Matthew Leingang, profile picture
Uploaded byMatthew Leingang
PDF, PPTX3,375 views

Lesson 18: Maximum and Minimum Values (slides)

The document outlines the concepts of maximum and minimum values in calculus, focusing on the Extreme Value Theorem and Fermat's Theorem. It explains the conditions under which a function can attain absolute maximum and minimum values, emphasizing the importance of continuity and closed intervals. Examples included illustrate methods for finding these extrema for both linear and quadratic functions.

Embed presentation

Download as PDF, PPTX
Sec on 4.1 Maximum and Minimum Values V63.0121.011: Calculus I Professor Ma hew Leingang New York University April 4, 2011 .
Announcements Quiz 4 on Sec ons 3.3, 3.4, 3.5, and 3.7 next week (April 14/15) Quiz 5 on Sec ons 4.1–4.4 April 28/29 Final Exam Monday May 12, 2:00–3:50pm
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 func on defined on a closed interval.
Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
Optimize .
Why go to the extremes? Ra onally speaking, it is advantageous to find the extreme values of a func on (maximize profit, minimize costs, etc.) Pierre-Louis Maupertuis (1698–1759)
Design .
Why go to the extremes? Ra onally speaking, it is advantageous to find the extreme values of a func on (maximize profit, minimize costs, etc.) Many laws of science are derived from minimizing principles. Pierre-Louis Maupertuis (1698–1759)
Optics .
Why go to the extremes? Ra onally speaking, it is advantageous to find the extreme values of a func on (maximize profit, minimize costs, etc.) Many laws of science are derived from minimizing principles. Maupertuis’ principle: “Ac on is minimized through the wisdom Pierre-Louis Maupertuis of God.” (1698–1759)
Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
Extreme points and values Defini on Let f have domain D. . Image credit: Patrick Q
Extreme points and values Defini on Let f have domain D. The func on f has an absolute maximum (or global maximum) (respec vely, absolute minimum) at c if f(c) ≥ f(x) (respec vely, f(c) ≤ f(x)) for all x in D . Image credit: Patrick Q
Extreme points and values Defini on Let f have domain D. The func on f has an absolute maximum (or global maximum) (respec vely, absolute minimum) at c if f(c) ≥ f(x) (respec vely, f(c) ≤ f(x)) for all x in D The number f(c) is called the maximum value (respec vely, minimum value) of f on D. . Image credit: Patrick Q
Extreme points and values Defini on Let f have domain D. The func on f has an absolute maximum (or global maximum) (respec vely, absolute minimum) at c if f(c) ≥ f(x) (respec vely, f(c) ≤ f(x)) for all x in D The number f(c) is called the maximum value (respec vely, 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
The Extreme Value Theorem Theorem (The Extreme Value Theorem) Let f be a func on which is con nuous on the closed interval [a, b]. Then f a ains an absolute maximum value f(c) and an absolute minimum value f(d) at numbers c and d in [a, b].
The Extreme Value Theorem Theorem (The Extreme Value Theorem) Let f be a func on which is con nuous on the closed interval [a, b]. Then f a ains an absolute maximum value f(c) and an absolute minimum value f(d) at numbers c and d . in [a, b]. a b
The Extreme Value Theorem Theorem (The Extreme Value Theorem) maximum value Let f be a func on which is f(c) con nuous on the closed interval [a, b]. Then f a ains an absolute maximum value f(c) and an absolute minimum value f(d) at numbers c and d . a c in [a, b]. b maximum
The Extreme Value Theorem Theorem (The Extreme Value Theorem) maximum value Let f be a func on which is f(c) con nuous on the closed interval [a, b]. Then f a ains minimum an absolute maximum value value f(c) and an absolute minimum f(d) value f(d) at numbers c and d . a d c in [a, b]. b minimum maximum
No proof of EVT forthcoming This theorem is very hard to prove without using technical facts about con nuous func ons and closed intervals. But we can show the importance of each of the hypotheses.
Bad Example #1 Example Consider the func on { x 0≤x<1 f(x) = x − 2 1 ≤ x ≤ 2.
Bad Example #1 Example Consider the func on { x 0≤x<1 . f(x) = | x − 2 1 ≤ x ≤ 2. 1
Bad Example #1 Example Consider the func on { 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 func on { 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 con nuous.
Bad Example #2 Example Consider the func on f(x) = x restricted to the interval [0, 1). . | 1
Bad Example #2 Example Consider the func on f(x) = x restricted to the interval [0, 1). There is s ll no maximum value (values get arbitrarily close to 1 but do not achieve it). . | 1
Bad Example #2 Example Consider the func on f(x) = x restricted to the interval [0, 1). There is s ll no maximum value (values get arbitrarily close to 1 but do not achieve it). This does not violate EVT . | because the domain is 1 not closed.
Final Bad Example Example 1 The func on f(x) = is con nuous on the closed interval [1, ∞). x . 1
Final Bad Example Example 1 The func on f(x) = is con nuous on the closed interval [1, ∞). x . 1 There is no minimum value (values get arbitrarily close to 0 but do not achieve it).
Final Bad Example Example 1 The func on f(x) = is con nuous 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.
Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
Local extrema Defini on A func on f has a local maximum or rela ve 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. |. | a b Similarly, f has a local minimum at c if f(c) ≤ f(x) when x is near c.
Local extrema Defini on A func on f has a local maximum or rela ve 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. |. | a local b Similarly, f has a local minimum maximum at c if f(c) ≤ f(x) when x is near c.
Local extrema Defini on A func on f has a local maximum or rela ve 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. |. | local local b a Similarly, f has a local minimum maximum minimum at c if f(c) ≤ f(x) when x is near c.
Local extrema A local extremum could be a global extremum, but not if there are more extreme values elsewhere. A global extremum could be a local extremum, but not if it is an endpoint. |. | a b local local and global maximum global max min
Fermat’s Theorem Theorem (Fermat’s Theorem) Suppose f has a local extremum at c and f is differen able at c. Then f′ (c) = 0. |. | a local local b maximum minimum
Fermat’s Theorem Theorem (Fermat’s Theorem) Suppose f has a local extremum at c and f is differen able at c. Then f′ (c) = 0. |. | a local local b maximum minimum
Proof of Fermat’s Theorem Suppose that f has a local maximum at c.
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
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
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
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
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 solu ons to x2 + y2 = z2 among posi ve whole numbers (e.g., x = 3, y = 4, z = 5)
Tangent: Fermat’s Last Theorem Plenty of solu ons to x2 + y2 = z2 among posi ve whole numbers (e.g., x = 3, y = 4, z = 5) No solu ons to x3 + y3 = z3 among posi ve whole numbers
Tangent: Fermat’s Last Theorem Plenty of solu ons to x2 + y2 = z2 among posi ve whole numbers (e.g., x = 3, y = 4, z = 5) No solu ons to x3 + y3 = z3 among posi ve whole numbers Fermat claimed no solu ons to xn + yn = zn but didn’t write down his proof
Tangent: Fermat’s Last Theorem Plenty of solu ons to x2 + y2 = z2 among posi ve whole numbers (e.g., x = 3, y = 4, z = 5) No solu ons to x3 + y3 = z3 among posi ve whole numbers Fermat claimed no solu ons to xn + yn = zn but didn’t write down his proof Not solved un l 1998! (Taylor–Wiles)
Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
Flowchart for placing extrema Thanks to Fermat Suppose f is a c is a . start con nuous local max func on on the closed, bounded Is c an no Is f diff’ble no f is not interval endpoint? at c? diff at c [a, b], and c is a global yes yes maximum c = a or point. 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 cri cal points or cri cal numbers x where either f′ (x) = 0 or f is not differen able at x. The points with the largest func on value are the global maximum points The points with the smallest or most nega ve func on value are the global minimum points.
Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
Extreme values of a linear function Example Find the extreme values of f(x) = 2x − 5 on [−1, 2].
Extreme values of a linear function Example Find the extreme values of f(x) = 2x − 5 on [−1, 2]. Solu on Since f′ (x) = 2, which is never zero, we have no cri cal points and we need only inves gate the endpoints: f(−1) = 2(−1) − 5 = −7 f(2) = 2(2) − 5 = −1
Extreme values of a linear function Example Find the extreme values of f(x) = 2x − 5 on [−1, 2]. Solu on So Since f′ (x) = 2, which is never zero, we have no cri cal points The absolute minimum and we need only inves gate (point) is at −1; the the endpoints: minimum value is −7. f(−1) = 2(−1) − 5 = −7 The absolute maximum (point) is at 2; the f(2) = 2(2) − 5 = −1 maximum value is −1.
Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on We have f′ (x) = 2x, which is zero when x = 0.
Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = f(0) = f(2) =
Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = 0 f(0) = f(2) =
Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on 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) =
Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on 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
Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on 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
Extreme values of a quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on 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)
Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1.
Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are
Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4
Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4 f(0) = 1
Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4 f(0) = 1 f(1) = 0
Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4 f(0) = 1 f(1) = 0 f(2) = 5
Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal 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
Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal 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)
Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal 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)
Extreme values of a cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal 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)
Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 .
Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . Then 5 4 1 f′ (x) = x2/3 + x−1/3 = x−1/3 (5x + 4) 3 3 3
Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . Then 5 4 1 f′ (x) = x2/3 + x−1/3 = x−1/3 (5x + 4) 3 3 3 Thus f′ (−4/5) = 0 and f is not differen able at 0. Thus there are two cri cal points.
Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = f(−4/5) = f(0) = f(2) =
Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = f(0) = f(2) =
Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = f(2) =
Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 f(2) =
Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 f(2) = 6.3496
Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 (absolute min) f(2) = 6.3496
Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 (absolute min) f(2) = 6.3496 (absolute max)
Extreme values of an algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 (rela ve max) f(0) = 0 (absolute min) f(2) = 6.3496 (absolute max)
Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1].
Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.)
Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) =
Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 f(0) =
Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 f(0) = 2 f(1) =
Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 f(0) = 2 √ f(1) = 3
Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 (absolute min) f(0) = 2 √ f(1) = 3
Extreme values of another algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 (absolute min) f(0) = 2 (absolute max) √ f(1) = 3
Summary The Extreme Value Theorem: a con nuous func on on a closed interval must achieve its max and min Fermat’s Theorem: local extrema are cri cal points The Closed Interval Method: an algorithm for finding global extrema Show your work unless you want to end up like Fermat!

