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Section 4.1Maximum and Minimum Values
V63.0121.041, Calculus I
New York University
November 8, 2010
Announcements
I Quiz 4 on Sections 3.3, 3.4, 3.5, and 3.7 next week (November16, 18, or 19)
. . . . . .
. . . . . .
Announcements
I Quiz 4 on Sections 3.3,3.4, 3.5, and 3.7 next week(November 16, 18, or 19)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 2 / 34
. . . . . .
Objectives
I Understand and be able toexplain the statement ofthe Extreme ValueTheorem.
I Understand and be able toexplain the statement ofFermat’s Theorem.
I Use the Closed IntervalMethod to find the extremevalues of a function definedon a closed interval.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 3 / 34
. . . . . .
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 4 / 34
.
.
Optimize
Why go to the extremes?
I Rationally speaking, it isadvantageous to find theextreme values of afunction (maximize profit,minimize costs, etc.)
I Many laws of science arederived from minimizingprinciples.
I Maupertuis’ principle:“Action is minimizedthrough the wisdom ofGod.”
Pierre-Louis Maupertuis(1698–1759)V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 6 / 34
Design
..Image credit: Jason TrommV63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 7 / 34
Why go to the extremes?
I Rationally speaking, it isadvantageous to find theextreme values of afunction (maximize profit,minimize costs, etc.)
I Many laws of science arederived from minimizingprinciples.
I Maupertuis’ principle:“Action is minimizedthrough the wisdom ofGod.”
Pierre-Louis Maupertuis(1698–1759)V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 8 / 34
Optics
..Image credit: jacreativeV63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 9 / 34
Why go to the extremes?
I Rationally speaking, it isadvantageous to find theextreme values of afunction (maximize profit,minimize costs, etc.)
I Many laws of science arederived from minimizingprinciples.
I Maupertuis’ principle:“Action is minimizedthrough the wisdom ofGod.”
Pierre-Louis Maupertuis(1698–1759)V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 10 / 34
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 11 / 34
Extreme points and values
DefinitionLet f have domain D.
I The function f has an absolutemaximum (or global maximum)(respectively, absolute minimum) at c iff(c) ≥ f(x) (respectively, f(c) ≤ f(x)) for allx in D
I The number f(c) is called the maximumvalue (respectively, 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
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 12 / 34
Extreme points and values
DefinitionLet f have domain D.
I The function f has an absolutemaximum (or global maximum)(respectively, absolute minimum) at c iff(c) ≥ f(x) (respectively, f(c) ≤ f(x)) for allx in D
I The number f(c) is called the maximumvalue (respectively, 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
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 12 / 34
Extreme points and values
DefinitionLet f have domain D.
I The function f has an absolutemaximum (or global maximum)(respectively, absolute minimum) at c iff(c) ≥ f(x) (respectively, f(c) ≤ f(x)) for allx in D
I The number f(c) is called the maximumvalue (respectively, 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
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 12 / 34
Extreme points and values
DefinitionLet f have domain D.
I The function f has an absolutemaximum (or global maximum)(respectively, absolute minimum) at c iff(c) ≥ f(x) (respectively, f(c) ≤ f(x)) for allx in D
I The number f(c) is called the maximumvalue (respectively, 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
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 12 / 34
The Extreme Value Theorem
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 absoluteminimum value f(d) at numbers c and d in [a,b].
.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 13 / 34
The Extreme Value Theorem
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 absoluteminimum value f(d) at numbers c and d in [a,b].
...a
..b
.
.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 13 / 34
The Extreme Value Theorem
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 absoluteminimum value f(d) at numbers c and d in [a,b].
...a
..b
.
.
.cmaximum
.maximum
value.f(c)
.
.d
minimum
.minimum
value.f(d)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 13 / 34
No proof of EVT forthcoming
I This theorem is very hard to prove without using technical factsabout continuous functions and closed intervals.
I But we can show the importance of each of the hypotheses.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 14 / 34
Bad Example #1
Example
Consider the function
f(x) =
{x 0 ≤ x < 1x− 2 1 ≤ x ≤ 2.
. .|.1
.
.
.
.
Then although values of f(x) get arbitrarily close to 1 and never biggerthan 1, 1 is not the maximum value of f on [0,1] because it is neverachieved. This does not violate EVT because f is not continuous.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 15 / 34
Bad Example #1
Example
Consider the function
f(x) =
{x 0 ≤ x < 1x− 2 1 ≤ x ≤ 2.
