1.
Section 4.1
Maximum and Minimum Values
V63.0121.002.2010Su, Calculus I
New York University
June 8, 2010
Announcements
Exams not graded yet
Assignment 4 is on the website
Quiz 3 on Thursday covering 3.3, 3.4, 3.5, 3.7
2.
Announcements
Exams not graded yet
Assignment 4 is on the
website
Quiz 3 on Thursday covering
3.3, 3.4, 3.5, 3.7
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 2 / 32
3.
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 3 / 32
4.
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 4 / 32
6.
Why go to the extremes?
Rationally speaking, it is
advantageous to ﬁnd the
extreme values of a function
(maximize proﬁt, minimize
costs, etc.)
PierreLouis Maupertuis
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values (1698–1759) 8, 2010
June 5 / 32
7.
Design
Image credit: Jason Tromm
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 6 / 32
8.
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.
PierreLouis Maupertuis
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values (1698–1759) 8, 2010
June 7 / 32
9.
Optics
Image credit: jacreative
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 8 / 32
10.
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.”
PierreLouis Maupertuis
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values (1698–1759) 8, 2010
June 9 / 32
11.
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 10 / 32
12.
Extreme points and values
Deﬁnition
Let f have domain D.
Image credit: Patrick Q
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 11 / 32
13.
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
Image credit: Patrick Q
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 11 / 32
14.
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.
Image credit: Patrick Q
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 11 / 32
15.
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
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 11 / 32
16.
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 absolute minimum
value f (d) at numbers c and d in [a, b].
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 12 / 32
17.
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 absolute minimum
value f (d) at numbers c and d in [a, b].
a b
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 12 / 32
18.
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 absolute minimum
value f (d) at numbers c and d in [a, b].
maximum f (c)
value
minimum f (d)
value
a d c
b
minimum maximum
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 12 / 32
19.
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 13 / 32
20.
Bad Example #1
Example
Consider the function
x 0≤x <1
f (x) =
x − 2 1 ≤ x ≤ 2.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 14 / 32
21.
Bad Example #1
Example
Consider the function
x 0≤x <1
f (x) = 
x − 2 1 ≤ x ≤ 2. 1
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 14 / 32
22.
Bad Example #1
Example
Consider the function
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 14 / 32
23.
Bad Example #1
Example
Consider the function
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 continuous.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 14 / 32
24.
Bad Example #2
Example
Consider the function f (x) = x restricted to the interval [0, 1).
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 15 / 32
25.
Bad Example #2
Example
Consider the function f (x) = x restricted to the interval [0, 1).

1
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 15 / 32
26.
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).
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 15 / 32
27.
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 15 / 32
28.
Final Bad Example
Example
1
Consider the function f (x) = is continuous on the closed interval [1, ∞).
x
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 16 / 32
29.
Final Bad Example
Example
1
Consider the function f (x) = is continuous on the closed interval [1, ∞).
x
1
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 16 / 32
30.
Final Bad Example
Example
1
Consider the function f (x) = is continuous on the closed interval [1, ∞).
x
1
There is no minimum value (values get arbitrarily close to 0 but do not
achieve it).
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 16 / 32
31.
Final Bad Example
Example
1
Consider the function f (x) = is continuous on the closed interval [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.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 16 / 32
32.
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 17 / 32
33.
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 18 / 32
34.
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 local local b
maximum minimum
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 18 / 32
35.
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 local local and global b global
maximum min max
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 18 / 32
36.
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 local local b
maximum minimum
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 19 / 32
37.
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 local local b
maximum minimum
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 19 / 32
38.
Sketch of proof of Fermat’s Theorem
Suppose that f has a local maximum at c.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 20 / 32
39.
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)
≤0
x −c
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 20 / 32
40.
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) f (x) − f (c)
≤ 0 =⇒ lim+ ≤0
x −c x→c x −c
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 20 / 32
41.
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) 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
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 20 / 32
42.