More Related Content

PPTX
Limits and continuity powerpoint
bycanalculus
 
PPT
3.1 Extreme Values of Functions
bySharon Henry
 
PPT
Application of derivative
byAmiraj College Of Engineering And Technology
 
PDF
Benginning Calculus Lecture notes 2 - limits and continuity
bybasyirstar
 
PDF
Fourier series 2
byDr. Nirav Vyas
 
PPTX
Limit of functions
byJuan Apolinario Reyes
 
PPTX
C.v.n.m (m.e. 130990119004-06)
byparth98796
 
PPT
Partial Differentiation & Application
byYana Qlah
 
Limits and continuity powerpoint
bycanalculus
 
3.1 Extreme Values of Functions
bySharon Henry
 
Application of derivative
byAmiraj College Of Engineering And Technology
 
Benginning Calculus Lecture notes 2 - limits and continuity
bybasyirstar
 
Fourier series 2
byDr. Nirav Vyas
 
Limit of functions
byJuan Apolinario Reyes
 
C.v.n.m (m.e. 130990119004-06)
byparth98796
 
Partial Differentiation & Application
byYana Qlah
 

What's hot

PPTX
Lagrange multiplier
byAnjali Devi J S
 
PPTX
Application of partial derivatives with two variables
bySagar Patel
 
PPTX
application of partial differentiation
byeteaching
 
PPTX
Limit of Function And Its Types
byAdeel Rasheed
 
PPT
Applications of maxima and minima
byrouwejan
 
PPTX
Limits and its theorem In Calculus
byUzair Ahmad
 
PPT
Application of derivatives 2 maxima and minima
bysudersana viswanathan
 
PDF
Langrange Interpolation Polynomials
bySohaib H. Khan
 
PPTX
Limit and continuity (2)
byDigvijaysinh Gohil
 
PDF
Eigenvalue eigenvector slides
byAmanSaeed11
 
PPT
Limits and continuity
byDigvijaysinh Gohil
 
PDF
Lesson 16: Derivatives of Exponential and Logarithmic Functions
byMatthew Leingang
 
PPTX
Types of function
bySanaullahMemon10
 
DOCX
Binomial theorem for any index
byindu psthakur
 
PPTX
Partial differential equations
byaman1894
 
PPTX
Complex integration
bySanthanam Krishnan
 
PPTX
The chain rule
byJ M
 
PPT
Functions
byDreams4school
 
PDF
Lesson 3: The Limit of a Function
byMatthew Leingang
 
PPTX
Newton's forward difference
byRaj Parekh
 
Lagrange multiplier
byAnjali Devi J S
 
Application of partial derivatives with two variables
bySagar Patel
 
application of partial differentiation
byeteaching
 
Limit of Function And Its Types
byAdeel Rasheed
 
Applications of maxima and minima
byrouwejan
 
Limits and its theorem In Calculus
byUzair Ahmad
 
Application of derivatives 2 maxima and minima
bysudersana viswanathan
 
Langrange Interpolation Polynomials
bySohaib H. Khan
 
Limit and continuity (2)
byDigvijaysinh Gohil
 
Eigenvalue eigenvector slides
byAmanSaeed11
 
Limits and continuity
byDigvijaysinh Gohil
 
Lesson 16: Derivatives of Exponential and Logarithmic Functions
byMatthew Leingang
 
Types of function
bySanaullahMemon10
 
Binomial theorem for any index
byindu psthakur
 
Partial differential equations
byaman1894
 
Complex integration
bySanthanam Krishnan
 
The chain rule
byJ M
 
Functions
byDreams4school
 
Lesson 3: The Limit of a Function
byMatthew Leingang
 
Newton's forward difference
byRaj Parekh
 

Viewers also liked

PDF
Lesson 14: Derivatives of Logarithmic and Exponential Functions (slides)
byMatthew Leingang
 
PDF
Lesson 5: Continuity (slides)
byMatthew Leingang
 
PDF
Lesson 16: Inverse Trigonometric Functions (slides)
byMatthew Leingang
 
PDF
Lesson 4: Calcuating Limits (slides)
byMatthew Leingang
 
PDF
Lesson 19: The Mean Value Theorem (slides)
byMatthew Leingang
 
PDF
Lesson 6: Limits Involving Infinity (slides)
byMatthew Leingang
 
PDF
Lesson 9: The Product and Quotient Rules (slides)
byMatthew Leingang
 
PDF
Lesson 11: Implicit Differentiation (slides)
byMatthew Leingang
 
PDF
Lesson 20: Derivatives and the Shapes of Curves (slides)
byMatthew Leingang
 
PDF
Lesson 8: Basic Differentation Rules (slides)
byMatthew Leingang
 
PDF
Lesson 1: Functions and their representations (slides)
byMatthew Leingang
 
PDF
Lesson 17: Indeterminate forms and l'Hôpital's Rule (slides)
byMatthew Leingang
 
PDF
Lesson 21: Curve Sketching (slides)
byMatthew Leingang
 
PDF
Lesson 12: Linear Approximations and Differentials (slides)
byMatthew Leingang
 
PDF
Lesson 2: A Catalog of Essential Functions (slides)
byMatthew Leingang
 
PDF
Lesson 15: Exponential Growth and Decay (slides)
byMatthew Leingang
 
PDF
Lesson 10: The Chain Rule (slides)
byMatthew Leingang
 
PDF
Lesson 7: The Derivative (slides)
byMatthew Leingang
 
PDF
Lesson 3: The Limit of a Function (slides)
byMatthew Leingang
 
DOC
Tugasan 4
byNoor Abidin
 
Lesson 14: Derivatives of Logarithmic and Exponential Functions (slides)
byMatthew Leingang
 