. .|.1
.
.
.
.
Then although values of f(x) get arbitrarily close to 1 and never biggerthan 1, 1 is not the maximum value of f on [0,1] because it is neverachieved. This does not violate EVT because f is not continuous.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 15 / 34
Bad Example #1
Example
Consider the function
f(x) =
{x 0 ≤ x < 1x− 2 1 ≤ x ≤ 2.
. .|.1
.
.
.
.
Then although values of f(x) get arbitrarily close to 1 and never biggerthan 1, 1 is not the maximum value of f on [0,1] because it is neverachieved.
This does not violate EVT because f is not continuous.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 15 / 34
Bad Example #1
Example
Consider the function
f(x) =
{x 0 ≤ x < 1x− 2 1 ≤ x ≤ 2.
. .|.1
.
.
.
.
Then although values of f(x) get arbitrarily close to 1 and never biggerthan 1, 1 is not the maximum value of f on [0,1] because it is neverachieved. This does not violate EVT because f is not continuous.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 15 / 34
Bad Example #2
Example
Consider the function f(x) = x restricted to the interval [0,1).
. .|.1
.
.
There is still no maximum value (values get arbitrarily close to 1 but donot achieve it). This does not violate EVT because the domain is notclosed.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 16 / 34
Bad Example #2
Example
Consider the function f(x) = x restricted to the interval [0,1).
. .|.1
.
.
There is still no maximum value (values get arbitrarily close to 1 but donot achieve it). This does not violate EVT because the domain is notclosed.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 16 / 34
Bad Example #2
Example
Consider the function f(x) = x restricted to the interval [0,1).
. .|.1
.
.
There is still no maximum value (values get arbitrarily close to 1 but donot achieve it).
This does not violate EVT because the domain is notclosed.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 16 / 34
Bad Example #2
Example
Consider the function f(x) = x restricted to the interval [0,1).
. .|.1
.
.
There is still no maximum value (values get arbitrarily close to 1 but donot achieve it). This does not violate EVT because the domain is notclosed.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 16 / 34
Final Bad Example
Example
Consider the function f(x) =1xis continuous on the closed interval
[1,∞).
. ..1
.
There is no minimum value (values get arbitrarily close to 0 but do notachieve it). This does not violate EVT because the domain is notbounded.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 17 / 34
Final Bad Example
Example
Consider the function f(x) =1xis continuous on the closed interval
[1,∞).
. ..1
.
There is no minimum value (values get arbitrarily close to 0 but do notachieve it). This does not violate EVT because the domain is notbounded.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 17 / 34
Final Bad Example
Example
Consider the function f(x) =1xis continuous on the closed interval
[1,∞).
. ..1
.
There is no minimum value (values get arbitrarily close to 0 but do notachieve it).
This does not violate EVT because the domain is notbounded.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 17 / 34
Final Bad Example
Example
Consider the function f(x) =1xis continuous on the closed interval
[1,∞).
. ..1
.
There is no minimum value (values get arbitrarily close to 0 but do notachieve it). This does not violate EVT because the domain is notbounded.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 17 / 34
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 18 / 34
Local extrema.
.
Definition
I A function f has a local maximum or relative 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 intervalcontaining c.
I Similarly, f has a local minimum at c if f(c) ≤ f(x) when x is near c.
..|.a
.|.b
.
.
.
.local
maximum
.
.local
minimum
.globalmax
.local and global
min
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 19 / 34
Local extrema.
.
Definition
I A function f has a local maximum or relative 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 intervalcontaining c.
I Similarly, f has a local minimum at c if f(c) ≤ f(x) when x is near c.
..|.a
.|.b
.
.
.
.local
maximum
.
.local
minimum
.globalmax
.local and global
min
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 19 / 34
Local extrema.
.
I So a local extremum must be inside the domain of f (not on the end).I A global extremum that is inside the domain is a local extremum.
..|.a
.|.b
.
.
.
.local
maximum
.
.local
minimum
.globalmax
.local and global
min
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 19 / 34
Fermat's Theorem
Theorem (Fermat’s Theorem)
Suppose f has a local extremum at c and f is differentiable at c. Thenf′(c) = 0.
..|.a
.|.b
.
.
.
.local
maximum
.
