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Lesson19 Maximum And Minimum Values 034 Slides

The document discusses the Extreme Value Theorem, which states that a continuous function on a closed interval [a, b] attains both an absolute maximum and minimum. It also covers definitions of local and global extrema, illustrates how to find extreme values using various examples, and references Fermat's Theorem and its implications. Additionally, it highlights the importance of identifying critical points and evaluating endpoints to determine extrema in a function.

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Section 4.1 Maximum and Minimum Values V63.0121.034, Calculus I November 4, 2009 Announcements Quiz next week on §§3.1–3.5 Final Exam Friday, December 18, 2:00–3:50pm . . Image credit: 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 . . . . . .
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. . . . . . .
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 . . .. a . d c b . minimum maximum . . . . . .
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 { x 0≤x<1 f (x ) = x − 2 1 ≤ x ≤ 2. . . . . . .
Bad Example #1 Example Consider the function . { x 0≤x<1 f (x ) = . . | . x − 2 1 ≤ x ≤ 2. 1 . . . . . . . .
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. . . . . . .
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, ∞) x but has no minimum value. . . . . . .
Final Bad Example Example 1 The function f(x) = is continuous on the closed interval [1, ∞) x but 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 . . |. b local and global . global 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→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→0 h 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→0 h 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→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→0 h 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→0 − h f(c + h) − f(c) Since the limit f′ (c) = lim exists, it must be 0. h→0 h . . . . . .
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 . . . . . .
Flowchart for placing extrema Thanks 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 c an Is f diff’ble f is not n .o n .o endpoint? at c? diff at c y . es y . es . c = a or . f′ (c) = 0 c = b . . . . . .
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. . . . . . .
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 . . . . . .
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) = 2x3 − 3x2 + 1 on [−1, 2]. . . . . . .
Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. . . . . . .
Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 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) = . . . . . .
Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 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) = . . . . . .
Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 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) = . . . . . .
Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 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) = . . . . . .
Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 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) = 5 . . . . . .
Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 (absolute min) f(0) = 1 f(1) = 0 f(2) = 5 . . . . . .
Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 (absolute min) f(0) = 1 f(1) = 0 f(2) = 5 (absolute max) . . . . . .
Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 (absolute min) f(0) = 1 (local max) f(1) = 0 f(2) = 5 (absolute max) . . . . . .
Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 (absolute min) f(0) = 1 (local max) f(1) = 0 (local min) f(2) = 5 (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 √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. . . . . . .
Example √ Find the extreme values of f(x) = 4 − x2 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.) . . . . . .
Example √ Find the extreme values of f(x) = 4 − x2 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) = . . . . . .
Example √ Find the extreme values of f(x) = 4 − x2 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) = . . . . . .
Example √ Find the extreme values of f(x) = 4 − x2 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) = . . . . . .
Example √ Find the extreme values of f(x) = 4 − x2 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) = 3 . . . . . .
Example √ Find the extreme values of f(x) = 4 − x2 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) = 3 . . . . . .
Example √ Find the extreme values of f(x) = 4 − x2 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) = 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: . . . . . .
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. . . . . . .
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. . . . . . .
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 . . . . . .

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Lesson19 Maximum And Minimum Values 034 Slides