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) 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
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 20 / 32
43.
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) 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
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 20 / 32
44.
Meet the Mathematician: Pierre de Fermat
1601–1665
Lawyer and number theorist
Proved many theorems,
didn’t quite prove his last
one
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 21 / 32
45.
Tangent: Fermat’s Last Theorem
Plenty of solutions to
x 2 + y 2 = z 2 among positive
whole numbers (e.g., x = 3,
y = 4, z = 5)
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 22 / 32
46.
Tangent: Fermat’s Last Theorem
Plenty of solutions to
x 2 + y 2 = z 2 among positive
whole numbers (e.g., x = 3,
y = 4, z = 5)
No solutions to
x 3 + y 3 = z 3 among positive
whole numbers
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 22 / 32
47.
Tangent: Fermat’s Last Theorem
Plenty of solutions to
x 2 + y 2 = z 2 among positive
whole numbers (e.g., x = 3,
y = 4, z = 5)
No solutions to
x 3 + y 3 = z 3 among positive
whole numbers
Fermat claimed no solutions
to x n + y n = z n but didn’t
write down his proof
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 22 / 32
48.
Tangent: Fermat’s Last Theorem
Plenty of solutions to
x 2 + y 2 = z 2 among positive
whole numbers (e.g., x = 3,
y = 4, z = 5)
No solutions to
x 3 + y 3 = z 3 among positive
whole numbers
Fermat claimed no solutions
to x n + y n = z n but didn’t
write down his proof
Not solved until 1998!
(Taylor–Wiles)
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 22 / 32
49.
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 23 / 32
50.
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.
c is a
start
local max
Is f
Is c an f is not
no diﬀ’ble at no
endpoint? diﬀ at c
c?
yes yes
c = a or
f (c) = 0
c = b
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 24 / 32
51.
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 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.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 25 / 32
52.
Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 26 / 32
53.
Extreme values of a linear function
Example
Find the extreme values of f (x) = 2x − 5 on [−1, 2].
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 27 / 32
54.
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
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 27 / 32
55.
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 27 / 32
56.
Extreme values of a quadratic function
Example
Find the extreme values of f (x) = x 2 − 1 on [−1, 2].
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
57.
Extreme values of a quadratic function
Example
Find the extreme values of f (x) = x 2 − 1 on [−1, 2].
Solution
We have f (x) = 2x, which is zero when x = 0.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
58.
Extreme values of a quadratic function
Example
Find the extreme values of f (x) = x 2 − 1 on [−1, 2].
Solution
We have f (x) = 2x, which is zero when x = 0. So our points to check
are:
f (−1) =
f (0) =
f (2) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
59.
Extreme values of a quadratic function
Example
Find the extreme values of f (x) = x 2 − 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) =
f (2) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
60.
Extreme values of a quadratic function
Example
Find the extreme values of f (x) = x 2 − 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
f (2) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
61.
Extreme values of a quadratic function
Example
Find the extreme values of f (x) = x 2 − 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
f (2) = 3
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
62.
Extreme values of a quadratic function
Example
Find the extreme values of f (x) = x 2 − 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
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
63.
Extreme values of a quadratic function
Example
Find the extreme values of f (x) = x 2 − 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)
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 28 / 32
64.
Extreme values of a cubic function
Example
Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2].
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
65.
Extreme values of a cubic function
Example
Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2].
Solution
Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and
x = 1.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
66.
Extreme values of a cubic function
Example
Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2].
Solution
Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and
x = 1. The values to check are
f (−1) =
f (0) =
f (1) =
f (2) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
67.
Extreme values of a cubic function
Example
Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2].
Solution
Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and
x = 1. The values to check are
f (−1) = − 4
f (0) =
f (1) =
f (2) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
68.
Extreme values of a cubic function
Example
Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2].
Solution
Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and
x = 1. The values to check are
f (−1) = − 4
f (0) = 1
f (1) =
f (2) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
69.