.local
minimum
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 21 / 34
Fermat's Theorem
Theorem (Fermat’s Theorem)
Suppose f has a local extremum at c and f is differentiable at c. Thenf′(c) = 0.
..|.a
.|.b
.
.
.
.local
maximum
.
.local
minimum
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 21 / 34
Sketch of proof of Fermat's Theorem
Suppose 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 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 Since the limit f′(c) = limx→c
f(x)− f(c)x− c
exists, it must be 0.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 22 / 34
Sketch of proof of Fermat's Theorem
Suppose 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 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 Since the limit f′(c) = limx→c
f(x)− f(c)x− c
exists, it must be 0.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 22 / 34
Sketch of proof of Fermat's Theorem
Suppose 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 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 Since the limit f′(c) = limx→c
f(x)− f(c)x− c
exists, it must be 0.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 22 / 34
Sketch of proof of Fermat's Theorem
Suppose 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 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 Since the limit f′(c) = limx→c
f(x)− f(c)x− c
exists, it must be 0.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 22 / 34
Sketch of proof of Fermat's Theorem
Suppose 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 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 Since the limit f′(c) = limx→c
f(x)− f(c)x− c
exists, it must be 0.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 22 / 34
Sketch of proof of Fermat's Theorem
Suppose 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 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 Since the limit f′(c) = limx→c
f(x)− f(c)x− c
exists, it must be 0.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 22 / 34
Meet the Mathematician: Pierre de Fermat
I 1601–1665I Lawyer and number
theoristI Proved many theorems,
didn’t quite prove his lastone
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 23 / 34
Tangent: Fermat's Last Theorem
I Plenty of solutions tox2 + y2 = z2 amongpositive whole numbers(e.g., x = 3, y = 4, z = 5)
I No solutions tox3 + y3 = z3 amongpositive whole numbers
I Fermat claimed nosolutions to xn + yn = zn
but didn’t write down hisproof
I Not solved until 1998!(Taylor–Wiles)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 24 / 34
Tangent: Fermat's Last Theorem
I Plenty of solutions tox2 + y2 = z2 amongpositive whole numbers(e.g., x = 3, y = 4, z = 5)
I No solutions tox3 + y3 = z3 amongpositive whole numbers
I Fermat claimed nosolutions to xn + yn = zn
but didn’t write down hisproof
I Not solved until 1998!(Taylor–Wiles)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 24 / 34
Tangent: Fermat's Last Theorem
I Plenty of solutions tox2 + y2 = z2 amongpositive whole numbers(e.g., x = 3, y = 4, z = 5)
I No solutions tox3 + y3 = z3 amongpositive whole numbers
I Fermat claimed nosolutions to xn + yn = zn
but didn’t write down hisproof
I Not solved until 1998!(Taylor–Wiles)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 24 / 34
Tangent: Fermat's Last Theorem
I Plenty of solutions tox2 + y2 = z2 amongpositive whole numbers(e.g., x = 3, y = 4, z = 5)
I No solutions tox3 + y3 = z3 amongpositive whole numbers
I Fermat claimed nosolutions to xn + yn = zn
but didn’t write down hisproof
I Not solved until 1998!(Taylor–Wiles)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 24 / 34
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 25 / 34
Flowchart for placing extremaThanks to Fermat
Suppose f is a continuous function on the closed, bounded interval[a,b], and c is a global maximum point.
..start
.Is c an
endpoint?
. c = a orc = b
.c is a
local max
.Is f diff’ble
at c?
.f is notdiff at c
.f′(c) = 0
.no
.yes
.no
.yes
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 26 / 34
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 critical points or critical numbers x where
either f′(x) = 0 or f is not differentiable at x.I The points with the largest function value are the global maximum
pointsI The points with the smallest or most negative function value are
the global minimum points.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 27 / 34
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 28 / 34
Extreme values of a linear function
Example
Find the extreme values of f(x) = 2x− 5 on [−1,2].
SolutionSince f′(x) = 2, which is never zero, we have no critical points and weneed only investigate the endpoints:
I f(−1) = 2(−1)− 5 = −7I f(2) = 2(2)− 5 = −1
SoI The absolute minimum (point) is at −1; the minimum value is −7.I The absolute maximum (point) is at 2; the maximum value is −1.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 29 / 34
Extreme values of a linear function
Example
Find the extreme values of f(x) = 2x− 5 on [−1,2].