Lesson 5: Continuity (slides)
byMatthew Leingang
 
Lesson 16: Inverse Trigonometric Functions (slides)
byMatthew Leingang
 
Lesson 4: Calcuating Limits (slides)
byMatthew Leingang
 
Lesson 19: The Mean Value Theorem (slides)
byMatthew Leingang
 
Lesson 6: Limits Involving Infinity (slides)
byMatthew Leingang
 
Lesson 9: The Product and Quotient Rules (slides)
byMatthew Leingang
 
Lesson 11: Implicit Differentiation (slides)
byMatthew Leingang
 
Lesson 20: Derivatives and the Shapes of Curves (slides)
byMatthew Leingang
 
Lesson 8: Basic Differentation Rules (slides)
byMatthew Leingang
 
Lesson 1: Functions and their representations (slides)
byMatthew Leingang
 
Lesson 17: Indeterminate forms and l'Hôpital's Rule (slides)
byMatthew Leingang
 
Lesson 21: Curve Sketching (slides)
byMatthew Leingang
 
Lesson 12: Linear Approximations and Differentials (slides)
byMatthew Leingang
 
Lesson 2: A Catalog of Essential Functions (slides)
byMatthew Leingang
 
Lesson 15: Exponential Growth and Decay (slides)
byMatthew Leingang
 
Lesson 10: The Chain Rule (slides)
byMatthew Leingang
 
Lesson 7: The Derivative (slides)
byMatthew Leingang
 
Lesson 3: The Limit of a Function (slides)
byMatthew Leingang
 
Tugasan 4
byNoor Abidin
 

Similar to Lesson 18: Maximum and Minimum Values (slides)

PDF
Lesson 19: Maximum and Minimum Values
byMatthew Leingang
 
PDF
Applications of Differentiation
byJoey Fontanilla Valdriz
 
PDF
Lesson19 Maximum And Minimum Values 034 Slides
byMatthew Leingang
 
PPT
derivatives. maximum and minimum value..
byayaulymsun
 
PPTX
Ap calculus extrema v2
bygregcross22
 
PPT
2301MaxMin.ppt
byAsadAltaf10
 
PDF
Lesson 18: Maximum and Minimum Values (Section 021 slides)
byMatthew Leingang
 
PPT
Lecture 15 max min - section 4.2
bynjit-ronbrown
 
PPT
derivatives. maximum and minimum value..
byayaulymsun
 
PDF
Lesson 18: Maximum and Minimum Vaues
byMatthew Leingang
 
PDF
Lesson 18: Maximum and Minimum Vaues
byMatthew Leingang
 
PDF
Lesson 18: Maximum and Minimum Values (Section 021 handout)
byMatthew Leingang
 
PDF
Lesson 18: Maximum and Minimum Values (Section 041 handout)
byMatthew Leingang
 
PDF
Lesson 18: Maximum and Minimum Values (Section 041 slides)
byMatthew Leingang
 
PDF
Lesson 18: Maximum and Minimum Values (Section 021 slides)
byMel Anthony Pepito
 
PDF
Lesson 18: Maximum and Minimum Values (Section 041 slides)
byMel Anthony Pepito
 
PDF
Lesson18 -maximum_and_minimum_values_slides
byMel Anthony Pepito
 
PDF
Lesson18 -maximum_and_minimum_values_slides
byMatthew Leingang
 
PDF
Lesson 18: Maximum and Minimum Values (handout)
byMatthew Leingang
 
PPTX
A07-derivarive-how to finde some values of
bydavi196568
 
Lesson 19: Maximum and Minimum Values
byMatthew Leingang
 
Applications of Differentiation
byJoey Fontanilla Valdriz
 
Lesson19 Maximum And Minimum Values 034 Slides
byMatthew Leingang
 
derivatives. maximum and minimum value..
byayaulymsun
 
Ap calculus extrema v2
bygregcross22
 
2301MaxMin.ppt
byAsadAltaf10
 
Lesson 18: Maximum and Minimum Values (Section 021 slides)
byMatthew Leingang
 
Lecture 15 max min - section 4.2
bynjit-ronbrown
 
derivatives. maximum and minimum value..
byayaulymsun
 
Lesson 18: Maximum and Minimum Vaues
byMatthew Leingang
 
Lesson 18: Maximum and Minimum Vaues
byMatthew Leingang
 
Lesson 18: Maximum and Minimum Values (Section 021 handout)
byMatthew Leingang
 
Lesson 18: Maximum and Minimum Values (Section 041 handout)
byMatthew Leingang
 
Lesson 18: Maximum and Minimum Values (Section 041 slides)
byMatthew Leingang
 
Lesson 18: Maximum and Minimum Values (Section 021 slides)
byMel Anthony Pepito
 
Lesson 18: Maximum and Minimum Values (Section 041 slides)
byMel Anthony Pepito
 
Lesson18 -maximum_and_minimum_values_slides
byMel Anthony Pepito
 
Lesson18 -maximum_and_minimum_values_slides
byMatthew Leingang
 
Lesson 18: Maximum and Minimum Values (handout)
byMatthew Leingang
 
A07-derivarive-how to finde some values of
bydavi196568
 

More from Matthew Leingang

PDF
Lesson 22: Optimization Problems (slides)
byMatthew Leingang
 
PDF
Lesson 27: Integration by Substitution (slides)
byMatthew Leingang
 
PDF
Lesson 26: The Fundamental Theorem of Calculus (slides)
byMatthew Leingang
 
PDF
Lesson 25: Evaluating Definite Integrals (slides)
byMatthew Leingang
 
PDF
Lesson 23: Antiderivatives (slides)
byMatthew Leingang
 
PDF
Lesson 16: Inverse Trigonometric Functions (handout)
byMatthew Leingang
 
PDF
Lesson 24: Areas and Distances, The Definite Integral (slides)
byMatthew Leingang
 
