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Section 4.1Maximum and Minimum Values
V63.0121.006/016, Calculus I
March 23, 2010
Announcements
Welcome back from Spring Break!
Quiz 3: April 2, Sections 2.63.5
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Announcements
Welcome back from Spring Break!
Quiz 3: April 2, Sections 2.63.5
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Computation of Midterm Letter Grades
HW
10%
WebAssign10%
Quizzes
15%
Midterm 25%
Final
40%
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Computation of Midterm Letter Grades
HW
10%
WebAssign10%
Quizzes
15%
Midterm 25%
Final (to be determined)
40%
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Computation of Midterm Letter Grades
HW
17%
WebAssign
17%Quizzes
25%
Midterm
41%
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Distribution of Midterm Averages and Letter Grades
Median = 81.35% (curved to B)
Average = 74.39% (curved to B-) Standard Deviation = 21%
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What can I do to improve my grade?
HW
10%
WebAssign
10%
Quizzes
15%
Midterm25%
Final
40%
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What can I do to improve my grade?
past HW
5%
future HW5% past WA
5% future WA
5%past Quizzes
6%future Quizzes9%
Midterm25%
Final
40%
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What can I do to improve my grade?
past HW
5%
past WA5% past Quizzes6%
Midterm
25%future HW 5%future WA 5%
future Quizzes
9%
Final
40%
59% of your grade is still in play!
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Help!
Free resources:
recitation
TAs office hours
my office hours
Math Tutoring Center(CIWW 524)
College Learning Center
http://learning.cas.nyu.edu/docs/CP/1928/FinalCLC_Schedule_jan25.pdfhttp://www.math.nyu.edu/degree/undergrad/tutor_schedule.html8/21/2019 14-max min
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Outline
Introduction
The Extreme Value Theorem
Fermats Theorem (not the last one)
Tangent: Fermats Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
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Optimize
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Why go to the extremes?
Rationally speaking, it isadvantageous to find theextreme values of afunction (maximizeprofit, minimize costs,
etc.)
Pierre-Louis Maupertuis(16981759)
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Design
Image credit: Jason Tromm
http://www.flickr.com/photos/trommetter/1338616263/8/21/2019 14-max min
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Why go to the extremes?
Rationally speaking, it isadvantageous to find theextreme values of afunction (maximizeprofit, minimize costs,
etc.) Many laws of science
are derived fromminimizing principles.
Pierre-Louis Maupertuis(16981759)
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Optics
Image credit: jacreative
h h
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Why go to the extremes?
Rationally speaking, it isadvantageous to find theextreme values of afunction (maximizeprofit, minimize costs,
etc.) Many laws of science
are derived fromminimizing principles.
Maupertuis principle:Action is minimizedthrough the wisdom ofGod. Pierre-Louis Maupertuis
(16981759)
O li
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Outline
Introduction
The Extreme Value Theorem
Fermats Theorem (not the last one)
Tangent: Fermats Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
E i d l
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Extreme points and values
Definition
Letfhave domainD.
Image credit: Patrick Q
E t i t d l
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Extreme points and values
Definition
Letfhave domainD. The functionfhas anabsolute
maximum(orglobal maximum)(respectively,absolute minimum) atc
iff(c) f(x)(respectively,f(c) f(x))for allxinD
Image credit: Patrick Q
E t i t d l
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Extreme points and values
Definition
Letfhave domainD. The functionfhas anabsolute
maximum(orglobal maximum)(respectively,absolute minimum) atc
iff(c) f(x)(respectively,f(c) f(x))for allxinD The numberf(c)is called the
maximum value(respectively,minimum value) offonD.
Image credit: Patrick Q
E treme points and al es
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Extreme points and values
Definition
Letfhave domainD. The functionfhas anabsolute
maximum(orglobal maximum)(respectively,absolute minimum) atc
iff(c) f(x)(respectively,f(c) f(x))for allxinD The numberf(c)is called the
maximum value(respectively,minimum value) offonD.
Anextremumis either a maximum ora minimum. Anextreme valueiseither a maximum value or minimumvalue.