SolutionSince f′(x) = 2, which is never zero, we have no critical points and weneed only investigate the endpoints:
I f(−1) = 2(−1)− 5 = −7I f(2) = 2(2)− 5 = −1
SoI The absolute minimum (point) is at −1; the minimum value is −7.I The absolute maximum (point) is at 2; the maximum value is −1.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 29 / 34
Extreme values of a linear function
Example
Find the extreme values of f(x) = 2x− 5 on [−1,2].
SolutionSince f′(x) = 2, which is never zero, we have no critical points and weneed only investigate the endpoints:
I f(−1) = 2(−1)− 5 = −7I f(2) = 2(2)− 5 = −1
SoI The absolute minimum (point) is at −1; the minimum value is −7.I The absolute maximum (point) is at 2; the maximum value is −1.
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 29 / 34
Extreme values of a quadratic function
Example
Find the extreme values of f(x) = x2 − 1 on [−1,2].
SolutionWe have f′(x) = 2x, which is zero when x = 0.
So our points to checkare:
I f(−1) =I f(0) =I f(2) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 30 / 34
Extreme values of a quadratic function
Example
Find the extreme values of f(x) = x2 − 1 on [−1,2].
SolutionWe have f′(x) = 2x, which is zero when x = 0.
So our points to checkare:
I f(−1) =I f(0) =I f(2) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 30 / 34
Extreme values of a quadratic function
Example
Find the extreme values of f(x) = x2 − 1 on [−1,2].
SolutionWe have f′(x) = 2x, which is zero when x = 0. So our points to checkare:
I f(−1) =I f(0) =I f(2) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 30 / 34
Extreme values of a quadratic function
Example
Find the extreme values of f(x) = x2 − 1 on [−1,2].
SolutionWe have f′(x) = 2x, which is zero when x = 0. So our points to checkare:
I f(−1) = 0I f(0) =I f(2) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 30 / 34
Extreme values of a quadratic function
Example
Find the extreme values of f(x) = x2 − 1 on [−1,2].
SolutionWe have f′(x) = 2x, which is zero when x = 0. So our points to checkare:
I f(−1) = 0I f(0) = − 1I f(2) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 30 / 34
Extreme values of a quadratic function
Example
Find the extreme values of f(x) = x2 − 1 on [−1,2].
SolutionWe have f′(x) = 2x, which is zero when x = 0. So our points to checkare:
I f(−1) = 0I f(0) = − 1I f(2) = 3
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 30 / 34
Extreme values of a quadratic function
Example
Find the extreme values of f(x) = x2 − 1 on [−1,2].
SolutionWe have f′(x) = 2x, which is zero when x = 0. So our points to checkare:
I f(−1) = 0I f(0) = − 1 (absolute min)I f(2) = 3
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 30 / 34
Extreme values of a quadratic function
Example
Find the extreme values of f(x) = x2 − 1 on [−1,2].
SolutionWe have f′(x) = 2x, which is zero when x = 0. So our points to checkare:
I f(−1) = 0I f(0) = − 1 (absolute min)I f(2) = 3 (absolute max)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 30 / 34
Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1,2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), we have critical points at x = 0 andx = 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)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 31 / 34
Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1,2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), we have critical points at x = 0 andx = 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)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 31 / 34
Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1,2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), we have critical points at x = 0 andx = 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)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 31 / 34
Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1,2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), we have critical points at x = 0 andx = 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)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 31 / 34
Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1,2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), we have critical points at x = 0 andx = 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)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 31 / 34
Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1,2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), we have critical points at x = 0 andx = 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)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 31 / 34
Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1,2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), we have critical points at x = 0 andx = 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)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 31 / 34
Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1,2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), we have critical points at x = 0 andx = 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)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 31 / 34
Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1,2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), we have critical points at x = 0 andx = 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)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 31 / 34
Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1,2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), we have critical points at x = 0 andx = 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)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 31 / 34
Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1,2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), we have critical points at x = 0 andx = 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)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 31 / 34
Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3(x+ 2) on [−1,2].
SolutionWrite 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 differentiable at 0.
So our points tocheck are:
I f(−1) =I f(−4/5) =
I f(0) =I f(2) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 32 / 34
Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3(x+ 2) on [−1,2].
SolutionWrite 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 differentiable at 0.
So our points tocheck are:
I f(−1) =I f(−4/5) =
I f(0) =I f(2) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 32 / 34
Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3(x+ 2) on [−1,2].
