Like this presentation? Why not share!

# Lesson 16: Maximum and Minimum Values

## by Matthew Leingang, Clinical Associate Professor of Mathematics at New York University on Mar 23, 2010

• 1,644 views

### Views

Total Views
1,644
Views on SlideShare
1,634
Embed Views
10

Likes
0
36
0

### 2 Embeds10

 http://www.slideshare.net 9 http://www.aoshu.my 1

## Lesson 16: Maximum and Minimum ValuesPresentation Transcript

• Section 4.1 Maximum and Minimum Values V63.0121.006/016, Calculus I March 23, 2010 Announcements Welcome back from Spring Break! Quiz 3: April 2, Sections 2.6–3.5 . . . . . .
• Announcements Welcome back from Spring Break! Quiz 3: April 2, Sections 2.6–3.5 . . . . . .
• Computation of Midterm Letter Grades Q . uizzes . W . ebAssign . 1 . 5% 1 . 0% . . W H M . idterm . 1 . 0% 2 . 5% . 4 . 0% . F . inal . . . . . .
• Computation of Midterm Letter Grades Q . uizzes . W . ebAssign . 1 . 5% 1 . 0% . . W H M . idterm . 1 . 0% 2 . 5% . 4 . 0% . F . inal (to be determined) . . . . . .
• Computation of Midterm Letter Grades W . ebAssign . 1 . 7% H . W . Q . uizzes 1 . 7% . 2 . 5% . 4 . 1% . M . idterm . . . . . .
• Distribution of Midterm Averages and Letter Grades Median = 81.35% (curved to B) Average = 74.39% (curved to B-) Standard Deviation = 21% . . . . . .
• What can I do to improve my grade? Q . uizzes . W . ebAssign . 1 . 5% 1 . 0% . . W H M . idterm . 1 . 0% 2 . 5% . 4 . 0% . F . inal . . . . . .
• What can I do to improve my grade? f .uture Quizzes p . ast Quizzes . . f .uture WA . p . ast WA 6 .% . 9 .% 5 .% 5 .% f .uture HW . 5 .% M . idterm . p . ast HW . 2 . 5% 5 .% . 4 . 0% . F . inal . . . . . .
• What can I do to improve my grade? M . idterm . p . ast Quizzes . 2 . 5% 6 .% p . ast WA f .uture HW . . 5 .% 5 .% p . ast HW f .uture WA . . 5 .% 5 .% . . .% 9 f .uture Quizzes 4 . 0% . F . inal 59% of your grade is still in play! . . . . . .
• Help! Free resources: recitation TAs’ ofﬁce hours my ofﬁce hours Math Tutoring Center (CIWW 524) College Learning Center . . . . . .
• 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. . . 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 . . 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 . . .. 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 #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. . . . . . .
• Bad Example #2 Example Consider the function f(x) = x restricted to the interval [0, 1). . . . . . .
• Bad Example #2 Example Consider the function f(x) = x restricted to the interval [0, 1). . . . | 1 . . . . . . .
• 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). . . . . . .
• 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. . . . . . .
• Final Bad Example Example 1 Consider the function f(x) = is continuous on the closed x interval [1, ∞). . . . . . .
• Final Bad Example Example 1 Consider the function f(x) = is continuous on the closed x interval [1, ∞). . . . 1 . . . . . . .
• Final Bad Example Example 1 Consider the function f(x) = is continuous on the closed x interval [1, ∞). . . . 1 . There is no minimum value (values get arbitrarily close to 0 but do not achieve it). . . . . . .
• Final Bad Example Example 1 Consider the function f(x) = is continuous on the closed x interval [1, ∞). . . . 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. . . . . . .
• 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 . . . . . .
• 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 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 . . . . . .
• 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 x is slightly greater than c, f(x) ≤ f(c). This means f(x) − f(c) ≤0 x−c . . . . . .
• 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 . . . . . .
• 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−c . . . . . .
• 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 . . . . . .
• 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−c . . . . . .
• 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 (global 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 (global min) f(0) = 1 f(1) = 0 f(2) = 5 (global 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 (global min) f(0) = 1 (local max) f(1) = 0 f(2) = 5 (global 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 (global min) f(0) = 1 (local max) f(1) = 0 (local min) f(2) = 5 (global 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. . . . . . .
• What have we learned today? 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! . . . . . .