Image credit: Patrick Q
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Theorem (The Extreme Value Theorem)Let f be a function which is continuous on the closed interval[a,b]. Then f attains an absolute maximum value f(c)and an
absolute minimum value f(d)at numbers c and d in[a,b].
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Theorem (The Extreme Value Theorem)Let f be a function which is continuous on the closed interval[a,b]. Then f attains an absolute maximum value f(c)and an
absolute minimum value f(d)at numbers c and d in[a,b].
a
b
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Theorem (The Extreme Value Theorem)Let f be a function which is continuous on the closed interval[a,b]. Then f attains an absolute maximum value f(c)and an
absolute minimum value f(d)at numbers c and d in[a,b].
a
b
cmaximum
maximum
value
f(c)
d
minimum
minimum
value
f(d)
No proof of EVT forthcoming
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No proof of EVT forthcoming
This theorem is very hard to prove without using technical
facts about continuous functions and closed intervals. But we can show the importance of each of the hypotheses.
Bad Example #1
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Bad Example #1
Example
Consider the function
f(x) =x 0 x
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Bad Example #1
Example
Consider the function
f(x) =x 0 x
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Bad Example #1
Example
Consider the function
f(x) =x 0 x
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Bad Example #1
Example
Consider the function
f(x) =x 0 x
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Bad Example #2
Example
Consider the functionf(x) =xrestricted to the interval[0, 1).
Bad Example #2
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ad a p e
Example
Consider the functionf(x) =xrestricted to the interval[0, 1).
|1
Bad Example #2
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p
Example
Consider the functionf(x) =xrestricted to the interval[0, 1).
|1
There is still no maximum value (values get arbitrarily close to 1but do not achieve it).
Bad Example #2
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p
Example
Consider the functionf(x) =xrestricted to the interval[0, 1).
|1
There is still no maximum value (values get arbitrarily close to 1but do not achieve it). This does not violate EVT because thedomain is not closed.
Final Bad Example
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p
ExampleConsider the functionf(x) =
1
xis continuous on the closed
interval [1,).
Final Bad Example
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p
ExampleConsider the functionf(x) =
1
xis continuous on the closed
interval [1,).
1
Final Bad Example
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p
ExampleConsider the functionf(x) =
1
xis continuous on the closed
interval [1,).
1
There is no minimum value (values get arbitrarily close to 0 butdo not achieve it).
Final Bad Example
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ExampleConsider the functionf(x) =
1
xis continuous on the closed
interval [1,).
1
There is no minimum value (values get arbitrarily close to 0 butdo not achieve it). This does not violate EVT because the domainis not bounded.
Outline
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Introduction
The Extreme Value Theorem
Fermats Theorem (not the last one)
Tangent: Fermats Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
Local extrema
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Definition
A functionfhas alocal maximumorrelative maximumatciff(c) f(x)whenxis nearc. This means thatf(c) f(x)forallxin some open interval containingc.
Similarly,fhas alocal minimumatciff(c) f(x)whenxisnearc.
Local extrema
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Definition
A functionfhas alocal maximumorrelative maximumatciff(c) f(x)whenxis nearc. This means thatf(c) f(x)forallxin some open interval containingc.
Similarly,fhas alocal minimumatciff(c) f(x)whenxisnearc.
|a |b
localmaximum
local
minimum
Local extrema
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So a local extremum must beinsidethe domain off(not onthe end).
A global extremum that is inside the domain is a localextremum.
|a |b
localmaximum
globalmax
local and global
min
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Theorem (Fermats Theorem)Suppose f has a local extremum at c and f is differentiable at c.
Then f(c) =0.
|a
|b
localmaximum
local
minimum
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Theorem (Fermats Theorem)Suppose f has a local extremum at c and f is differentiable at c.
Then f(c) =0.
|a
|b
localmaximum
local
minimum
Sketch of proof of Fermats Theorem
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Suppose thatfhas a local maximum atc.
Sketch of proof of Fermats Theorem
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Suppose thatfhas a local maximum atc.
