..
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
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)
..
Optics
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)
OutlineIntroduc 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 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.
..
Image credit: Patrick Q
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.
..
Image credit: Patrick Q
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.
..
Image credit: Patrick Q
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.
..
Image credit: Patrick Q
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].
.
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..
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...
cmaximum
.
maximumvaluef(c)
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...
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.
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.
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.
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.
Bad Example #2Example
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
..
Bad Example #2Example
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
..
Bad Example #2Example
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
..
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.
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.
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.
OutlineIntroduc 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
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..
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....
localmaximum
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....
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.
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.
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.
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.
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.
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.
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)
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)
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)
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
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 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
The Closed Interval Method
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
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 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.
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.
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.
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) =
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) =
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) =
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) = 0I f(0) =I f(2) =
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) = 0I f(0) = − 1I f(2) =
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) = 0I f(0) = − 1I f(2) = 3
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) = 0I f(0) = − 1 (absolute min)I f(2) = 3
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) = 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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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.
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.
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. Then
f′(x) =53x2/3 +
43x−1/3 =
13x−1/3(5x+ 4)
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. 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.
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.
I f(−1) =I f(−4/5) =
I f(0) =I f(2) =
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.
I f(−1) = 1I f(−4/5) =
I f(0) =I f(2) =
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.
I f(−1) = 1I f(−4/5) = 1.0341I f(0) =I f(2) =
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.
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0I f(2) =
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.
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0I f(2) = 6.3496
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.
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0 (absolute min)I f(2) = 6.3496
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.
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0 (absolute min)I f(2) = 6.3496 (absolute 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.
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) =
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) =
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) =
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) = 0I f(0) =
I f(1) =
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) = 0I f(0) = 2I f(1) =
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) = 0I f(0) = 2I f(1) =
√3
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) = 0 (absolute min)I f(0) = 2I f(1) =
√3
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) = 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!