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Lesson 18: Maximum and Minimum Values (slides)
 

Lesson 18: Maximum and Minimum Values (slides)

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There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. ...

There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.

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    Lesson 18: Maximum and Minimum Values (slides) Lesson 18: Maximum and Minimum Values (slides) Presentation Transcript

    • 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 extremaDefini 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 extremaDefini 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 extremaDefini 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 extremaThanks to FermatSuppose f is a c is a . startcon nuous local maxfunc on onthe closed,bounded Is c an no Is f diff’ble no f is notinterval endpoint? at c? diff at c[a, b], and c isa global yes yesmaximum c = a orpoint. 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 quadraticfunction Example Find the extreme values of f(x) = x2 − 1 on [−1, 2].
    • Extreme values of a quadraticfunction 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 quadraticfunction 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 quadraticfunction 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 quadraticfunction 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 quadraticfunction 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 quadraticfunction 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 quadraticfunction 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!