Ifxis slightly greater thanc,f(x) f(c). This meansf(x) f(c)
x c 0
Sketch of proof of Fermats Theorem
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Suppose thatfhas a local maximum atc.
Ifxis slightly greater thanc,f(x) f(c). This meansf(x) f(c)
x c 0 = limxc+f(x) f(c)
x c 0
Sketch of proof of Fermats Theorem
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Suppose thatfhas a local maximum atc.
Ifxis slightly greater thanc,f(x) f(c). This meansf(x) f(c)
x c 0 = limxc+f(x) f(c)
x c 0
The same will be true on the other end: ifxis slightly lessthanc,f(x) f(c). This means
f(x) f(c)x
c 0
Sketch of proof of Fermats Theorem
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Suppose thatfhas a local maximum atc.
Ifxis slightly greater thanc,f(x) f(c). This meansf(x) f(c)
x c 0 = limxc+f(x) f(c)
x c 0
The same will be true on the other end: ifxis slightly lessthanc,f(x) f(c). This means
f(x) f(c)x
c 0 = lim
xc
f(x) f(c)x
c 0
Sketch of proof of Fermats Theorem
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Suppose thatfhas a local maximum atc.
Ifxis slightly greater thanc,f(x) f(c). This meansf(x) f(c)
x c 0 = limxc+f(x) f(c)
x c 0
The same will be true on the other end: ifxis slightly lessthanc,f(x) f(c). This means
f(x) f(c)x
c 0 = lim
xc
f(x) f(c)x
c 0
Since the limitf(c) =limxc
f(x) f(c)x c exists, it must be 0.
Meet the Mathematician: Pierre de Fermat
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16011665
Lawyer and number
theorist Proved many theorems,
didnt quite prove hislast one
Tangent: Fermats Last Theorem
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Plenty of solutions tox2 +y2 =z2 amongpositive whole numbers(e.g.,x=3,y=4,z=5)
Tangent: Fermats Last Theorem
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Plenty of solutions tox2 +y2 =z2 amongpositive whole numbers(e.g.,x=3,y=4,z=5)
No solutions to
x3 +y3 =z3 amongpositive whole numbers
Tangent: Fermats Last Theorem
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Plenty of solutions tox2 +y2 =z2 amongpositive whole numbers(e.g.,x=3,y=4,z=5)
No solutions to
x3 +y3 =z3 amongpositive whole numbers
Fermat claimed nosolutions toxn +yn =zn
but didnt write downhis proof
Tangent: Fermats Last Theorem
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Plenty of solutions tox2 +y2 =z2 amongpositive whole numbers(e.g.,x=3,y=4,z=5)
No solutions to
x3 +y3 =z3 amongpositive whole numbers
Fermat claimed nosolutions toxn +yn =zn
but didnt write downhis proof
Not solved until 1998!(TaylorWiles)
Outline
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Introduction
The Extreme Value Theorem
Fermats Theorem (not the last one)
Tangent: Fermats Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
Flowchart for placing extremaThanks to Fermat
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Supposefis a continuous function on the closed, boundedinterval [a, b], andcis a global maximum point.
start
Iscan
endpoint?
c =aorc =b
cis a
local max
Isfdiffble
atc?
fis notdiff atc
f(c) =0
no
yes
no
yes
The Closed Interval Method
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This means to find the maximum value offon[a,b], we need to:
Evaluatefat theendpointsaandb
Evaluatefat thecritical pointsorcritical numbersxwhereeitherf(x) =0 orfis not differentiable atx.
The points with the largest function value are the globalmaximum points
The points with the smallest or most negative function valueare the global minimum points.
Outline
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Introduction
The Extreme Value Theorem
Fermats Theorem (not the last one)
Tangent: Fermats Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
E l
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Example
Find the extreme values off(x) =2x 5 on[1, 2].
Example
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Example
Find the extreme values off(x) =2x 5 on[1, 2].
