Lesson 18: Maximum and Minimum Values (slides)

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Sec on 4.1Maximum and Minimum Values

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

April 4, 2011

Announcements

I Quiz 4 on Sec ons 3.3, 3.4, 3.5,and 3.7 next week (April 14/15)

I Quiz 5 on Sec ons 4.1–4.4April 28/29

I Final Exam Monday May 12,2:00–3:50pm

ObjectivesI Understand and be able toexplain the statement of theExtreme Value Theorem.

I Understand and be able toexplain the statement ofFermat’s Theorem.

I Use the Closed Interval Methodto find the extreme values of afunc on defined on a closedinterval.

OutlineIntroduc 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?I Ra onally speaking, it isadvantageous to find theextreme values of a func on(maximize profit, minimize costs,etc.)

I Many laws of science arederived from minimizingprinciples.

I Maupertuis’ principle: “Ac on isminimized through the wisdomof God.”

Pierre-Louis Maupertuis(1698–1759)

..

Design





..

Optics





OutlineIntroduc on




Examples

Extreme points and valuesDefini onLet f have domain D.

I 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

I The number f(c) is called themaximumvalue (respec vely,minimum value) of fon D.

I An extremum is either a maximum or aminimum. An extreme value is either amaximum value or minimum value.

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Image credit: Patrick Q

http://www.flickr.com/photos/patrick_q/204252734/





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..







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The Extreme Value TheoremTheorem (The Extreme ValueTheorem)

Let f be a func on which iscon nuous on the closedinterval [a, b]. Then f a ainsan absolute maximum valuef(c) and an absolute minimumvalue f(d) at numbers c and din [a, b].

.



...a..

b..



...a..

b...

cmaximum

.

maximumvaluef(c)



...a..

b...

cmaximum

.

maximumvaluef(c)

..d

minimum

.

minimumvaluef(d)

No proof of EVT forthcoming

I This theorem is very hard to prove without using technical factsabout con nuous func ons and closed intervals.

I But we can show the importance of each of the hypotheses.

Bad Example #1Example

Consider the func on

f(x) =

{x 0 ≤ x < 1x− 2 1 ≤ x ≤ 2.

.. |.1

....

Then although values of f(x) get arbitrarily close to 1 and neverbigger than 1, 1 is not the maximum value of f on [0, 1] because it isnever achieved. This does not violate EVT because f is notcon nuous.



f(x) =

{x 0 ≤ x < 1x− 2 1 ≤ x ≤ 2.

.. |.1

....




f(x) =

{x 0 ≤ x < 1x− 2 1 ≤ x ≤ 2.

.. |.1

....

Then although values of f(x) get arbitrarily close to 1 and neverbigger than 1, 1 is not the maximum value of f on [0, 1] because it isnever achieved.

This does not violate EVT because f is notcon nuous.



f(x) =

{x 0 ≤ x < 1x− 2 1 ≤ x ≤ 2.

.. |.1

....



Consider the func on f(x) = x restricted to the interval [0, 1).

I There is s ll no maximumvalue (values getarbitrarily close to 1 butdo not achieve it).

I This does not violate EVTbecause the domain isnot closed.

.. |.1

..





.. |.1

..





.. |.1

..

Final Bad ExampleExample

The func on f(x) =1xis con nuous on the closed interval [1,∞).

...1

.

There is no minimum value (values get arbitrarily close to 0 but donot achieve it). This does not violate EVT because the domain is notbounded.



...1

.

There is no minimum value (values get arbitrarily close to 0 but donot achieve it).

This does not violate EVT because the domain is notbounded.



...1

.

There is no minimum value (values get arbitrarily close to 0 but donot achieve it). This does not violate EVT because the domain is notbounded.

OutlineIntroduc on




Examples

Local extremaDefini on

I A func on f has a localmaximum or rela ve maximumat c if f(c) ≥ f(x) when x is nearc. This means that f(c) ≥ f(x)for all x in some open intervalcontaining c.

I Similarly, f has a local minimumat c if f(c) ≤ f(x) when x is nearc.

..|.a. |.

b..




..|.a. |.

b....

localmaximum




..|.a. |.

b....

localmaximum

..local

minimum

Local extremaI A local extremum could be aglobal extremum, but not ifthere are more extreme valueselsewhere.

I A global extremum could be alocal extremum, but not if it isan endpoint.

..|.a. |.

b....

localmaximum

..globalmax

.local andglobalmin

Fermat’s TheoremTheorem (Fermat’s Theorem)

Suppose f has alocal extremum at cand f isdifferen able at c.Then f′(c) = 0. ..|.

a. |.

b....local

maximum

..localminimum

Fermat’s TheoremTheorem (Fermat’s Theorem)

Suppose f has alocal extremum at cand f isdifferen able at c.Then f′(c) = 0. ..|.

a. |.

b....local

maximum

..localminimum

Proof of Fermat’s TheoremSuppose that f has a local maximum at c.

