Post on 18-May-2015
description
. . . . . .
Section 4.1Maximum and Minimum Values
V63.0121, Calculus I
March 24, 2009
Announcements
I Homework due ThursdayI Quiz April 2, on Sections 2.5–3.5I Final Exam Friday, May 8, 2:00–3:50pm
..Image: Flickr user Karen with a K
. . . . . .
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
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)
. . . . . .
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)
. . . . . .
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)
. . . . . .
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
. . . . . .
Extreme points and values
DefinitionLet f have domain D.
I The function f has an absolutemaximum (or global maximum)(respectively, absolute minimum)at c if f(c) ≥ f(x) (respectively,f(c) ≤ f(x)) for all x in D
I The number f(c) is called themaximum value (respectively,minimum value) of f on D.
I An extremum is either a maximumor a minimum. An extreme value iseither a maximum value or minimumvalue.
.
.Image credit: Patrick Q
. . . . . .
Extreme points and values
DefinitionLet f have domain D.
I The function f has an absolutemaximum (or global maximum)(respectively, absolute minimum)at c if f(c) ≥ f(x) (respectively,f(c) ≤ f(x)) for all x in D
I The number f(c) is called themaximum value (respectively,minimum value) of f on D.
I An extremum is either a maximumor a minimum. An extreme value iseither a maximum value or minimumvalue.
.
.Image credit: Patrick Q
. . . . . .
Extreme points and values
DefinitionLet f have domain D.
I The function f has an absolutemaximum (or global maximum)(respectively, absolute minimum)at c if f(c) ≥ f(x) (respectively,f(c) ≤ f(x)) for all x in D
I The number f(c) is called themaximum value (respectively,minimum value) of f on D.
I An extremum is either a maximumor a minimum. An extreme value iseither a maximum value or minimumvalue.
.
.Image credit: Patrick Q
. . . . . .
Theorem (The Extreme Value Theorem)Let f be a function which is continuous on the closed interval [a, b]. Then fattains an absolute maximum value f(c) and an absolute minimum valuef(d) at numbers c and d in [a, b].
.
. . . . . .
Theorem (The Extreme Value Theorem)Let f be a function which is continuous on the closed interval [a, b]. Then fattains an absolute maximum value f(c) and an absolute minimum valuef(d) at numbers c and d in [a, b].
...a
..b
.
.
. . . . . .
Theorem (The Extreme Value Theorem)Let f be a function which is continuous on the closed interval [a, b]. Then fattains an absolute maximum value f(c) and an absolute minimum valuef(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
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.
. . . . . .
Bad Example #1
ExampleConsider the function
f(x) =
{x 0 ≤ x < 1
x − 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.
. . . . . .
Bad Example #1
ExampleConsider the function
f(x) =
{x 0 ≤ x < 1
x − 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.
. . . . . .
Bad Example #1
ExampleConsider the function
f(x) =
{x 0 ≤ x < 1
x − 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.
. . . . . .
Bad Example #2
ExampleThe function f(x) = x restricted to the interval [0, 1) still has nomaximum value.
. .|.1
.
.
. . . . . .
Bad Example #2
ExampleThe function f(x) = x restricted to the interval [0, 1) still has nomaximum value.
. .|.1
.
.
. . . . . .
Final Bad Example
Example
The function f(x) =1x
is continuous on the closed interval [1,∞) but
has no minimum value.
. ..1
.
. . . . . .
Final Bad Example
Example
The function f(x) =1x
is continuous on the closed interval [1,∞) but
has no minimum value.
. ..1
.
. . . . . .
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
. . . . . .
Local extremaDefinition
I A function f has a local maximum or relative maximumat c if f(c) ≥ f(x) when x is near c. This means that f(c) ≥ f(x)for all x in some open interval containing c.
I Similarly, f has a local minimum at c if f(c) ≤ f(x) when x isnear c.
..|.a
.|.b
.
.
.
.local
maximum
.
.local
minimum
. . . . . .
Local extremaDefinition
I A function f has a local maximum or relative maximumat c if f(c) ≥ f(x) when x is near c. This means that f(c) ≥ f(x)for all x in some open interval containing c.
I Similarly, f has a local minimum at c if f(c) ≤ f(x) when x isnear c.
..|.a
.|.b
.
.
.
.local
maximum
.
.local
minimum
. . . . . .
I So a local extremum must be inside the domain of f (not on theend).
I A global extremum that is inside the domain is a local extremum.
..|.a
.|.b
.
.
.
.globalmax
.localmax
.
.local and global
min
. . . . . .
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
. . . . . .
Sketch of proof of Fermat’s Theorem
Suppose that f has a local maximum at c.
I If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). Thismeans
f(c + h) − f(c)h
≤ 0 =⇒ limh→0+
f(c + h) − f(c)h
≤ 0
I The same will be true on the other end: if h is close enough to 0but less than 0, f(c + h) ≤ f(c). This means
f(c + h) − f(c)h
≥ 0 =⇒ limh→0−
f(c + h) − f(c)h
≥ 0
I Since the limit f′(c) = limh→0
f(c + h) − f(c)h
exists, it must be 0.
. . . . . .
Sketch of proof of Fermat’s Theorem
Suppose that f has a local maximum at c.I If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This
means
f(c + h) − f(c)h
≤ 0
=⇒ limh→0+
f(c + h) − f(c)h
≤ 0
I The same will be true on the other end: if h is close enough to 0but less than 0, f(c + h) ≤ f(c). This means
f(c + h) − f(c)h
≥ 0 =⇒ limh→0−
f(c + h) − f(c)h
≥ 0
I Since the limit f′(c) = limh→0
f(c + h) − f(c)h
exists, it must be 0.
. . . . . .
Sketch of proof of Fermat’s Theorem
Suppose that f has a local maximum at c.I If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This
means
f(c + h) − f(c)h
≤ 0 =⇒ limh→0+
f(c + h) − f(c)h
≤ 0
I The same will be true on the other end: if h is close enough to 0but less than 0, f(c + h) ≤ f(c). This means
f(c + h) − f(c)h
≥ 0 =⇒ limh→0−
f(c + h) − f(c)h
≥ 0
I Since the limit f′(c) = limh→0
f(c + h) − f(c)h
exists, it must be 0.
. . . . . .
Sketch of proof of Fermat’s Theorem
Suppose that f has a local maximum at c.I If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This
means
f(c + h) − f(c)h
≤ 0 =⇒ limh→0+
f(c + h) − f(c)h
≤ 0
I The same will be true on the other end: if h is close enough to 0but less than 0, f(c + h) ≤ f(c). This means
f(c + h) − f(c)h
≥ 0
=⇒ limh→0−
f(c + h) − f(c)h
≥ 0
I Since the limit f′(c) = limh→0
f(c + h) − f(c)h
exists, it must be 0.
. . . . . .
Sketch of proof of Fermat’s Theorem
Suppose that f has a local maximum at c.I If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This
means
f(c + h) − f(c)h
≤ 0 =⇒ limh→0+
f(c + h) − f(c)h
≤ 0
I The same will be true on the other end: if h is close enough to 0but less than 0, f(c + h) ≤ f(c). This means
f(c + h) − f(c)h
≥ 0 =⇒ limh→0−
f(c + h) − f(c)h
≥ 0
I Since the limit f′(c) = limh→0
f(c + h) − f(c)h
exists, it must be 0.
. . . . . .
Sketch of proof of Fermat’s Theorem
Suppose that f has a local maximum at c.I If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This
means
f(c + h) − f(c)h
≤ 0 =⇒ limh→0+
f(c + h) − f(c)h
≤ 0
I The same will be true on the other end: if h is close enough to 0but less than 0, f(c + h) ≤ f(c). This means
f(c + h) − f(c)h
≥ 0 =⇒ limh→0−
f(c + h) − f(c)h
≥ 0
I Since the limit f′(c) = limh→0
f(c + h) − f(c)h
exists, it must be 0.
. . . . . .
Meet the Mathematician: Pierre de Fermat
I 1601–1665I Lawyer and number
theoristI Proved many theorems,
didn’t quite prove his lastone
. . . . . .
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)
. . . . . .
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)
. . . . . .
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)
. . . . . .
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)
. . . . . .
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
. . . . . .
The Closed Interval Method
Let’s put this together logically. Let f be a continuous functiondefined on a closed interval [a, b]. We are in search of its globalmaximum, call it c. Then:
I Either the maximumoccurs at an endpoint ofthe interval, i.e., c = a orc = b,
I Or the maximum occursinside (a, b). In this case, cis also a local maximum.
I Either f is differentiableat c, in which casef′(c) = 0 by Fermat’sTheorem.
I Or f is notdifferentiable at c.
This means to find themaximum value of f on [a, b],we need to check:
I a and bI Points x where f′(x) = 0I Points x where f is not
differentiable.
The latter two are both calledcritical points of f. Thistechnique is called the ClosedInterval Method.
. . . . . .
The Closed Interval Method
Let’s put this together logically. Let f be a continuous functiondefined on a closed interval [a, b]. We are in search of its globalmaximum, call it c. Then:
I Either the maximumoccurs at an endpoint ofthe interval, i.e., c = a orc = b,
I Or the maximum occursinside (a, b). In this case, cis also a local maximum.
I Either f is differentiableat c, in which casef′(c) = 0 by Fermat’sTheorem.
I Or f is notdifferentiable at c.
This means to find themaximum value of f on [a, b],we need to check:
I a and bI Points x where f′(x) = 0I Points x where f is not
differentiable.
The latter two are both calledcritical points of f. Thistechnique is called the ClosedInterval Method.
. . . . . .
The Closed Interval Method
Let’s put this together logically. Let f be a continuous functiondefined on a closed interval [a, b]. We are in search of its globalmaximum, call it c. Then:
I Either the maximumoccurs at an endpoint ofthe interval, i.e., c = a orc = b,
I Or the maximum occursinside (a, b). In this case, cis also a local maximum.
I Either f is differentiableat c, in which casef′(c) = 0 by Fermat’sTheorem.
I Or f is notdifferentiable at c.
This means to find themaximum value of f on [a, b],we need to check:
I a and bI Points x where f′(x) = 0I Points x where f is not
differentiable.
The latter two are both calledcritical points of f. Thistechnique is called the ClosedInterval Method.
. . . . . .
The Closed Interval Method
Let’s put this together logically. Let f be a continuous functiondefined on a closed interval [a, b]. We are in search of its globalmaximum, call it c. Then:
I Either the maximumoccurs at an endpoint ofthe interval, i.e., c = a orc = b,
I Or the maximum occursinside (a, b). In this case, cis also a local maximum.
I Either f is differentiableat c, in which casef′(c) = 0 by Fermat’sTheorem.
I Or f is notdifferentiable at c.
This means to find themaximum value of f on [a, b],we need to check:
I a and bI Points x where f′(x) = 0I Points x where f is not
differentiable.
The latter two are both calledcritical points of f. Thistechnique is called the ClosedInterval Method.
. . . . . .
The Closed Interval Method
Let’s put this together logically. Let f be a continuous functiondefined on a closed interval [a, b]. We are in search of its globalmaximum, call it c. Then:
I Either the maximumoccurs at an endpoint ofthe interval, i.e., c = a orc = b,
I Or the maximum occursinside (a, b). In this case, cis also a local maximum.
I Either f is differentiableat c, in which casef′(c) = 0 by Fermat’sTheorem.
I Or f is notdifferentiable at c.
This means to find themaximum value of f on [a, b],we need to check:
I a and bI Points x where f′(x) = 0I Points x where f is not
differentiable.
The latter two are both calledcritical points of f. Thistechnique is called the ClosedInterval Method.
. . . . . .
The Closed Interval Method
Let’s put this together logically. Let f be a continuous functiondefined on a closed interval [a, b]. We are in search of its globalmaximum, call it c. Then:
I Either the maximumoccurs at an endpoint ofthe interval, i.e., c = a orc = b,
I Or the maximum occursinside (a, b). In this case, cis also a local maximum.
I Either f is differentiableat c, in which casef′(c) = 0 by Fermat’sTheorem.
I Or f is notdifferentiable at c.
This means to find themaximum value of f on [a, b],we need to check:
I a and bI Points x where f′(x) = 0I Points x where f is not
differentiable.
The latter two are both calledcritical points of f. Thistechnique is called the ClosedInterval Method.
. . . . . .
The Closed Interval Method
Let’s put this together logically. Let f be a continuous functiondefined on a closed interval [a, b]. We are in search of its globalmaximum, call it c. Then:
I Either the maximumoccurs at an endpoint ofthe interval, i.e., c = a orc = b,
I Or the maximum occursinside (a, b). In this case, cis also a local maximum.
I Either f is differentiableat c, in which casef′(c) = 0 by Fermat’sTheorem.
I Or f is notdifferentiable at c.
This means to find themaximum value of f on [a, b],we need to check:
I a and b
I Points x where f′(x) = 0I Points x where f is not
differentiable.
The latter two are both calledcritical points of f. Thistechnique is called the ClosedInterval Method.
. . . . . .
The Closed Interval Method
Let’s put this together logically. Let f be a continuous functiondefined on a closed interval [a, b]. We are in search of its globalmaximum, call it c. Then:
I Either the maximumoccurs at an endpoint ofthe interval, i.e., c = a orc = b,
I Or the maximum occursinside (a, b). In this case, cis also a local maximum.
I Either f is differentiableat c, in which casef′(c) = 0 by Fermat’sTheorem.
I Or f is notdifferentiable at c.
This means to find themaximum value of f on [a, b],we need to check:
I a and bI Points x where f′(x) = 0
I Points x where f is notdifferentiable.
The latter two are both calledcritical points of f. Thistechnique is called the ClosedInterval Method.
. . . . . .
The Closed Interval Method
Let’s put this together logically. Let f be a continuous functiondefined on a closed interval [a, b]. We are in search of its globalmaximum, call it c. Then:
I Either the maximumoccurs at an endpoint ofthe interval, i.e., c = a orc = b,
I Or the maximum occursinside (a, b). In this case, cis also a local maximum.
I Either f is differentiableat c, in which casef′(c) = 0 by Fermat’sTheorem.
I Or f is notdifferentiable at c.
This means to find themaximum value of f on [a, b],we need to check:
I a and bI Points x where f′(x) = 0I Points x where f is not
differentiable.
The latter two are both calledcritical points of f. Thistechnique is called the ClosedInterval Method.
. . . . . .
The Closed Interval Method
Let’s put this together logically. Let f be a continuous functiondefined on a closed interval [a, b]. We are in search of its globalmaximum, call it c. Then:
I Either the maximumoccurs at an endpoint ofthe interval, i.e., c = a orc = b,
I Or the maximum occursinside (a, b). In this case, cis also a local maximum.
I Either f is differentiableat c, in which casef′(c) = 0 by Fermat’sTheorem.
I Or f is notdifferentiable at c.
This means to find themaximum value of f on [a, b],we need to check:
I a and bI Points x where f′(x) = 0I Points x where f is not
differentiable.
The latter two are both calledcritical points of f. Thistechnique is called the ClosedInterval Method.
. . . . . .
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
. . . . . .
ExampleFind 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.
. . . . . .
ExampleFind 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.
. . . . . .
ExampleFind 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) =
. . . . . .
ExampleFind 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) =
. . . . . .
ExampleFind 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) =
. . . . . .
ExampleFind 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) =
. . . . . .
ExampleFind 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) =
. . . . . .
ExampleFind 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
. . . . . .
ExampleFind 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
. . . . . .
ExampleFind 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)
. . . . . .
ExampleFind the extreme values of f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53
x2/3 +43
x−1/3 =13
x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f is not differentiable at 0.
So our points to checkare:
I f(−1) =
I f(−4/5) =
I f(0) =
I f(2) =
. . . . . .
ExampleFind the extreme values of f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53
x2/3 +43
x−1/3 =13
x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f is not differentiable at 0.
So our points to checkare:
I f(−1) =
I f(−4/5) =
I f(0) =
I f(2) =
. . . . . .
ExampleFind the extreme values of f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53
x2/3 +43
x−1/3 =13
x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f is not differentiable at 0. So our points to checkare:
I f(−1) =
I f(−4/5) =
I f(0) =
I f(2) =
. . . . . .
ExampleFind the extreme values of f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53
x2/3 +43
x−1/3 =13
x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f is not differentiable at 0. So our points to checkare:
I f(−1) = 1I f(−4/5) =
I f(0) =
I f(2) =
. . . . . .
ExampleFind the extreme values of f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53
x2/3 +43
x−1/3 =13
x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f is not differentiable at 0. So our points to checkare:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) =
I f(2) =
. . . . . .
ExampleFind the extreme values of f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53
x2/3 +43
x−1/3 =13
x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f is not differentiable at 0. So our points to checkare:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0I f(2) =
. . . . . .
ExampleFind the extreme values of f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53
x2/3 +43
x−1/3 =13
x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f is not differentiable at 0. So our points to checkare:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0I f(2) = 6.3496
. . . . . .
ExampleFind the extreme values of f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53
x2/3 +43
x−1/3 =13
x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f is not differentiable at 0. So our points to checkare:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0 (absolute min)I f(2) = 6.3496
. . . . . .
ExampleFind the extreme values of f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53
x2/3 +43
x−1/3 =13
x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f is not differentiable at 0. So our points to checkare:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0 (absolute min)I f(2) = 6.3496 (absolute max)
. . . . . .
ExampleFind the extreme values of f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53
x2/3 +43
x−1/3 =13
x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f is not differentiable at 0. So our points to checkare:
I f(−1) = 1I f(−4/5) = 1.0341 (relative max)I f(0) = 0 (absolute min)I f(2) = 6.3496 (absolute max)
. . . . . .
ExampleFind 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) =
. . . . . .
ExampleFind 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) =
. . . . . .
ExampleFind 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) =
. . . . . .
ExampleFind 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) =
. . . . . .
ExampleFind 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) =
. . . . . .
ExampleFind 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
. . . . . .
ExampleFind 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
. . . . . .
ExampleFind 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
. . . . . .
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
. . . . . .
Challenge: Cubic functions
ExampleHow many critical points can a cubic function
f(x) = ax3 + bx2 + cx + d
have?
. . . . . .
SolutionIf f′(x) = 0, we have
3ax2 + 2bx + c = 0,
and so
x =−2b ±
√4b2 − 12ac6a
=−b ±
√b2 − 3ac
3a,
and so we have three possibilities:I 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.I b2 − 3ac < 0, in which case there are no real roots to the quadratic,
hence no critical points. An example would be f(x) = x3 + x2 + x,where a = b = c = 1.
I b2 − 3ac = 0, in which case there is a single critical point. Example:x3, where a = 1 and b = c = 0.
. . . . . .
Review
I Concept: absolute (global) and relative (local) maxima/minimaI Fact: Fermat’s theorem: f′(x) = 0 at local extremaI Technique: the Closed Interval Method