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Lesson 18: Maximum and Minimum Vaues
 

Lesson 18: Maximum and Minimum Vaues

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We define what it means for a function to have a maximum or minimum value, and explain the Extreme Value Theorem, which indicates these maxima and minima must be there under certain conditions.

We define what it means for a function to have a maximum or minimum value, and explain the Extreme Value Theorem, which indicates these maxima and minima must be there under certain conditions.

Fermat's Theorem says that at differentiable extreme points, the derivative should be zero, and thus we arrive at a technique for finding extrema: look among the endpoints of the domain of definition and the critical points of the function.

There's also a little digression on Fermat's Last theorem, which is not related to calculus but is a big deal in recent mathematical history.

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    Lesson 18: Maximum and Minimum Vaues Lesson 18: Maximum and Minimum Vaues Presentation Transcript

    • Section 4.1 Maximum and Minimum Values V63.0121, Calculus I March 24, 2009 Announcements Homework due Thursday Quiz April 2, on Sections 2.5–3.5 Final Exam Friday, May 8, 2:00–3:50pm . . Image: Flickr user Karen with a K . . . . . .
    • Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples Challenge: Cubic functions . . . . . .
    • Optimize . . . . . .
    • 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 Maupertuis (1698–1759) . . . . . .
    • Design . . Image credit: Jason Tromm . . . . . .
    • 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 Maupertuis (1698–1759) . . . . . .
    • Optics . . Image credit: jacreative . . . . . .
    • 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 Maupertuis (1698–1759) . . . . . .
    • Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples Challenge: Cubic functions . . . . . .
    • Extreme points and values Definition Let 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 . . Image credit: Patrick Q . . . . . .
    • Extreme points and values Definition Let 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 Q . . . . . .
    • Extreme points and values Definition Let 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 Q . . . . . .
    • 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]. . . . . . .
    • 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 . b . . . . . . .
    • 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 . (c) f . value . . minimum . (d) f . value . .. . c d a . b . maximum minimum . . . . . .
    • 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. . . . . . .
    • Bad Example #1 Example Consider the function { 0≤x<1 x f(x) = x−2 1 ≤ x ≤ 2. . . . . . .
    • Bad Example #1 Example Consider the function { 0≤x<1 x f(x) = x−2 1 ≤ x ≤ 2. . . . . | 1 . . . . . . . .
    • Bad Example #1 Example Consider the function { 0≤x<1 x 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. . . . . . .
    • Bad Example #2 Example The function f(x) = x restricted to the interval [0, 1) still has no maximum value. . . . . . .
    • Bad Example #2 Example The function f(x) = x restricted to the interval [0, 1) still has no maximum value. . . . | 1 . . . . . . .
    • Final Bad Example Example 1 The function f(x) = is continuous on the closed interval [1, ∞) but x has no minimum value. . . . . . .
    • Final Bad Example Example 1 The function f(x) = is continuous on the closed interval [1, ∞) but x has no minimum value. . . . 1 . . . . . . .
    • Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples Challenge: Cubic functions . . . . . .
    • 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. . . . . . .
    • 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 minimum . . . . . .
    • 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 . global b max min max . . . . . .
    • 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 minimum . . . . . .
    • Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. . . . . . .
    • Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This means f(c + h) − f(c) ≤0 h . . . . . .
    • Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This means f(c + h) − f(c) f(c + h) − f(c) ≤ 0 =⇒ lim+ ≤0 h h h→0 . . . . . .
    • Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This means f(c + h) − f(c) f(c + h) − f(c) ≤ 0 =⇒ lim+ ≤0 h h h→0 The same will be true on the other end: if h is close enough to 0 but less than 0, f(c + h) ≤ f(c). This means f(c + h) − f(c) ≥0 h . . . . . .
    • Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This means f(c + h) − f(c) f(c + h) − f(c) ≤ 0 =⇒ lim+ ≤0 h h h→0 The same will be true on the other end: if h is close enough to 0 but less than 0, f(c + h) ≤ f(c). This means f(c + h) − f(c) f(c + h) − f(c) ≥ 0 =⇒ lim ≥0 h − h h→0 . . . . . .
    • Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This means f(c + h) − f(c) f(c + h) − f(c) ≤ 0 =⇒ lim+ ≤0 h h h→0 The same will be true on the other end: if h is close enough to 0 but less than 0, f(c + h) ≤ f(c). This means f(c + h) − f(c) f(c + h) − f(c) ≥ 0 =⇒ lim ≥0 h − h h→0 f(c + h) − f(c) Since the limit f′ (c) = lim exists, it must be 0. h h→0 . . . . . .