  • 1.
    Section 4.1 Maximum and Minimum Values V63.0121.034, Calculus I November 4, 2009 Announcements Quiz next week on §§3.1–3.5 Final Exam Friday, December 18, 2:00–3:50pm . . Image credit: Karen with a K . . . . . .
  • 2.
    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 . . . . . .
  • 3.
    Optimize . . . . . .
  • 4.
    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) . . . . . .
  • 5.
  • 6.
    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) . . . . . .
  • 7.
    Optics . . Image credit: jacreative . . . . . .
  • 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. Maupertuis’ principle: “Action is minimized through the wisdom of God.” Pierre-Louis Maupertuis (1698–1759) . . . . . .
  • 9.
    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 . . . . . .
  • 10.
    Extreme points and values Definition Let f have domainD. 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 . . . . . .
  • 11.
    Extreme points and values Definition Let f have domainD. 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. . . . . . .
  • 12.
    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 . . . . . .
  • 13.
    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]. . . . . . .
  • 14.
    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 . . . . . . .
  • 15.
    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 . . .. a . d c b . minimum maximum . . . . . .
  • 16.
    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. . . . . . .
  • 17.
    Bad Example #1 Example Consider the function { x 0≤x<1 f (x ) = x − 2 1 ≤ x ≤ 2. . . . . . .
  • 18.
    Bad Example #1 Example Consider the function . { x 0≤x<1 f (x ) = . . | . x − 2 1 ≤ x ≤ 2. 1 . . . . . . . .
  • 19.
    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. . . . . . .
  • 20.
    Bad Example #2 Example The function f(x) = x restricted to the interval [0, 1) still has no maximum value. . . . . . .
  • 21.
    Bad Example #2 Example The function f(x) = x restricted to the interval [0, 1) still has no maximum value. . . . | 1 . . . . . . .
  • 22.
    Final Bad Example Example 1 The function f(x) = is continuous on the closed interval [1, ∞) x but has no minimum value. . . . . . .
  • 23.
    Final Bad Example Example 1 The function f(x) = is continuous on the closed interval [1, ∞) x but has no minimum value. . . . 1 . . . . . . .
  • 24.
    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 . . . . . .
  • 25.
    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. . . . . . .
  • 26.
    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 . . . . . .
  • 27.
    So a local extremum must be inside the domain off (not on the end). A global extremum that is inside the domain is a local extremum. . . . . .... | a local . . |. b local and global . global max min max . . . . . .
  • 28.
    Theorem (Fermat’s Theorem) Suppose f has a local extremum atc and f is differentiable at c. Then f′ (c) = 0. . . . . .... | a local . . | local b . maximum minimum . . . . . .
  • 29.
    Sketch of proof of Fermat’s Theorem Suppose that f has a local maximum at c. . . . . . .
  • 30.
    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 . . . . . .
  • 31.
    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→0 h . . . . . .
  • 32.
    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→0 h 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 . . . . . .
  • 33.
    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→0 h 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→0 − h . . . . . .
  • 34.
    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→0 h 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→0 − h f(c + h) − f(c) Since the limit f′ (c) = lim exists, it must be 0. h→0 h . . . . . .
  • 35.
    Meet the Mathematician: Pierre de Fermat 1601–1665 Lawyer and number theorist Proved many theorems, didn’t quite prove his last one . . . . . .
  • 36.
    Tangent: Fermat’s Last Theorem Plenty of solutions to x2 + y2 = z2 among positive whole numbers (e.g., x = 3, y = 4, z = 5) . . . . . .
  • 37.
    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 . . . . . .
  • 38.
    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 . . . . . .
  • 39.
    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) . . . . . .
  • 40.
    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 . . . . . .
  • 41.
    Flowchart for placing extrema Thanks 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 c an Is f diff’ble f is not n .o n .o endpoint? at c? diff at c y . es y . es . c = a or . f′ (c) = 0 c = b . . . . . .
  • 42.
    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. . . . . . .
  • 43.
    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 . . . . . .
  • 44.
    Example Find the extreme values of f(x) =2x − 5 on [−1, 2]. . . . . . .
  • 45.
    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 . . . . . .
  • 46.
    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. . . . . . .
  • 47.
    Example Find the extreme values of f(x) =x2 − 1 on [−1, 2]. . . . . . .
  • 48.
    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. . . . . . .
  • 49.
    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) = . . . . . .
  • 50.
    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) = . . . . . .
  • 51.
    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) = . . . . . .
  • 52.
    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 . . . . . .
  • 53.
    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 . . . . . .
  • 54.
    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) . . . . . .
  • 55.
    Example Find the extreme values of f(x) =2x3 − 3x2 + 1 on [−1, 2]. . . . . . .
  • 56.
    Example Find the extreme values of f(x) =2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. . . . . . .
  • 57.
    Example Find the extreme values of f(x) =2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 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) = . . . . . .
  • 58.
    Example Find the extreme values of f(x) =2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 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) = . . . . . .
  • 59.
    Example Find the extreme values of f(x) =2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 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) = . . . . . .
  • 60.
    Example Find the extreme values of f(x) =2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 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) = . . . . . .
  • 61.
    Example Find the extreme values of f(x) =2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 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) = 5 . . . . . .
  • 62.
    Example Find the extreme values of f(x) =2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 (absolute min) f(0) = 1 f(1) = 0 f(2) = 5 . . . . . .
  • 63.
    Example Find the extreme values of f(x) =2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 (absolute min) f(0) = 1 f(1) = 0 f(2) = 5 (absolute max) . . . . . .
  • 64.
    Example Find the extreme values of f(x) =2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 (absolute min) f(0) = 1 (local max) f(1) = 0 f(2) = 5 (absolute max) . . . . . .
  • 65.
    Example Find the extreme values of f(x) =2x3 − 3x2 + 1 on [−1, 2]. Solution Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and x = 1. The values to check are f(−1) = − 4 (absolute min) f(0) = 1 (local max) f(1) = 0 (local min) f(2) = 5 (absolute max) . . . . . .
  • 66.
    Example Find the extreme values of f(x) =x2/3 (x + 2) on [−1, 2]. . . . . . .
  • 67.
    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. . . . . . .
  • 68.
    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) = . . . . . .
  • 69.
    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) = . . . . . .
  • 70.
    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) = . . . . . .
  • 71.
    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) = . . . . . .
  • 72.
    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 . . . . . .
  • 73.
    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 . . . . . .
  • 74.
    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) . . . . . .
  • 75.
    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) . . . . . .
  • 76.
    Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. . . . . . .
  • 77.
    Example √ Find the extreme values of f(x) = 4 − x2 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.) . . . . . .
  • 78.
    Example √ Find the extreme values of f(x) = 4 − x2 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) = . . . . . .
  • 79.
    Example √ Find the extreme values of f(x) = 4 − x2 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) = . . . . . .
  • 80.
    Example √ Find the extreme values of f(x) = 4 − x2 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) = . . . . . .
  • 81.
    Example √ Find the extreme values of f(x) = 4 − x2 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) = 3 . . . . . .
  • 82.
    Example √ Find the extreme values of f(x) = 4 − x2 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) = 3 . . . . . .
  • 83.
    Example √ Find the extreme values of f(x) = 4 − x2 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) = 3 . . . . . .
  • 84.
    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 . . . . . .
  • 85.
    Challenge: Cubic functions Example How many critical points can a cubic function f(x) = ax3 + bx2 + cx + d have? . . . . . .
  • 86.
    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: . . . . . .
  • 87.
    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. . . . . . .
  • 88.
    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. . . . . . .
  • 89.
    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. . . . . . .
  • 90.
    Review Concept: absolute (global) and relative (local) maxima/minima Fact: Fermat’s theorem: f′ (x) = 0 at local extrema Technique: the Closed Interval Method . . . . . .