I If x is slightly greater than c, f(x) ≤ f(c). This means

f(x)− f(c)x− c

≤ 0 =⇒ limx→c+

f(x)− f(c)x− c

≤ 0

I The same will be true on the other end: if x is slightly less thanc, f(x) ≤ f(c). This means

f(x)− f(c)x− c

≥ 0 =⇒ limx→c−

f(x)− f(c)x− c

≥ 0

I Since the limit f′(c) = limx→c

f(x)− f(c)x− c

exists, it must be 0.



f(x)− f(c)x− c

≤ 0

=⇒ limx→c+

f(x)− f(c)x− c

≤ 0


f(x)− f(c)x− c

≥ 0 =⇒ limx→c−

f(x)− f(c)x− c

≥ 0


f(x)− f(c)x− c




f(x)− f(c)x− c

≤ 0 =⇒ limx→c+

f(x)− f(c)x− c

≤ 0


f(x)− f(c)x− c

≥ 0 =⇒ limx→c−

f(x)− f(c)x− c

≥ 0


f(x)− f(c)x− c




f(x)− f(c)x− c

≤ 0 =⇒ limx→c+

f(x)− f(c)x− c

≤ 0


f(x)− f(c)x− c

≥ 0

=⇒ limx→c−

f(x)− f(c)x− c

≥ 0


f(x)− f(c)x− c




f(x)− f(c)x− c

≤ 0 =⇒ limx→c+

f(x)− f(c)x− c

≤ 0


f(x)− f(c)x− c

≥ 0 =⇒ limx→c−

f(x)− f(c)x− c

≥ 0


f(x)− f(c)x− c




f(x)− f(c)x− c

≤ 0 =⇒ limx→c+

f(x)− f(c)x− c

≤ 0


f(x)− f(c)x− c

≥ 0 =⇒ limx→c−

f(x)− f(c)x− c

≥ 0


f(x)− f(c)x− c


Meet the Mathematician: Pierre de Fermat

I 1601–1665I Lawyer and numbertheorist

I Proved many theorems,didn’t quite prove his lastone

Tangent: Fermat’s Last TheoremI Plenty of solu ons to

x2 + y2 = z2 among posi vewhole numbers (e.g., x = 3,y = 4, z = 5)

I No solu ons to x3 + y3 = z3among posi ve whole numbers

I Fermat claimed no solu ons toxn + yn = zn but didn’t writedown his proof

I Not solved un l 1998!(Taylor–Wiles)
















OutlineIntroduc on




Examples

Flowchart for placing extremaThanks to FermatSuppose f is acon nuousfunc on onthe closed,boundedinterval[a, b], and c isa globalmaximumpoint.

..start.

Is c anendpoint?

.

c = a orc = b

. c is alocal max

.

Is f diff’bleat c?

.

f is notdiff at c

.

f′(c) = 0

.

no

.

yes

.

no

.

yes


This means to find the maximum value of f on [a, b], we need to:I Evaluate f at the endpoints a and bI Evaluate f at the cri cal points or cri cal numbers x whereeither f′(x) = 0 or f is not differen able at x.

I The points with the largest func on value are the globalmaximum points

I The points with the smallest or most nega ve func on valueare the global minimum points.

OutlineIntroduc on




Examples

Extreme values of a linear functionExample

Find the extreme values of f(x) = 2x− 5 on [−1, 2].

Solu on

Since f′(x) = 2, which is neverzero, we have no cri cal pointsand we need only inves gatethe endpoints:

I f(−1) = 2(−1)− 5 = −7I f(2) = 2(2)− 5 = −1

SoI The absolute minimum

(point) is at−1; theminimum value is−7.

I The absolute maximum(point) is at 2; themaximum value is−1.



Solu on


I f(−1) = 2(−1)− 5 = −7I f(2) = 2(2)− 5 = −1






Solu on


I f(−1) = 2(−1)− 5 = −7I f(2) = 2(2)− 5 = −1




Extreme values of a quadraticfunction

Example

Find the extreme values of f(x) = x2 − 1 on [−1, 2].

Solu onWe have f′(x) = 2x, which is zero when x = 0.

So our points tocheck are:

I f(−1) =I f(0) =I f(2) =


Example


Solu onWe have f′(x) = 2x, which is zero when x = 0.

So our points tocheck are:

I f(−1) =I f(0) =I f(2) =


Example


Solu onWe have f′(x) = 2x, which is zero when x = 0. So our points tocheck are:

I f(−1) =I f(0) =I f(2) =


Example



I f(−1) = 0I f(0) =I f(2) =


Example



I f(−1) = 0I f(0) = − 1I f(2) =


Example



I f(−1) = 0I f(0) = − 1I f(2) = 3


Example



I f(−1) = 0I f(0) = − 1 (absolute min)I f(2) = 3


Example



I f(−1) = 0I f(0) = − 1 (absolute min)I f(2) = 3 (absolute max)

Extreme values of a cubic functionExample

Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

Solu onSince f′(x) = 6x2 − 6x = 6x(x− 1), we have cri cal points at x = 0and x = 1.

