Lesson 18: Maximum and Minimum Values (Section 021 handout)

Section 4.1
Maximum and Minimum Values
V63.0121.021, Calculus I
New York University
November 9, 2010
Announcements
Quiz 4 on Sections 3.3, 3.4, 3.5, and 3.7 next week (November 16,
18, or 19)
Announcements
Quiz 4 on Sections 3.3, 3.4,
3.5, and 3.7 next week
(November 16, 18, or 19)
V63.0121.021, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 9, 2010 2 / 34
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 function deﬁned
on a closed interval.
Notes
Notes
Notes
1
Section 4.1 : Maximum and Minimum ValuesV63.0121.021, Calculus I November 9, 2010

Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
Optimize
Why go to the extremes?
Rationally speaking, it is
advantageous to ﬁnd the
extreme values of a function
(maximize proﬁt, minimize
costs, etc.)
Many laws of science are
derived from minimizing
principles.
Maupertuis’ principle:
“Action is minimized
through the wisdom of
God.”
Pierre-Louis Maupertuis
(1698–1759)V63.0121.021, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 9, 2010 6 / 34
Notes
Notes
Notes
2

Design
Image credit: Jason Tromm
Optics
Image credit: jacreative
Outline
Introduction
Examples
Notes
Notes
Notes
3

Extreme points and values
Deﬁnition
Let f have domain D.
The function f has an absolute maximum
(or global maximum) (respectively,
absolute minimum) at c if f (c) ≥ f (x)
(respectively, f (c) ≤ f (x)) for all x in D
The number f (c) is called the maximum
value (respectively, 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.
Image credit: Patrick Q
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
c
maximum
maximum
value
f (c)
d
minimum
minimum
value
f (d)
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.
Notes
Notes
Notes
4

Bad Example #1
Example
Consider 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 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 continuous.
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 do not
achieve it). This does not violate EVT because the domain is not closed.
Final Bad Example
Example
Consider the function f (x) =
1
x
is continuous on the closed interval [1, ∞).
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.
Notes
Notes
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5

Outline
Introduction
Examples
Local extrema
Deﬁnition
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 interval
containing c.
Similarly, f has a local minimum at c if f (c) ≤ f (x) when x is near c.
|
a
|
blocal
maximum
local
minimum
global
max
local and global
min
Local extrema
So a local extremum must be inside the domain of f (not on the end).
A global extremum that is inside the domain is a local extremum.
|
a
|
blocal
maximum
local
minimum
global
max
local and global
min
Notes
Notes
Notes
6

Fermat’s Theorem
Theorem (Fermat’s Theorem)
Suppose f has a local extremum at c and f is diﬀerentiable at c. Then
f (c) = 0.
|
a
|
blocal
maximum
local
minimum
Sketch of 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)
x − c
≤ 0 =⇒ lim
x→c+
f (x) − f (c)
x − c
≤ 0
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 =⇒ lim
x→c−
f (x) − f (c)
x − c
≥ 0
Since the limit f (c) = lim
x→c
f (x) − f (c)
x − c
exists, it must be 0.
Meet the Mathematician: Pierre de Fermat
1601–1665
Lawyer and number theorist
Proved many theorems,
didn’t quite prove his last
one
Notes
Notes
Notes
7

Outline
Introduction
Examples
Flowchart for placing extrema
Thanks 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 or
c = b
c is a
local max
Is f
diff’ble at
c?
f is not
diff at c
f (c) = 0
no
yes
no
yes
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 critical points or critical numbers x where either
f (x) = 0 or f is not differentiable at x.
The points with the largest function value are the global maximum
points
The points with the smallest or most negative function value are the
global minimum points.
Notes
Notes
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8

Outline
Introduction
Examples
Extreme values of a linear function
Example
Find the extreme values of f (x) = 2x − 5 on [−1, 2].
Solution
Since f (x) = 2, which is never zero, we have no critical points and we
need only investigate the endpoints:
f (−1) = 2(−1) − 5 = −7
f (2) = 2(2) − 5 = −1
So
The absolute minimum (point) is at −1; the minimum value is −7.
The absolute maximum (point) is at 2; the maximum value is −1.
Extreme values of a quadratic function
Example
Find the extreme values of f (x) = x2
− 1 on [−1, 2].
Solution
We have f (x) = 2x, which is zero when x = 0. So our points to check
are:
f (−1) = 0
f (0) = − 1 (absolute min)
f (2) = 3 (absolute max)
Notes
Notes
Notes
9

Extreme values of a cubic function
Example
Find the extreme values of f (x) = 2x3
− 3x2
+ 1 on [−1, 2].
Solution
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = x2/3
(x + 2) on [−1, 2].
Solution
Extreme values of another algebraic function
Example
Find the extreme values of f (x) = 4 − x2 on [−2, 1].
Solution
Notes
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10

Summary
The Extreme Value Theorem: a continuous function on a closed
interval must achieve its max and min
Fermat’s Theorem: local extrema are critical points
The Closed Interval Method: an algorithm for ﬁnding global extrema
Notes
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11

Lesson 18: Maximum and Minimum Values (Section 021 handout)

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Lesson 18: Maximum and Minimum Values (Section 021 handout)