PDF
Lesson 17: Indeterminate forms and l'Hôpital's Rule (handout)
byMatthew Leingang
 
PPT
Making Lesson Plans
byMatthew Leingang
 
PDF
Lesson 21: Curve Sketching (handout)
byMatthew Leingang
 
PDF
Lesson 22: Optimization Problems (handout)
byMatthew Leingang
 
PDF
Lesson 27: Integration by Substitution (handout)
byMatthew Leingang
 
PPT
Streamlining assessment, feedback, and archival with auto-multiple-choice
byMatthew Leingang
 
PDF
Lesson 26: The Fundamental Theorem of Calculus (slides)
byMatthew Leingang
 
PDF
Lesson 26: The Fundamental Theorem of Calculus (handout)
byMatthew Leingang
 
PDF
Lesson 20: Derivatives and the Shapes of Curves (handout)
byMatthew Leingang
 
PDF
Lesson 25: Evaluating Definite Integrals (handout)
byMatthew Leingang
 
PDF
Electronic Grading of Paper Assessments
byMatthew Leingang
 
PDF
Lesson 23: Antiderivatives (slides)
byMatthew Leingang
 
PDF
Lesson 24: Areas and Distances, The Definite Integral (handout)
byMatthew Leingang
 
Lesson 22: Optimization Problems (slides)
byMatthew Leingang
 
Lesson 27: Integration by Substitution (slides)
byMatthew Leingang
 
Lesson 26: The Fundamental Theorem of Calculus (slides)
byMatthew Leingang
 
Lesson 25: Evaluating Definite Integrals (slides)
byMatthew Leingang
 
Lesson 23: Antiderivatives (slides)
byMatthew Leingang
 
Lesson 16: Inverse Trigonometric Functions (handout)
byMatthew Leingang
 
Lesson 24: Areas and Distances, The Definite Integral (slides)
byMatthew Leingang
 
Lesson 17: Indeterminate forms and l'Hôpital's Rule (handout)
byMatthew Leingang
 
Making Lesson Plans
byMatthew Leingang
 
Lesson 21: Curve Sketching (handout)
byMatthew Leingang
 
Lesson 22: Optimization Problems (handout)
byMatthew Leingang
 
Lesson 27: Integration by Substitution (handout)
byMatthew Leingang
 
Streamlining assessment, feedback, and archival with auto-multiple-choice
byMatthew Leingang
 
Lesson 26: The Fundamental Theorem of Calculus (slides)
byMatthew Leingang
 
Lesson 26: The Fundamental Theorem of Calculus (handout)
byMatthew Leingang
 
Lesson 20: Derivatives and the Shapes of Curves (handout)
byMatthew Leingang
 
Lesson 25: Evaluating Definite Integrals (handout)
byMatthew Leingang
 
Electronic Grading of Paper Assessments
byMatthew Leingang
 
Lesson 23: Antiderivatives (slides)
byMatthew Leingang
 
Lesson 24: Areas and Distances, The Definite Integral (handout)
byMatthew Leingang
 

Recently uploaded

PDF
Digit Expo 2025 - EICC Edinburgh 27th November
byRay Bugg
 
PDF
8 LLM Surveys https://tinyurl.com/bdz8e6fp
byBob Marcus
 
PDF
Zero Trust & Defense-in-Depth: The Future of Critical Infrastructure Security
bysleepandremedies
 
PPTX
The Future of IT Service Management AI Automation & Beyond.pptx
byChetu
 
PDF
Innovative AI Solutions for Business Growth.pdf
byVox Works
 
PDF
AI and Computer Architecture: 200 Years Together
byDmitry Zinoviev
 
PDF
HCI and Its Impact on individual quality of life.pdf
byParham Abolghasemi
 
PDF
Igniting the Future: Copilot trends, agentic transformation and product roadm...
byUni Systems S.M.S.A.
 
PDF
Cleveland, OH Developer Group Dec 2, 2025 Slides for December Demos
byLynda Kane
 
PDF
Generative AI in UiPath: Mastering the Generative Extractor for Intelligent D...
byDianaGray10
 
PPTX
Ritesh_kumar_Aatmanirbhar Bharat: Make in India, Make for the World.pptx
byriteshrkgs2008
 
PDF
How Mobile Apps Are Shaping the Future of Digital Innovation
byWilliam Taylor
 
PPTX
The Landscape of Computer Software Development Companies
byYash Singh
 
PDF
Agentic Automation for Testers – A Hands-On Deep Dive
byUiPathCommunity
 
PDF
Top Cybersecurity Threats 2025 Guide by Sureshdas
byPrintersassist
 
PPTX
RISE with SAP for Automotive - S4 HANA Value.pptx
byAmira Jemi
 
PDF
Top 7 Manufacturing Software for Small Businesses Boosting Growth in 2025
byEnvertis Software Solutions
 