    • Meet the Mathematician: Pierre de Fermat 1601–1665 Lawyer and number theorist Proved many theorems, didn’t quite prove his last one . . . . . .
    • Tangent: Fermat’s Last Theorem Plenty of solutions to x2 + y2 = z2 among positive whole numbers (e.g., x = 3, y = 4, z = 5) . . . . . .
    • Tangent: Fermat’s Last Theorem Plenty of solutions to x2 + y2 = z2 among positive whole numbers (e.g., x = 3, y = 4, z = 5) No solutions to x3 + y3 = z3 among positive whole numbers . . . . . .
    • Tangent: Fermat’s Last Theorem Plenty of solutions to x2 + y2 = z2 among positive whole numbers (e.g., x = 3, y = 4, z = 5) No solutions to x3 + y3 = z3 among positive whole numbers Fermat claimed no solutions to xn + yn = zn but didn’t write down his proof . . . . . .
    • Tangent: Fermat’s Last Theorem Plenty of solutions to x2 + y2 = z2 among positive whole numbers (e.g., x = 3, y = 4, z = 5) No solutions to x3 + y3 = z3 among positive whole numbers Fermat claimed no solutions to xn + yn = zn but didn’t write down his proof Not solved until 1998! (Taylor–Wiles) . . . . . .
    • Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples Challenge: Cubic functions . . . . . .
    • The Closed Interval Method Let’s put this together logically. Let f be a continuous function defined on a closed interval [a, b]. We are in search of its global maximum, call it c. Then: . . . . . .
    • The Closed Interval Method Let’s put this together logically. Let f be a continuous function defined on a closed interval [a, b]. We are in search of its global maximum, call it c. Then: Either the maximum occurs at an endpoint of the interval, i.e., c = a or c = b, . . . . . .
    • The Closed Interval Method Let’s put this together logically. Let f be a continuous function defined on a closed interval [a, b]. We are in search of its global maximum, call it c. Then: Either the maximum occurs at an endpoint of the interval, i.e., c = a or c = b, Or the maximum occurs inside (a, b). In this case, c is also a local maximum. . . . . . .
    • The Closed Interval Method Let’s put this together logically. Let f be a continuous function defined on a closed interval [a, b]. We are in search of its global maximum, call it c. Then: Either the maximum occurs at an endpoint of the interval, i.e., c = a or c = b, Or the maximum occurs inside (a, b). In this case, c is also a local maximum. Either f is differentiable at c, in which case f′ (c) = 0 by Fermat’s Theorem. . . . . . .
    • The Closed Interval Method Let’s put this together logically. Let f be a continuous function defined on a closed interval [a, b]. We are in search of its global maximum, call it c. Then: Either the maximum occurs at an endpoint of the interval, i.e., c = a or c = b, Or the maximum occurs inside (a, b). In this case, c is also a local maximum. Either f is differentiable at c, in which case f′ (c) = 0 by Fermat’s Theorem. Or f is not differentiable at c. . . . . . .
    • The Closed Interval Method Let’s put this together logically. Let f be a continuous function defined on a closed interval [a, b]. We are in search of its global maximum, call it c. Then: This means to find the Either the maximum maximum value of f on [a, b], occurs at an endpoint of we need to check: the interval, i.e., c = a or c = b, Or the maximum occurs inside (a, b). In this case, c is also a local maximum. Either f is differentiable at c, in which case f′ (c) = 0 by Fermat’s Theorem. Or f is not differentiable at c. . . . . . .
    • The Closed Interval Method Let’s put this together logically. Let f be a continuous function defined on a closed interval [a, b]. We are in search of its global maximum, call it c. Then: This means to find the Either the maximum maximum value of f on [a, b], occurs at an endpoint of we need to check: the interval, i.e., c = a or a and b c = b, Or the maximum occurs inside (a, b). In this case, c is also a local maximum. Either f is differentiable at c, in which case f′ (c) = 0 by Fermat’s Theorem. Or f is not differentiable at c. . . . . . .
    • The Closed Interval Method Let’s put this together logically. Let f be a continuous function defined on a closed interval [a, b]. We are in search of its global maximum, call it c. Then: This means to find the Either the maximum maximum value of f on [a, b], occurs at an endpoint of we need to check: the interval, i.e., c = a or a and b c = b, Points x where f′ (x) = 0 Or the maximum occurs inside (a, b). In this case, c is also a local maximum. Either f is differentiable at c, in which case f′ (c) = 0 by Fermat’s Theorem. Or f is not differentiable at c. . . . . . .
    • The Closed Interval Method Let’s put this together logically. Let f be a continuous function defined on a closed interval [a, b]. We are in search of its global maximum, call it c. Then: This means to find the Either the maximum maximum value of f on [a, b], occurs at an endpoint of we need to check: the interval, i.e., c = a or a and b c = b, Points x where f′ (x) = 0 Or the maximum occurs inside (a, b). In this case, c Points x where f is not is also a local maximum. differentiable. Either f is differentiable at c, in which case f′ (c) = 0 by Fermat’s Theorem. Or f is not differentiable at c. . . . . . .