SolutionSince f(x) =2, which is never zero, we have no critical pointsand we need only investigate the endpoints:
f(
1) =2(
1)
5=
7
f(2) =2(2) 5= 1
Example
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Example
Find the extreme values off(x) =2x 5 on[1, 2].
SolutionSince f(x) =2, which is never zero, we have no critical pointsand we need only investigate the endpoints:
f(
1) =2(
1)
5=
7
f(2) =2(2) 5= 1So
The absolute minimum (point) is at1; the minimum valueis
7.
The absolute maximum (point) is at2; the maximum value is1.
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Example
Find the extreme values off(x) =x2 1 on[1,2].
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Example
Find the extreme values off(x) =x2 1 on[1,2].
SolutionWe have f(x) =2x, which is zero when x =0.
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Example
Find the extreme values off(x) =x2 1 on[1,2].
SolutionWe have f(x) =2x, which is zero when x =0. So our points to
check are: f(1) = f(0) =
f(2) =
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Example
Find the extreme values off(x) =x2 1 on[1,2].
SolutionWe have f(x) =2x, which is zero when x =0. So our points to
check are: f(1) =0 f(0) =
f(2) =
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Example
Find the extreme values off(x) =x2 1 on[1,2].
SolutionWe have f(x) =2x, which is zero when x =0. So our points to
check are: f(1) =0 f(0) = 1 f(2) =
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Example
Find the extreme values off(x) =x2 1 on[1,2].
SolutionWe have f(x) =2x, which is zero when x =0. So our points tocheck are:
f(1) =0 f(0) = 1 f(2) =3
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Example
Find the extreme values off(x) =x2 1 on[1,2].
SolutionWe have f(x) =2x, which is zero when x =0. So our points tocheck are:
f(1) =0 f(0) = 1(absolute min) f(2) =3
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Example
Find the extreme values off(x) =x2 1 on[1,2].
SolutionWe have f(x) =2x, which is zero when x =0. So our points tocheck are:
f(1) =0 f(0) = 1(absolute min) f(2) =3(absolute max)
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Example
Find the extreme values off(x) =2x3 3x2 +1 on[1,2].
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Example
Find the extreme values off(x) =2x3 3x2 +1 on[1,2].SolutionSince f(x) =6x2 6x=6x(x 1), we have critical points atx=0and x=1.
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Example
Find the extreme values off(x) =2x3 3x2 +1 on[1,2].SolutionSince f(x) =6x2 6x=6x(x 1), we have critical points atx=0and x=1. The values to check are
f(1) = f(0) =
f(1) =
f(2) =
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Example
Find the extreme values off(x) =2x3 3x2 +1 on[1,2].SolutionSince f(x) =6x2 6x=6x(x 1), we have critical points atx=0and x=1. The values to check are
f(1) = 4 f(0) =
f(1) =
f(2) =
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Example
Find the extreme values off(x) =2x3 3x2 +1 on[1,2].SolutionSince f(x) =6x2 6x=6x(x 1), we have critical points atx=0and x=1. The values to check are
f(1) = 4 f(0) =1
f(1) =
f(2) =
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Example
Find the extreme values off(x) =2x3 3x2 +1 on[1,2].SolutionSince f(x) =6x2 6x=6x(x 1), we have critical points atx=0and x=1. The values to check are
f(1) = 4 f(0) =1
f(1) =0
f(2) =
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Example
Find the extreme values off(x) =2x3 3x2 +1 on[1,2].SolutionSince f(x) =6x2 6x=6x(x 1), we have critical points atx=0and x=1. The values to check are
f(1) = 4 f(0) =1
f(1) =0
f(2) =5
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Example
Find the extreme values off(x) =2x3 3x2 +1 on[1,2].SolutionSince f(x) =6x2 6x=6x(x 1), we have critical points atx=0and x=1. The values to check are
f(1) = 4(global min) f(0) =1
f(1) =0
f(2) =5
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Example
Find the extreme values off(x) =2x3 3x2 +1 on[1,2].SolutionSince f(x) =6x2 6x=6x(x 1), we have critical points atx=0and x=1. The values to check are
f(1) = 4(global min) f(0) =1
f(1) =0
f(2) =5(global max)
l
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Example
Find the extreme values off(x) =2x3 3x2 +1 on[1,2].SolutionSince f(x) =6x2 6x=6x(x 1), we have critical points atx=0and 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)
l
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Example
Find the extreme values off(x) =2x3 3x2 +1 on[1,2].SolutionSince f(x) =6x2 6x=6x(x 1), we have critical points atx=0and x=1. The values to check are
f(1) = 4(global min) f(0) =1(local max)
f(1) =0(local min)
f(2) =5(global max)
Example
Find the extreme values of f(x) = x2/3(x + 2) on [1,2].