PDF
Recursive Self Improvement vs Continuous Learning
byBob Marcus
 
PDF
AI Technology important knowledge ......
byutkarshj2007
 
PDF
The Ultimate Guide to Problem Management Dashboards for IT Teams.pdf
byomnex systems
 
Digit Expo 2025 - EICC Edinburgh 27th November
byRay Bugg
 
8 LLM Surveys https://tinyurl.com/bdz8e6fp
byBob Marcus
 
Zero Trust & Defense-in-Depth: The Future of Critical Infrastructure Security
bysleepandremedies
 
The Future of IT Service Management AI Automation & Beyond.pptx
byChetu
 
Innovative AI Solutions for Business Growth.pdf
byVox Works
 
AI and Computer Architecture: 200 Years Together
byDmitry Zinoviev
 
HCI and Its Impact on individual quality of life.pdf
byParham Abolghasemi
 
Igniting the Future: Copilot trends, agentic transformation and product roadm...
byUni Systems S.M.S.A.
 
Cleveland, OH Developer Group Dec 2, 2025 Slides for December Demos
byLynda Kane
 
Generative AI in UiPath: Mastering the Generative Extractor for Intelligent D...
byDianaGray10
 
Ritesh_kumar_Aatmanirbhar Bharat: Make in India, Make for the World.pptx
byriteshrkgs2008
 
How Mobile Apps Are Shaping the Future of Digital Innovation
byWilliam Taylor
 
The Landscape of Computer Software Development Companies
byYash Singh
 
Agentic Automation for Testers – A Hands-On Deep Dive
byUiPathCommunity
 
Top Cybersecurity Threats 2025 Guide by Sureshdas
byPrintersassist
 
RISE with SAP for Automotive - S4 HANA Value.pptx
byAmira Jemi
 
Top 7 Manufacturing Software for Small Businesses Boosting Growth in 2025
byEnvertis Software Solutions
 
Recursive Self Improvement vs Continuous Learning
byBob Marcus
 
AI Technology important knowledge ......
byutkarshj2007
 
The Ultimate Guide to Problem Management Dashboards for IT Teams.pdf
byomnex systems
 

Lesson 18: Maximum and Minimum Values (slides)

