The document contains lecture notes for a Calculus I class focusing on maximum and minimum values, covering the extreme value theorem and Fermat's theorem. Key methods for finding extrema on closed intervals are discussed, including the closed interval method and evaluations at critical points and endpoints. Examples illustrate the application of these concepts, reinforcing the importance of continuous functions on closed intervals to achieve absolute extremum values.

1. . V63.0121.001: Calculus I
. Sec on 4.1: Max/Min .Values April 4, 2011
Notes
Sec on 4.1
Maximum and Minimum Values
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University
April 4, 2011
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Notes
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
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Notes
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.
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2. . V63.0121.001: Calculus I
. Sec on 4.1: Max/Min .Values April 4, 2011
Notes
Outline
Introduc 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
Optimize
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Notes
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.)
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)
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3. . V63.0121.001: Calculus I
. Sec on 4.1: Max/Min .Values April 4, 2011
Notes
Design
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Notes
Optics
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Notes
Outline
Introduc 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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4. . V63.0121.001: Calculus I
. Sec on 4.1: Max/Min .Values April 4, 2011
Notes
Extreme points and values
Defini on
Let f have domain D.
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
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.
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Image credit: Patrick Q
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Notes
The Extreme Value Theorem
Theorem (The Extreme Value
Theorem) maximum
value
Let f be a func on which is f(c)
con nuous on the closed
interval [a, b]. Then f a ains minimum
an absolute maximum value value
f(c) and an absolute minimum f(d)
value f(d) at numbers c and d .
a d c
in [a, b]. b
minimum maximum
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Notes
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.
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5. . V63.0121.001: Calculus I
. Sec on 4.1: Max/Min .Values April 4, 2011
Notes
Bad Example #1
Example
Consider the func on
{
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.
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Notes
Bad Example #2
Example
Consider the func on f(x) = x restricted to the interval [0, 1).
There is s ll no maximum
value (values get
arbitrarily close to 1 but
do not achieve it).
This does not violate EVT
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because the domain is 1
not closed.
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Notes
Final Bad Example
Example
1
The func on f(x) = is con nuous on the closed interval [1, ∞).
x
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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.
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6. . V63.0121.001: Calculus I
. Sec on 4.1: Max/Min .Values April 4, 2011
Notes
Outline
Introduc 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
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. |. |
local local b
a
Similarly, f has a local minimum maximum minimum
at c if f(c) ≤ f(x) when x is near
c.
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Notes
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 b
local local and global
maximum global max
min
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7. . V63.0121.001: Calculus I
. Sec on 4.1: Max/Min .Values April 4, 2011
Notes
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
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Notes
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) 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) 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
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Notes
Meet the Mathematician: Pierre de Fermat
1601–1665
Lawyer and number
theorist
Proved many theorems,
didn’t quite prove his last
one
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8. . V63.0121.001: Calculus I
. Sec on 4.1: Max/Min .Values April 4, 2011
Notes
Outline
Introduc 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 extrema Notes
Thanks to Fermat
Suppose f is a c is a
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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
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Notes
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.
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9. . V63.0121.001: Calculus I
. Sec on 4.1: Max/Min .Values April 4, 2011
Notes
Outline
Introduc 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
Extreme values of a linear function
Example
Find the extreme values of f(x) = 2x − 5 on [−1, 2].
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.
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Extreme values of a quadratic Notes
function
Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Solu on
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)
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10. . V63.0121.001: Calculus I
. Sec on 4.1: Max/Min .Values April 4, 2011
Notes
Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solu on
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Notes
Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solu on
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Notes
Extreme values of another algebraic function
Example
√
Find the extreme values of f(x) = 4 − x2 on [−2, 1].
Solu on
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11. . V63.0121.001: Calculus I
. Sec on 4.1: Max/Min .Values April 4, 2011
Notes
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
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Notes
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Notes
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