example Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2]. Solution Write f (x) = x 5/3 + 2x 2/3 , then 5 4 1 f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4) 3 3 3 Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check are: f (−1) = 1 f (−4/5) = f (0) = f (2) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
79. 79. Extreme values of an algebraic function Example Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2]. Solution Write f (x) = x 5/3 + 2x 2/3 , then 5 4 1 f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4) 3 3 3 Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check are: f (−1) = 1 f (−4/5) = 1.0341 f (0) = f (2) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
80. 80. Extreme values of an algebraic function Example Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2]. Solution Write f (x) = x 5/3 + 2x 2/3 , then 5 4 1 f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4) 3 3 3 Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check are: f (−1) = 1 f (−4/5) = 1.0341 f (0) = 0 f (2) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
81. 81. Extreme values of an algebraic function Example Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2]. Solution Write f (x) = x 5/3 + 2x 2/3 , then 5 4 1 f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4) 3 3 3 Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check are: f (−1) = 1 f (−4/5) = 1.0341 f (0) = 0 f (2) = 6.3496V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
82. 82. Extreme values of an algebraic function Example Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2]. Solution Write f (x) = x 5/3 + 2x 2/3 , then 5 4 1 f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4) 3 3 3 Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check are: f (−1) = 1 f (−4/5) = 1.0341 f (0) = 0 (absolute min) f (2) = 6.3496V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
83. 83. Extreme values of an algebraic function Example Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2]. Solution Write f (x) = x 5/3 + 2x 2/3 , then 5 4 1 f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4) 3 3 3 Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check are: f (−1) = 1 f (−4/5) = 1.0341 f (0) = 0 (absolute min) f (2) = 6.3496 (absolute max)V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
84. 84. Extreme values of an algebraic function Example Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2]. Solution Write f (x) = x 5/3 + 2x 2/3 , then 5 4 1 f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4) 3 3 3 Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check are: f (−1) = 1 f (−4/5) = 1.0341 (relative max) f (0) = 0 (absolute min) f (2) = 6.3496 (absolute max)V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
85. 85. Extreme values of an algebraic function Example Find the extreme values of f (x) = 4 − x 2 on [−2, 1].V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
86. 86. Extreme values of an algebraic function Example Find the extreme values of f (x) = 4 − x 2 on [−2, 1]. Solution x We have f (x) = − √ , which is zero when x = 0. (f is not 4 − x2 diﬀerentiable at ±2 as well.)V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
87. 87. Extreme values of an algebraic function Example Find the extreme values of f (x) = 4 − x 2 on [−2, 1]. Solution x We have f (x) = − √ , which is zero when x = 0. (f is not 4 − x2 diﬀerentiable at ±2 as well.) So our points to check are: f (−2) = f (0) = f (1) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
88. 88. Extreme values of an algebraic function Example Find the extreme values of f (x) = 4 − x 2 on [−2, 1]. Solution x We have f (x) = − √ , which is zero when x = 0. (f is not 4 − x2 diﬀerentiable at ±2 as well.) So our points to check are: f (−2) = 0 f (0) = f (1) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
89. 89. Extreme values of an algebraic function Example Find the extreme values of f (x) = 4 − x 2 on [−2, 1]. Solution x We have f (x) = − √ , which is zero when x = 0. (f is not 4 − x2 diﬀerentiable at ±2 as well.) So our points to check are: f (−2) = 0 f (0) = 2 f (1) =V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
90. 90. Extreme values of an algebraic function Example Find the extreme values of f (x) = 4 − x 2 on [−2, 1]. Solution x We have f (x) = − √ , which is zero when x = 0. (f is not 4 − x2 diﬀerentiable at ±2 as well.) So our points to check are: f (−2) = 0 f (0) = 2 √ f (1) = 3V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
91. 91. Extreme values of an algebraic function Example Find the extreme values of f (x) = 4 − x 2 on [−2, 1]. Solution x We have f (x) = − √ , which is zero when x = 0. (f is not 4 − x2 diﬀerentiable at ±2 as well.) So our points to check are: f (−2) = 0 (absolute min) f (0) = 2 √ f (1) = 3V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
92. 92. Extreme values of an algebraic function Example Find the extreme values of f (x) = 4 − x 2 on [−2, 1]. Solution x We have f (x) = − √ , which is zero when x = 0. (f is not 4 − x2 diﬀerentiable at ±2 as well.) So our points to check are: f (−2) = 0 (absolute min) f (0) = 2 (absolute max) √ f (1) = 3V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
93. 93. Summary The Extreme Value Theorem: a continuous function on a closed interval must achieve its max and min Fermat’s Theorem: local extrema are critical points The Closed Interval Method: an algorithm for ﬁnding global extrema Show your work unless you want to end up like Fermat!V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 32 / 32