  • 1.
    Sec on 4.1 Maximum and Minimum Values V63.0121.011: Calculus I Professor Ma hew Leingang New York University April 4, 2011 .
  • 2.
    Announcements Quiz4 on Sec ons 3.3, 3.4, 3.5, and 3.7 next week (April 14/15) Quiz 5 on Sec ons 4.1–4.4 April 28/29 Final Exam Monday May 12, 2:00–3:50pm
  • 3.
    Objectives Understandand 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 func on defined on a closed interval.
  • 4.
    Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
  • 5.
  • 6.
    Why go tothe extremes? Ra onally speaking, it is advantageous to find the extreme values of a func on (maximize profit, minimize costs, etc.) Pierre-Louis Maupertuis (1698–1759)
  • 7.
  • 8.
    Why go tothe extremes? Ra onally speaking, it is advantageous to find the extreme values of a func on (maximize profit, minimize costs, etc.) Many laws of science are derived from minimizing principles. Pierre-Louis Maupertuis (1698–1759)
  • 9.
  • 10.
    Why go tothe extremes? Ra onally speaking, it is advantageous to find the extreme values of a func on (maximize profit, minimize costs, etc.) Many laws of science are derived from minimizing principles. Maupertuis’ principle: “Ac on is minimized through the wisdom Pierre-Louis Maupertuis of God.” (1698–1759)
  • 11.
    Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
  • 12.
    Extreme points andvalues Defini on Let f have domain D. . Image credit: Patrick Q
  • 13.
    Extreme points andvalues Defini on Let f have domain D. The func on f has an absolute maximum (or global maximum) (respec vely, absolute minimum) at c if f(c) ≥ f(x) (respec vely, f(c) ≤ f(x)) for all x in D . Image credit: Patrick Q
  • 14.
    Extreme points andvalues Defini on Let f have domain D. The func on f has an absolute maximum (or global maximum) (respec vely, absolute minimum) at c if f(c) ≥ f(x) (respec vely, f(c) ≤ f(x)) for all x in D The number f(c) is called the maximum value (respec vely, minimum value) of f on D. . Image credit: Patrick Q
  • 15.
    Extreme points andvalues Defini on Let f have domain D. The func on f has an absolute maximum (or global maximum) (respec vely, absolute minimum) at c if f(c) ≥ f(x) (respec vely, f(c) ≤ f(x)) for all x in D The number f(c) is called the maximum value (respec vely, 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
  • 16.
    The Extreme ValueTheorem Theorem (The Extreme Value Theorem) Let f be a func on which is con nuous on the closed interval [a, b]. Then f a ains an absolute maximum value f(c) and an absolute minimum value f(d) at numbers c and d in [a, b].
  • 17.
    The Extreme ValueTheorem Theorem (The Extreme Value Theorem) Let f be a func on which is con nuous on the closed interval [a, b]. Then f a ains an absolute maximum value f(c) and an absolute minimum value f(d) at numbers c and d . in [a, b]. a b
  • 18.
    The Extreme ValueTheorem Theorem (The Extreme Value Theorem) maximum value Let f be a func on which is f(c) con nuous on the closed interval [a, b]. Then f a ains an absolute maximum value f(c) and an absolute minimum value f(d) at numbers c and d . a c in [a, b]. b maximum
  • 19.
    The Extreme ValueTheorem Theorem (The Extreme Value Theorem) maximum value Let f be a func on which is f(c) con nuous on the closed interval [a, b]. Then f a ains minimum an absolute maximum value value f(c) and an absolute minimum f(d) value f(d) at numbers c and d . a d c in [a, b]. b minimum maximum
  • 20.
    No proof ofEVT forthcoming This theorem is very hard to prove without using technical facts about con nuous func ons and closed intervals. But we can show the importance of each of the hypotheses.
  • 21.
    Bad Example #1 Example Consider the func on { x 0≤x<1 f(x) = x − 2 1 ≤ x ≤ 2.
  • 22.
    Bad Example #1 Example Consider the func on { x 0≤x<1 . f(x) = | x − 2 1 ≤ x ≤ 2. 1
  • 23.
    Bad Example #1 Example Consider the func on { 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.
  • 24.
    Bad Example #1 Example Consider the func on { 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 con nuous.
  • 25.
    Bad Example #2 Example Consider the func on f(x) = x restricted to the interval [0, 1). . | 1
  • 26.
    Bad Example #2 Example Consider the func on f(x) = x restricted to the interval [0, 1). There is s ll no maximum value (values get arbitrarily close to 1 but do not achieve it). . | 1
  • 27.
    Bad Example #2 Example Consider the func on f(x) = x restricted to the interval [0, 1). There is s ll no maximum value (values get arbitrarily close to 1 but do not achieve it). This does not violate EVT . | because the domain is 1 not closed.
  • 28.
    Final Bad Example Example 1 The func on f(x) = is con nuous on the closed interval [1, ∞). x . 1
  • 29.
    Final Bad Example Example 1 The func on f(x) = is con nuous on the closed interval [1, ∞). x . 1 There is no minimum value (values get arbitrarily close to 0 but do not achieve it).
  • 30.
    Final Bad Example Example 1 The func on f(x) = is con nuous 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.
  • 31.
    Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
  • 32.
    Local extrema Defini on A func on f has a local maximum or rela ve 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. |. | a b Similarly, f has a local minimum at c if f(c) ≤ f(x) when x is near c.
  • 33.
    Local extrema Defini on A func on f has a local maximum or rela ve 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. |. | a local b Similarly, f has a local minimum maximum at c if f(c) ≤ f(x) when x is near c.
  • 34.
    Local extrema Defini on A func on f has a local maximum or rela ve 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. |. | local local b a Similarly, f has a local minimum maximum minimum at c if f(c) ≤ f(x) when x is near c.
  • 35.
    Local extrema A local extremum could be a global extremum, but not if there are more extreme values elsewhere. A global extremum could be a local extremum, but not if it is an endpoint. |. | a b local local and global maximum global max min
  • 36.
    Fermat’s Theorem Theorem(Fermat’s Theorem) Suppose f has a local extremum at c and f is differen able at c. Then f′ (c) = 0. |. | a local local b maximum minimum
  • 37.
    Fermat’s Theorem Theorem(Fermat’s Theorem) Suppose f has a local extremum at c and f is differen able at c. Then f′ (c) = 0. |. | a local local b maximum minimum
  • 38.
    Proof of Fermat’sTheorem Suppose that f has a local maximum at c.
  • 39.
    Proof of Fermat’sTheorem 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
  • 40.
    Proof of Fermat’sTheorem 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
  • 41.
    Proof of Fermat’sTheorem 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
  • 42.
    Proof of Fermat’sTheorem 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
  • 43.
    Proof of Fermat’sTheorem 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
  • 44.
    Meet the Mathematician:Pierre de Fermat 1601–1665 Lawyer and number theorist Proved many theorems, didn’t quite prove his last one
  • 45.
    Tangent: Fermat’s LastTheorem Plenty of solu ons to x2 + y2 = z2 among posi ve whole numbers (e.g., x = 3, y = 4, z = 5)
  • 46.
    Tangent: Fermat’s LastTheorem Plenty of solu ons to x2 + y2 = z2 among posi ve whole numbers (e.g., x = 3, y = 4, z = 5) No solu ons to x3 + y3 = z3 among posi ve whole numbers
  • 47.
