. . SecƟon 4.1 Maximum and Minimum Values V63.0121.001: Calculus I Professor MaƩhew Leingang New York University April 4, 2011 . Announcements I Quiz 4 on SecƟons 3.3, 3.4, 3.5, and 3.7 next week (April 14/15) I Quiz 5 on SecƟons 4.1–4.4 April 28/29 I Final Exam Monday May 12, 2:00–3:50pm . Objectives I Understand and be able to explain the statement of the Extreme Value Theorem. I Understand and be able to explain the statement of Fermat’s Theorem. I Use the Closed Interval Method to find the extreme values of a funcƟon defined on a closed interval. . Notes . Notes . Notes . 1 . . SecƟon 4.1: Max/Min Values . V63.0121.001: Calculus I . April 4, 2011
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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Sec on 4.1Maximum and Minimum Values
V63.0121.001: Calculus IProfessor Ma hew Leingang
New York University
April 4, 2011
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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
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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.
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Notes
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Notes
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Notes
. 1.
. Sec on 4.1: Max/Min Values. V63.0121.001: Calculus I . April 4, 2011
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OutlineIntroduc on
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
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..
Optimize
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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)
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Notes
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Notes
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Notes
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. Sec on 4.1: Max/Min Values. V63.0121.001: Calculus I . April 4, 2011
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Design
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Optics
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OutlineIntroduc on
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
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Notes
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Notes
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Notes
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. Sec on 4.1: Max/Min Values. V63.0121.001: Calculus I . April 4, 2011
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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
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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...
cmaximum
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maximumvaluef(c)
..d
minimum
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minimumvaluef(d)
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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.
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Notes
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Notes
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Notes
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. Sec on 4.1: Max/Min Values. V63.0121.001: Calculus I . April 4, 2011
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.
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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
..
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Final Bad ExampleExample
The func on f(x) =1xis con nuous on the closed interval [1,∞).
...1
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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.
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Notes
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Notes
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Notes
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. Sec on 4.1: Max/Min Values. V63.0121.001: Calculus I . April 4, 2011
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OutlineIntroduc on
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
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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
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Local extremaI So a local extremum must be
inside the domain of f (not onthe end).
I A global extremum that is insidethe domain is a local extremum.
..|.a. |.
b....
localmaximum
..globalmax
.local andglobalmin
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Notes
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Notes
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Notes
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. Sec on 4.1: Max/Min Values. V63.0121.001: Calculus I . April 4, 2011
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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
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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.
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Meet the Mathematician: Pierre de Fermat
I 1601–1665I Lawyer and numbertheorist
I Proved many theorems,didn’t quite prove his lastone
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Notes
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Notes
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Notes
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. Sec on 4.1: Max/Min Values. V63.0121.001: Calculus I . April 4, 2011
.
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OutlineIntroduc on
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
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Flowchart for placing extremaThanks to FermatSuppose f is acon nuousfunc on onthe closed,boundedinterval[a, b], and c isa globalmaximumpoint.
..start.
Is c anendpoint?
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c = a orc = b
. c is alocal max
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Is f diff’bleat c?
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f is notdiff at c
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f′(c) = 0
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no
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yes
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no
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yes
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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.
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Notes
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Notes
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Notes
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. Sec on 4.1: Max/Min Values. V63.0121.001: Calculus I . April 4, 2011
.
.
OutlineIntroduc on
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
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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.
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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: