Lesson 18: Maximum and Minimum Values (slides)

Sec on 4.1
Maximum and Minimum Values
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University

April 4, 2011

.

Announcements

Quiz 4 on Sec ons 3.3, 3.4, 3.5,
and 3.7 next week (April 14/15)
Quiz 5 on Sec ons 4.1–4.4
April 28/29
Final Exam Monday May 12,
2:00–3:50pm

Objectives
Understand and be able to
explain the statement of the
Extreme Value Theorem.
Understand and be able to
explain the statement of
Fermat’s Theorem.
Use the Closed Interval Method
to ﬁnd the extreme values of a
func on deﬁned on a closed
interval.

Outline
Introduc on

The Extreme Value Theorem

Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem

The Closed Interval Method

Examples

Why go to the extremes?
Ra onally speaking, it is
advantageous to ﬁnd the
extreme values of a func on
(maximize proﬁt, minimize costs,
etc.)

Pierre-Louis Maupertuis
(1698–1759)

etc.)
Many laws of science are
derived from minimizing
principles.

Pierre-Louis Maupertuis
(1698–1759)

etc.)
Many laws of science are
derived from minimizing
principles.
Maupertuis’ principle: “Ac on is
minimized through the wisdom Pierre-Louis Maupertuis
of God.” (1698–1759)

Extreme points and values
Deﬁni on
Let f have domain D.

.

Image credit: Patrick Q

Deﬁni on
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

.


Deﬁni on
The number f(c) is called the maximum
value (respec vely, minimum value) of f
on D.
.


Deﬁni on
The number f(c) is called the maximum
value (respec vely, minimum value) of f
on D.
An extremum is either a maximum or a .
minimum. An extreme value is either a
maximum value or minimum value.

Theorem (The Extreme Value
Theorem)
Let f be a func on which is
con nuous on the closed
interval [a, b]. Then f a ains
an absolute maximum value
f(c) and an absolute minimum
value f(d) at numbers c and d
in [a, b].

Theorem)
Let f be a func on which is
value f(d) at numbers c and d .
in [a, b]. a b

Theorem) maximum
value
Let f be a func on which is f(c)
a c
in [a, b]. b
maximum

Theorem) maximum
value
Let f be a func on which is f(c)
interval [a, b]. Then f a ains minimum
an absolute maximum value value
f(c) and an absolute minimum f(d)
a d c
in [a, b]. b
minimum maximum

No proof of EVT forthcoming

This theorem is very hard to prove without using technical facts
about con nuous func ons and closed intervals.
But we can show the importance of each of the hypotheses.

Bad Example #1
Example
Consider the func on
{
x 0≤x<1
f(x) =
x − 2 1 ≤ x ≤ 2.

Bad Example #1
Example
{
x 0≤x<1 .
f(x) = |
x − 2 1 ≤ x ≤ 2. 1

Bad Example #1
Example
{
x 0≤x<1 .
f(x) = |
x − 2 1 ≤ x ≤ 2. 1
Then although values of f(x) get arbitrarily close to 1 and never
bigger than 1, 1 is not the maximum value of f on [0, 1] because it is
never achieved.

Bad Example #1
Example
{
x 0≤x<1 .
f(x) = |
x − 2 1 ≤ x ≤ 2. 1
Then although values of f(x) get arbitrarily close to 1 and never
bigger than 1, 1 is not the maximum value of f on [0, 1] because it is
never achieved. This does not violate EVT because f is not
con nuous.

Bad Example #2
Example
Consider the func on f(x) = x restricted to the interval [0, 1).

. |
1

Bad Example #2
Example
There is s ll no maximum
value (values get
arbitrarily close to 1 but
do not achieve it).
. |
1

Bad Example #2
Example
There is s ll no maximum
value (values get
arbitrarily close to 1 but
do not achieve it).
This does not violate EVT
. |
because the domain is 1
not closed.

Final Bad Example
Example
1
The func on f(x) = is con nuous on the closed interval [1, ∞).
x

.
1

Final Bad Example
Example
1
x

.
1

There is no minimum value (values get arbitrarily close to 0 but do
not achieve it).

Final Bad Example
Example
1
x

.
1

There is no minimum value (values get arbitrarily close to 0 but do
not achieve it). This does not violate EVT because the domain is not
bounded.

Local extrema
Deﬁni on
A func on f has a local
maximum or rela ve 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 interval
containing c. |. |
a b
Similarly, f has a local minimum
at c if f(c) ≤ f(x) when x is near
c.

Local extrema
Deﬁni on
containing c. |. |
a
local b
Similarly, f has a local minimum maximum
c.

Local extrema
Deﬁni on
containing c. |. |
local local b
a
Similarly, f has a local minimum maximum minimum
c.

Local extrema
A local extremum could be a
global extremum, but not if
there are more extreme values
elsewhere.
A global extremum could be a
local extremum, but not if it is
an endpoint.
|. |
a b
local local and global
maximum global max
min

Fermat’s Theorem
Theorem (Fermat’s Theorem)

Suppose f has a
local extremum at c
and f is
diﬀeren able at c.
Then f′ (c) = 0. |. |
a local local b
maximum minimum

Proof of Fermat’s Theorem
Suppose that f has a local maximum at c.

If x is slightly greater than c, f(x) ≤ f(c). This means
f(x) − f(c)
≤0
x−c

f(x) − f(c) f(x) − f(c)
≤ 0 =⇒ lim ≤0
x−c x→c+ x−c

f(x) − f(c) f(x) − f(c)
≤ 0 =⇒ lim ≤0
x−c x→c+ x−c

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)
≥0
x−c

f(x) − f(c) f(x) − f(c)
≤ 0 =⇒ lim ≤0
x−c x→c+ x−c

f(x) − f(c) f(x) − f(c)
≥ 0 =⇒ lim ≥0
x−c x→c− x−c

f(x) − f(c) f(x) − f(c)
≤ 0 =⇒ lim ≤0
x−c x→c+ x−c

f(x) − f(c) f(x) − f(c)
≥ 0 =⇒ lim ≥0
x−c x→c− x−c
f(x) − f(c)
Since the limit f′ (c) = lim exists, it must be 0.
x→c x−c

Meet the Mathematician: Pierre de Fermat

1601–1665
Lawyer and number
theorist
Proved many theorems,
didn’t quite prove his last
one

Plenty of solu ons to
x2 + y2 = z2 among posi ve
whole numbers (e.g., x = 3,
y = 4, z = 5)

y = 4, z = 5)
No solu ons to x3 + y3 = z3
among posi ve whole numbers

y = 4, z = 5)
Fermat claimed no solu ons to
xn + yn = zn but didn’t write
down his proof

y = 4, z = 5)
Fermat claimed no solu ons to
xn + yn = zn but didn’t write
down his proof
Not solved un l 1998!
(Taylor–Wiles)

Flowchart for placing extrema
Thanks to Fermat
Suppose f is a c is a
.
start
con nuous local max
func on on
the closed,
bounded Is c an
no Is f diﬀ’ble no f is not
interval endpoint? at c? diﬀ at c
[a, b], and c is
a global
yes yes
maximum
c = a or
point. f′ (c) = 0
c = b

The Closed Interval Method
This means to ﬁnd the maximum value of f on [a, b], we need to:
Evaluate f at the endpoints a and b
Evaluate f at the cri cal points or cri cal numbers x where
either f′ (x) = 0 or f is not diﬀeren able at x.
The points with the largest func on value are the global
maximum points
The points with the smallest or most nega ve func on value
are the global minimum points.

Extreme values of a linear function
Example
Find the extreme values of f(x) = 2x − 5 on [−1, 2].

Example

Solu on
Since f′ (x) = 2, which is never
zero, we have no cri cal points
and we need only inves gate
the endpoints:
f(−1) = 2(−1) − 5 = −7
f(2) = 2(2) − 5 = −1

Example

Solu on
So
Since f′ (x) = 2, which is never
zero, we have no cri cal points The absolute minimum
and we need only inves gate (point) is at −1; the
the endpoints: minimum value is −7.
f(−1) = 2(−1) − 5 = −7 The absolute maximum
(point) is at 2; the
f(2) = 2(2) − 5 = −1
maximum value is −1.

Extreme values of a quadratic
function
Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].

function
Example

Solu on
We have f′ (x) = 2x, which is zero when x = 0.

function
Example

Solu on
We have f′ (x) = 2x, which is zero when x = 0. So our points to
check are:
f(−1) =
f(0) =
f(2) =

function
Example

Solu on
check are:
f(−1) = 0
f(0) =
f(2) =

function
Example

Solu on
check are:
f(−1) = 0
f(0) = − 1
f(2) =

function
Example

Solu on
check are:
f(−1) = 0
f(0) = − 1
f(2) = 3

function
Example

Solu on
check are:
f(−1) = 0
f(0) = − 1 (absolute min)
f(2) = 3

function
Example

Solu on
check are:
f(−1) = 0
f(0) = − 1 (absolute min)
f(2) = 3 (absolute max)

Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

Example

Solu on
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
and x = 1.

Example

Solu on
and x = 1. The values to check are

Example

Solu on
f(−1) = − 4

Example

Solu on
f(−1) = − 4
f(0) = 1

Example

Solu on
f(−1) = − 4
f(0) = 1
f(1) = 0

Example

Solu on
f(−1) = − 4
f(0) = 1
f(1) = 0
f(2) = 5

Example

Solu on
f(−1) = − 4 (global min)
f(0) = 1
f(1) = 0
f(2) = 5

Example

Solu on
f(0) = 1
f(1) = 0
f(2) = 5 (global max)

Example

Solu on
f(0) = 1 (local max)
f(1) = 0

Example

Solu on
f(0) = 1 (local max)
f(1) = 0 (local min)

Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].

Example

Solu on
Write f(x) = x5/3 + 2x2/3 .

Example

Solu on
Write f(x) = x5/3 + 2x2/3 . Then
5 4 1
f′ (x) = x2/3 + x−1/3 = x−1/3 (5x + 4)
3 3 3

Example

Solu on
Write f(x) = x5/3 + 2x2/3 . Then
5 4 1
f′ (x) = x2/3 + x−1/3 = x−1/3 (5x + 4)
3 3 3
Thus f′ (−4/5) = 0 and f is not diﬀeren able at 0. Thus there are two
cri cal points.

Example

Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) =
f(−4/5) =
f(0) =
f(2) =

Example

Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) =
f(0) =
f(2) =

Example

Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) = 1.0341
f(0) =
f(2) =

Example

Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0
f(2) =

Example

Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0
f(2) = 6.3496

Example

Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0 (absolute min)
f(2) = 6.3496

Example

Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) = 1.0341
f(2) = 6.3496 (absolute max)

Example

Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) = 1.0341 (rela ve max)
f(2) = 6.3496 (absolute max)

Extreme values of another algebraic function
Example
√
Find the extreme values of f(x) = 4 − x2 on [−2, 1].

Example
√

Solu on
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
diﬀeren able at ±2 as well.)

Example
√

Solu on
x
4 − x2
diﬀeren able at ±2 as well.) So our points to check are:
f(−2) =

Example
√

Solu on
x
4 − x2
f(−2) = 0
f(0) =

Example
√

Solu on
x
4 − x2
f(−2) = 0
f(0) = 2
f(1) =

Example
√

Solu on
x
4 − x2
f(−2) = 0
f(0) = 2
√
f(1) = 3

Example
√

Solu on
x
4 − x2
f(−2) = 0 (absolute min)
f(0) = 2
√
f(1) = 3

Example
√

Solu on
x
4 − x2
f(−2) = 0 (absolute min)
f(0) = 2 (absolute max)
√
f(1) = 3

Summary

The Extreme Value Theorem: a con nuous func on on a closed
interval must achieve its max and min
Fermat’s Theorem: local extrema are cri cal points
The Closed Interval Method: an algorithm for ﬁnding global
extrema
Show your work unless you want to end up like Fermat!

Lesson 18: Maximum and Minimum Values (slides)

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