    Tangent: Fermat’s LastTheorem Plenty of solu ons to x2 + y2 = z2 among posi ve whole numbers (e.g., x = 3, y = 4, z = 5) No solu ons to x3 + y3 = z3 among posi ve whole numbers Fermat claimed no solu ons to xn + yn = zn but didn’t write down his proof
  • 48.
    Tangent: Fermat’s LastTheorem Plenty of solu ons to x2 + y2 = z2 among posi ve whole numbers (e.g., x = 3, y = 4, z = 5) No solu ons to x3 + y3 = z3 among posi ve whole numbers Fermat claimed no solu ons to xn + yn = zn but didn’t write down his proof Not solved un l 1998! (Taylor–Wiles)
  • 49.
    Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
  • 50.
    Flowchart for placingextrema Thanks to Fermat Suppose f is a c is a . start con nuous local max func on on the closed, bounded Is c an no Is f diff’ble no f is not interval endpoint? at c? diff at c [a, b], and c is a global yes yes maximum c = a or point. f′ (c) = 0 c = b
  • 51.
    The Closed IntervalMethod 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 cri cal points or cri cal numbers x where either f′ (x) = 0 or f is not differen able at x. The points with the largest func on value are the global maximum points The points with the smallest or most nega ve func on value are the global minimum points.
  • 52.
    Outline Introduc on The Extreme Value Theorem Fermat’s Theorem (not the last one) Tangent: Fermat’s Last Theorem The Closed Interval Method Examples
  • 53.
    Extreme values ofa linear function Example Find the extreme values of f(x) = 2x − 5 on [−1, 2].
  • 54.
    Extreme values ofa linear function Example Find the extreme values of f(x) = 2x − 5 on [−1, 2]. Solu on Since f′ (x) = 2, which is never zero, we have no cri cal points and we need only inves gate the endpoints: f(−1) = 2(−1) − 5 = −7 f(2) = 2(2) − 5 = −1
  • 55.
    Extreme values ofa linear function Example Find the extreme values of f(x) = 2x − 5 on [−1, 2]. Solu on So Since f′ (x) = 2, which is never zero, we have no cri cal points The absolute minimum and we need only inves gate (point) is at −1; the the endpoints: minimum value is −7. f(−1) = 2(−1) − 5 = −7 The absolute maximum (point) is at 2; the f(2) = 2(2) − 5 = −1 maximum value is −1.
  • 56.
    Extreme values ofa quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2].
  • 57.
    Extreme values ofa quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on We have f′ (x) = 2x, which is zero when x = 0.
  • 58.
    Extreme values ofa quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = f(0) = f(2) =
  • 59.
    Extreme values ofa quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on We have f′ (x) = 2x, which is zero when x = 0. So our points to check are: f(−1) = 0 f(0) = f(2) =
  • 60.
    Extreme values ofa quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on 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) =
  • 61.
    Extreme values ofa quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on 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
  • 62.
    Extreme values ofa quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on 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
  • 63.
    Extreme values ofa quadratic function Example Find the extreme values of f(x) = x2 − 1 on [−1, 2]. Solu on 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)
  • 64.
    Extreme values ofa cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
  • 65.
    Extreme values ofa cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1.
  • 66.
    Extreme values ofa cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are
  • 67.
    Extreme values ofa cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4
  • 68.
    Extreme values ofa cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4 f(0) = 1
  • 69.
    Extreme values ofa cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4 f(0) = 1 f(1) = 0
  • 70.
    Extreme values ofa cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0 and x = 1. The values to check are f(−1) = − 4 f(0) = 1 f(1) = 0 f(2) = 5
  • 71.
    Extreme values ofa cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal 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
  • 72.
    Extreme values ofa cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal 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)
  • 73.
    Extreme values ofa cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal 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)
  • 74.
    Extreme values ofa cubic function Example Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2]. Solu on Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal 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)
  • 75.
    Extreme values ofan algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
  • 76.
    Extreme values ofan algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 .
  • 77.
    Extreme values ofan algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . Then 5 4 1 f′ (x) = x2/3 + x−1/3 = x−1/3 (5x + 4) 3 3 3
  • 78.
    Extreme values ofan algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . Then 5 4 1 f′ (x) = x2/3 + x−1/3 = x−1/3 (5x + 4) 3 3 3 Thus f′ (−4/5) = 0 and f is not differen able at 0. Thus there are two cri cal points.
  • 79.
    Extreme values ofan algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = f(−4/5) = f(0) = f(2) =
  • 80.
    Extreme values ofan algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = f(0) = f(2) =
  • 81.
    Extreme values ofan algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = f(2) =
  • 82.
    Extreme values ofan algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 f(2) =
  • 83.
    Extreme values ofan algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 f(2) = 6.3496
  • 84.
    Extreme values ofan algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 (absolute min) f(2) = 6.3496
  • 85.
    Extreme values ofan algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 f(0) = 0 (absolute min) f(2) = 6.3496 (absolute max)
  • 86.
    Extreme values ofan algebraic function Example Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2]. Solu on Write f(x) = x5/3 + 2x2/3 . f(−1) = 1 f(−4/5) = 1.0341 (rela ve max) f(0) = 0 (absolute min) f(2) = 6.3496 (absolute max)
  • 87.
    Extreme values ofanother algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1].
  • 88.
    Extreme values ofanother algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.)
  • 89.
    Extreme values ofanother algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) =
  • 90.
    Extreme values ofanother algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 f(0) =
  • 91.
    Extreme values ofanother algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 f(0) = 2 f(1) =
  • 92.
    Extreme values ofanother algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 f(0) = 2 √ f(1) = 3
  • 93.
    Extreme values ofanother algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 (absolute min) f(0) = 2 √ f(1) = 3
  • 94.
    Extreme values ofanother algebraic function Example √ Find the extreme values of f(x) = 4 − x2 on [−2, 1]. Solu on x We have f′ (x) = − √ , which is zero when x = 0. (f is not 4 − x2 differen able at ±2 as well.) So our points to check are: f(−2) = 0 (absolute min) f(0) = 2 (absolute max) √ f(1) = 3
  • 95.
    Summary TheExtreme Value Theorem: a con nuous func on on a closed interval must achieve its max and min Fermat’s Theorem: local extrema are cri cal points The Closed Interval Method: an algorithm for finding global extrema Show your work unless you want to end up like Fermat!