The values to check are

I f(−1) =

− 4 (global min)

I f(0) =

1 (local max)

I f(1) =

0 (local min)

I f(2) =

5 (global max)



Solu onSince f′(x) = 6x2 − 6x = 6x(x− 1), we have cri cal points at x = 0and x = 1.

The values to check are

I f(−1) =

− 4 (global min)

I f(0) =

1 (local max)

I f(1) =

0 (local min)

I f(2) =

5 (global max)



Solu onSince f′(x) = 6x2 − 6x = 6x(x− 1), we have cri cal points at x = 0and x = 1. The values to check are

I f(−1) =

− 4 (global min)

I f(0) =

1 (local max)

I f(1) =

0 (local min)

I f(2) =

5 (global max)




I f(−1) = − 4

(global min)I f(0) =

1 (local max)

I f(1) =

0 (local min)

I f(2) =

5 (global max)




I f(−1) = − 4

(global min)

I f(0) = 1

(local max)I f(1) =

0 (local min)

I f(2) =

5 (global max)




I f(−1) = − 4

(global min)

I f(0) = 1

(local max)

I f(1) = 0

(local min)I f(2) =

5 (global max)




I f(−1) = − 4

(global min)

I f(0) = 1

(local max)

I f(1) = 0

(local min)

I f(2) = 5

(global max)




I f(−1) = − 4 (global min)I f(0) = 1

(local max)

I f(1) = 0

(local min)

I f(2) = 5

(global max)




I f(−1) = − 4 (global min)I f(0) = 1

(local max)

I f(1) = 0

(local min)

I f(2) = 5 (global max)




I f(−1) = − 4 (global min)I f(0) = 1 (local max)I f(1) = 0

(local min)

I f(2) = 5 (global max)




I f(−1) = − 4 (global min)I f(0) = 1 (local max)I f(1) = 0 (local min)I f(2) = 5 (global max)

Extreme values of an algebraic functionExample

Find the extreme values of f(x) = x2/3(x+ 2) on [−1, 2].

Solu onWrite f(x) = x5/3 + 2x2/3.






Solu onWrite f(x) = x5/3 + 2x2/3. Then

f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x+ 4)



Solu onWrite f(x) = x5/3 + 2x2/3. Then

f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x+ 4)

Thus f′(−4/5) = 0 and f is not differen able at 0. Thus there are twocri cal points.




I f(−1) =I f(−4/5) =

I f(0) =I f(2) =




I f(−1) = 1I f(−4/5) =

I f(0) =I f(2) =




I f(−1) = 1I f(−4/5) = 1.0341I f(0) =I f(2) =




I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0I f(2) =




I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0I f(2) = 6.3496




I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0 (absolute min)I f(2) = 6.3496




I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0 (absolute min)I f(2) = 6.3496 (absolute max)




I f(−1) = 1I f(−4/5) = 1.0341 (rela ve max)I f(0) = 0 (absolute min)I 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].

Solu onWe have f′(x) = − x√

4− x2, which is zero when x = 0. (f is not

differen able at±2 as well.)

So our points to check are:

I f(−2) =I f(0) =I f(1) =


Example


4− x2 on [−2, 1].



differen able at±2 as well.)

So our points to check are:

I f(−2) =I f(0) =I f(1) =


Example


4− x2 on [−2, 1].



differen able at±2 as well.) So our points to check are:I f(−2) =

I f(0) =I f(1) =


Example


4− x2 on [−2, 1].



differen able at±2 as well.) So our points to check are:I f(−2) = 0I f(0) =

I f(1) =


Example


4− x2 on [−2, 1].



differen able at±2 as well.) So our points to check are:I f(−2) = 0I f(0) = 2I f(1) =


Example


4− x2 on [−2, 1].



differen able at±2 as well.) So our points to check are:I f(−2) = 0I f(0) = 2I f(1) =

√3


Example


4− x2 on [−2, 1].



differen able at±2 as well.) So our points to check are:I f(−2) = 0 (absolute min)I f(0) = 2I f(1) =

√3


Example


4− x2 on [−2, 1].



differen able at±2 as well.) So our points to check are:I f(−2) = 0 (absolute min)I f(0) = 2 (absolute max)I f(1) =

√3

Summary

I The Extreme Value Theorem: a con nuous func on on a closedinterval must achieve its max and min

I Fermat’s Theorem: local extrema are cri cal pointsI The Closed Interval Method: an algorithm for finding globalextrema

I Show your work unless you want to end up like Fermat!

Lesson 18: Maximum and Minimum Values (slides)

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