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Find the extreme values off(x) x (x+2)on[ 1,2].
Example
Find the extreme values off(x) =x2/3(x+2)on[1,2].
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d t e e t e e a ues o ( ) ( + ) o [ , ].
SolutionWrite f(x) =x5/3 +2x2/3, then
f(x) =5
3x2/3 +
4
3x1/3 =
1
3x1/3(5x+4)
Thus f(4/5) =0and f is not differentiable at0.
Example
Find the extreme values off(x) =x2/3(x+2)on[1,2].
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( ) ( + ) [ , ]
SolutionWrite f(x) =x5/3 +2x2/3, then
f(x) =5
3x2/3 +
4
3x1/3 =
1
3x1/3(5x+4)
Thus f(4/5) =0and f is not differentiable at0. So our points tocheck are:
f(1) = f(
4/5) =
f(0) =
f(2) =
Example
Find the extreme values off(x) =x2/3(x+2)on[1,2].
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( ) ( ) [ , ]
SolutionWrite f(x) =x5/3 +2x2/3, then
f(x) =5
3x2/3 +
4
3x1/3 =
1
3x1/3(5x+4)
Thus f(4/5) =0and f is not differentiable at0. So our points tocheck are:
f(1) =1 f(
4/5) =
f(0) =
f(2) =
Example
Find the extreme values off(x) =x2/3(x+2)on[1,2].
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( ) ( ) [ ]
SolutionWrite f(x) =x5/3 +2x2/3, then
f(x) =5
3x2/3 +
4
3x1/3 =
1
3x1/3(5x+4)
Thus f(4/5) =0and f is not differentiable at0. So our points tocheck are:
f(1) =1 f(
4/5) =1.0341
f(0) =
f(2) =
Example
Find the extreme values off(x) =x2/3(x+2)on[1,2].
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SolutionWrite f(x) =x5/3 +2x2/3, then
f(x) =5
3x2/3 +
4
3x1/3 =
1
3x1/3(5x+4)
Thus f(4/5) =0and f is not differentiable at0. So our points tocheck are:
f(1) =1 f(
4/5) =1.0341
f(0) =0
f(2) =
Example
Find the extreme values off(x) =x2/3(x+2)on[1,2].
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SolutionWrite f(x) =x5/3 +2x2/3, then
f(x) =5
3x2/3 +
4
3x1/3 =
1
3x1/3(5x+4)
Thus f(4/5) =0and f is not differentiable at0. So our points tocheck are:
f(1) =1 f(
4/5) =1.0341
f(0) =0
f(2) =6.3496
Example
Find the extreme values off(x) =x2/3(x+2)on[1,2].
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SolutionWrite f(x) =x5/3 +2x2/3, then
f(x) =5
3x2/3 +
4
3x1/3 =
1
3x1/3(5x+4)
Thus f(4/5) =0and f is not differentiable at0. So our points tocheck are:
f(1) =1 f(
4/5) =1.0341
f(0) =0(absolute min)
f(2) =6.3496
Example
Find the extreme values off(x) =x2/3(x+2)on[1,2].
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SolutionWrite f(x) =x5/3 +2x2/3, then
f(x) =5
3x2/3 +
4
3x1/3 =
1
3x1/3(5x+4)
Thus f(4/5) =0and f is not differentiable at0. So our points tocheck are:
f(1) =1 f(
4/5) =1.0341
f(0) =0(absolute min)
f(2) =6.3496(absolute max)
Example
Find the extreme values off(x) =x2/3(x+2)on[1,2].