    • The Closed Interval Method Let’s put this together logically. Let f be a continuous function defined on a closed interval [a, b]. We are in search of its global maximum, call it c. Then: This means to find the Either the maximum maximum value of f on [a, b], occurs at an endpoint of we need to check: the interval, i.e., c = a or a and b c = b, Points x where f′ (x) = 0 Or the maximum occurs inside (a, b). In this case, c Points x where f is not is also a local maximum. differentiable. Either f is differentiable The latter two are both called at c, in which case f′ (c) = 0 by Fermat’s critical points of f. This technique is called the Closed Theorem. Interval Method. Or f is not differentiable at c. . . . . . .
    • Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples Challenge: Cubic functions . . . . . .
    • Example Find the extreme values of f(x) = 2x − 5 on [−1, 2]. . . . . . .
    • 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. . . . . . .
    • Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. . . . . . .
    • Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solution We have f′ (x) = 2x, which is zero when x = 0. . . . . . .
    • Example Find the extreme values of f(x) = x2 − 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) = . . . . . .
    • Example Find the extreme values of f(x) = x2 − 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) = . . . . . .
    • Example Find the extreme values of f(x) = x2 − 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) = . . . . . .
    • Example Find the extreme values of f(x) = x2 − 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) = 3 . . . . . .
    • Example Find the extreme values of f(x) = x2 − 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 . . . . . .
    • Example Find the extreme values of f(x) = x2 − 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) . . . . . .
    • Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. . . . . . .
    • Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x +x =x (5x + 4) 3 3 3 Thus f′ (−4/5) = 0 and f is not differentiable at 0. . . . . . .
    • Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x +x =x (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) = . . . . . .
    • Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x +x =x (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) = . . . . . .
    • Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x +x =x (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) = . . . . . .
    • Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x +x =x (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) = . . . . . .
    • Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x +x =x (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) = 6.3496 . . . . . .
    • Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x +x =x (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 . . . . . .
    • Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x +x =x (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) . . . . . .
    • Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solution Write f(x) = x5/3 + 2x2/3 , then 5 2/3 4 −1/3 1 −1/3 f′ (x) = x +x =x (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) . . . . . .
    • Example √ 4 − x2 on [−2, 1]. Find the extreme values of f(x) = . . . . . .
    • Example √ 4 − x2 on [−2, 1]. Find the extreme values of f(x) = Solution x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differentiable at ±2 as well.) . . . . . .
    • Example √ 4 − x2 on [−2, 1]. Find the extreme values of f(x) = 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) = . . . . . .
    • Example √ 4 − x2 on [−2, 1]. Find the extreme values of f(x) = 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) = . . . . . .
    • Example √ 4 − x2 on [−2, 1]. Find the extreme values of f(x) = 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) = . . . . . .
    • Example √ 4 − x2 on [−2, 1]. Find the extreme values of f(x) = 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) = 3 . . . . . .
    • Example √ 4 − x2 on [−2, 1]. Find the extreme values of f(x) = 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) = 3 . . . . . .
    • Example √ 4 − x2 on [−2, 1]. Find the extreme values of f(x) = 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) = 3 . . . . . .
    • Outline Introduction The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples Challenge: Cubic functions . . . . . .
    • Challenge: Cubic functions Example How many critical points can a cubic function f(x) = ax3 + bx2 + cx + d have? . . . . . .
    • Solution If f′ (x) = 0, we have 3ax2 + 2bx + c = 0, and so √ √ −2b ± 4b2 − 12ac −b ± b2 − 3ac x= = , 6a 3a and so we have three possibilities: b2 − 3ac > 0, in which case there are two distinct critical points. An example would be f(x) = x3 + x2 , where a = 1, b = 1, and c = 0. b2 − 3ac < 0, in which case there are no real roots to the quadratic, hence no critical points. An example would be f(x) = x3 + x2 + x, where a = b = c = 1. b2 − 3ac = 0, in which case there is a single critical point. Example: x3 , where a = 1 and b = c = 0. . . . . . .
    • Review Concept: absolute (global) and relative (local) maxima/minima Fact: Fermat’s theorem: f′ (x) = 0 at local extrema Technique: the Closed Interval Method . . . . . .