SolutionWrite 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 differentiable at 0. So our points tocheck are:
I f(−1) =
I f(−4/5) =
I f(0) =I f(2) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 32 / 34
Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3(x+ 2) on [−1,2].
SolutionWrite 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 differentiable at 0. So our points tocheck are:
I f(−1) = 1I f(−4/5) =
I f(0) =I f(2) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 32 / 34
Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3(x+ 2) on [−1,2].
SolutionWrite 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 differentiable at 0. So our points tocheck are:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) =
I f(2) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 32 / 34
Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3(x+ 2) on [−1,2].
SolutionWrite 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 differentiable at 0. So our points tocheck are:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0I f(2) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 32 / 34
Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3(x+ 2) on [−1,2].
SolutionWrite 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 differentiable at 0. So our points tocheck are:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0I f(2) = 6.3496
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 32 / 34
Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3(x+ 2) on [−1,2].
SolutionWrite 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 differentiable at 0. So our points tocheck are:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0 (absolute min)I f(2) = 6.3496
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 32 / 34
Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3(x+ 2) on [−1,2].
SolutionWrite 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 differentiable at 0. So our points tocheck are:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0 (absolute min)I f(2) = 6.3496 (absolute max)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 32 / 34
Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3(x+ 2) on [−1,2].
SolutionWrite 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 differentiable at 0. So our points tocheck are:
I f(−1) = 1I f(−4/5) = 1.0341 (relative max)I f(0) = 0 (absolute min)I f(2) = 6.3496 (absolute max)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 32 / 34
Extreme values of another algebraic function
Example
Find the extreme values of f(x) =√4− x2 on [−2,1].
SolutionWe have f′(x) = − x√
4− x2, which is zero when x = 0. (f is not
differentiable at ±2 as well.)
So our points to check are:
I f(−2) =I f(0) =I f(1) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 33 / 34
Extreme values of another algebraic function
Example
Find the extreme values of f(x) =√4− x2 on [−2,1].
SolutionWe have f′(x) = − x√
4− x2, which is zero when x = 0. (f is not
differentiable at ±2 as well.)
So our points to check are:
I f(−2) =I f(0) =I f(1) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 33 / 34
Extreme values of another algebraic function
Example
Find the extreme values of f(x) =√4− x2 on [−2,1].
SolutionWe have f′(x) = − x√
4− x2, which is zero when x = 0. (f is not
differentiable at ±2 as well.) So our points to check are:I f(−2) =
I f(0) =I f(1) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 33 / 34
Extreme values of another algebraic function
Example
Find the extreme values of f(x) =√4− x2 on [−2,1].
SolutionWe have f′(x) = − x√
4− x2, which is zero when x = 0. (f is not
differentiable at ±2 as well.) So our points to check are:I f(−2) = 0I f(0) =
I f(1) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 33 / 34
Extreme values of another algebraic function
Example
Find the extreme values of f(x) =√4− x2 on [−2,1].
SolutionWe have f′(x) = − x√
4− x2, which is zero when x = 0. (f is not
differentiable at ±2 as well.) So our points to check are:I f(−2) = 0I f(0) = 2I f(1) =
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 33 / 34
Extreme values of another algebraic function
Example
Find the extreme values of f(x) =√4− x2 on [−2,1].
SolutionWe have f′(x) = − x√
4− x2, which is zero when x = 0. (f is not
differentiable at ±2 as well.) So our points to check are:I f(−2) = 0I f(0) = 2I f(1) =
√3
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 33 / 34
Extreme values of another algebraic function
Example
Find the extreme values of f(x) =√4− x2 on [−2,1].
SolutionWe have f′(x) = − x√
4− x2, which is zero when x = 0. (f is not
differentiable at ±2 as well.) So our points to check are:I f(−2) = 0 (absolute min)I f(0) = 2I f(1) =
√3
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 33 / 34
Extreme values of another algebraic function
Example
Find the extreme values of f(x) =√4− x2 on [−2,1].
SolutionWe have f′(x) = − x√
4− x2, which is zero when x = 0. (f is not
differentiable 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
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 33 / 34
Summary
I The Extreme Value Theorem: a continuous function on a closedinterval must achieve its max and min
I Fermat’s Theorem: local extrema are critical pointsI The Closed Interval Method: an algorithm for finding global
extremaI Show your work unless you want to end up like Fermat!
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 34 / 34