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SolutionWrite f(x) =x5/3 +2x2/3, then
f(x) =5
3x2/3 +
4
3x1/3 =
1
3x1/3(5x+4)
Thus f(4/5) =0and f is not differentiable at0. So our points tocheck are:
f(1) =1 f(
4/5) =1.0341(relative max)
f(0) =0(absolute min)
f(2) =6.3496(absolute max)
Example
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Example
Find the extreme values off(x) =
4x2 on[2,1].
Example
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Example
Find the extreme values off(x) =
4x2 on[2,1].SolutionWe have f(x) = x
4x2, which is zero when x=0. (f is not
differentiable at
2as well.)
Example
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Example
Find the extreme values off(x) =
4x2 on[2,1].SolutionWe have f(x) = x
4x2, which is zero when x=0. (f is not
differentiable at
2as well.) So our points to check are:
f(2) = f(0) =
f(1) =
Example
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Example
Find the extreme values off(x) =
4x2 on[2,1].SolutionWe have f(x) = x
4x2, which is zero when x=0. (f is not
differentiable at
2as well.) So our points to check are:
f(2) =0 f(0) =
f(1) =
Example
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Example
Find the extreme values off(x) =
4x2 on[2,1].SolutionWe have f(x) = x
4x2, which is zero when x=0. (f is not
differentiable at
2as well.) So our points to check are:
f(2) =0 f(0) =2
f(1) =
Example
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Example
Find the extreme values off(x) =
4x2 on[2,1].SolutionWe have f(x) = x
4x2, which is zero when x=0. (f is not
differentiable at
2as well.) So our points to check are:
f(2) =0 f(0) =2
f(1) =
3
Example
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Example
Find the extreme values off(x) =
4x2 on[2,1].SolutionWe have f(x) = x
4x2, which is zero when x=0. (f is not
differentiable at
2as well.) So our points to check are:
f(2) =0(absolute min) f(0) =2
f(1) =
3
Example
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Example
Find the extreme values off(x) =
4x2 on[2,1].SolutionWe have f(x) = x
4x2, which is zero when x=0. (f is not
differentiable at
2as well.) So our points to check are:
f(2) =0(absolute min) f(0) =2(absolute max)
f(1) =
3
Outline
Introduction
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The Extreme Value Theorem
Fermats Theorem (not the last one)Tangent: Fermats Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
Challenge: Cubic functions
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Example
How many critical points can a cubic function
f(x) =ax3
+bx2
+cx+d
have?
SolutionIf f(x) =0, we have
3ax2 +2bx+c=0,
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and so
x=2b
4b2 12ac6a
=b
b2 3ac
3a ,
and so we have three possibilities:
SolutionIf f(x) =0, we have
3ax2 +2bx+c=0,
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and so
x=2b
4b2 12ac6a
=b
b2 3ac
3a ,
and so we have three possibilities: b2 3ac>0, in which case there are two distinct critical
points. An example would be f(x) =x3 +x2, where a=1,b=1, and c =0.
SolutionIf f(x) =0, we have
3ax2 +2bx+c=0,
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and so
x=2b
4b2 12ac6a
=b
b2 3ac
3a ,
and so we have three possibilities: b2 3ac>0, in which case there are two distinct critical
points. An example would be f(x) =x3 +x2, where a=1,b=1, and c =0.
b2
3ac
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and so
x=2b
4b2 12ac6a
=b
b2 3ac
3a ,
and so we have three possibilities: b2 3ac>0, in which case there are two distinct critical
points. An example would be f(x) =x3 +x2, where a=1,b=1, and c =0.
b2
3ac
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The Extreme Value Theorem: a continuous function on aclosed interval must achieve its max and min
Fermats Theorem: local extrema are critical points
The Closed Interval Method: an algorithm for finding globalextrema
Show your work unless you want to end up like Fermat!