Extreme values of a cubic function
Example
Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2].
Solution
Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and
x = 1. The values to check are
f (−1) = − 4
f (0) = 1
f (1) = 0
f (2) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
70.
Extreme values of a cubic function
Example
Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2].
Solution
Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and
x = 1. The values to check are
f (−1) = − 4
f (0) = 1
f (1) = 0
f (2) = 5
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
71.
Extreme values of a cubic function
Example
Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2].
Solution
Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and
x = 1. The values to check are
f (−1) = − 4 (global min)
f (0) = 1
f (1) = 0
f (2) = 5
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
72.
Extreme values of a cubic function
Example
Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2].
Solution
Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and
x = 1. The values to check are
f (−1) = − 4 (global min)
f (0) = 1
f (1) = 0
f (2) = 5 (global max)
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
73.
Extreme values of a cubic function
Example
Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2].
Solution
Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and
x = 1. The values to check are
f (−1) = − 4 (global min)
f (0) = 1 (local max)
f (1) = 0
f (2) = 5 (global max)
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
74.
Extreme values of a cubic function
Example
Find the extreme values of f (x) = 2x 3 − 3x 2 + 1 on [−1, 2].
Solution
Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and
x = 1. The values to check are
f (−1) = − 4 (global min)
f (0) = 1 (local max)
f (1) = 0 (local min)
f (2) = 5 (global max)
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 29 / 32
75.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2].
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
76.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2].
Solution
Write f (x) = x 5/3 + 2x 2/3 , then
5 4 1
f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4)
3 3 3
Thus f (−4/5) = 0 and f is not diﬀerentiable at 0.
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
77.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2].
Solution
Write f (x) = x 5/3 + 2x 2/3 , then
5 4 1
f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4)
3 3 3
Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check
are:
f (−1) =
f (−4/5) =
f (0) =
f (2) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
78.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2].
Solution
Write f (x) = x 5/3 + 2x 2/3 , then
5 4 1
f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4)
3 3 3
Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check
are:
f (−1) = 1
f (−4/5) =
f (0) =
f (2) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
79.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2].
Solution
Write f (x) = x 5/3 + 2x 2/3 , then
5 4 1
f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4)
3 3 3
Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check
are:
f (−1) = 1
f (−4/5) = 1.0341
f (0) =
f (2) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
80.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2].
Solution
Write f (x) = x 5/3 + 2x 2/3 , then
5 4 1
f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4)
3 3 3
Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check
are:
f (−1) = 1
f (−4/5) = 1.0341
f (0) = 0
f (2) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
81.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2].
Solution
Write f (x) = x 5/3 + 2x 2/3 , then
5 4 1
f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4)
3 3 3
Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check
are:
f (−1) = 1
f (−4/5) = 1.0341
f (0) = 0
f (2) = 6.3496
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
82.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2].
Solution
Write f (x) = x 5/3 + 2x 2/3 , then
5 4 1
f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4)
3 3 3
Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check
are:
f (−1) = 1
f (−4/5) = 1.0341
f (0) = 0 (absolute min)
f (2) = 6.3496
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
83.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2].
Solution
Write f (x) = x 5/3 + 2x 2/3 , then
5 4 1
f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4)
3 3 3
Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check
are:
f (−1) = 1
f (−4/5) = 1.0341
f (0) = 0 (absolute min)
f (2) = 6.3496 (absolute max)
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
84.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = x 2/3 (x + 2) on [−1, 2].
Solution
Write f (x) = x 5/3 + 2x 2/3 , then
5 4 1
f (x) = x 2/3 + x −1/3 = x −1/3 (5x + 4)
3 3 3
Thus f (−4/5) = 0 and f is not diﬀerentiable at 0. So our points to check
are:
f (−1) = 1
f (−4/5) = 1.0341 (relative max)
f (0) = 0 (absolute min)
f (2) = 6.3496 (absolute max)
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 30 / 32
85.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = 4 − x 2 on [−2, 1].
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
86.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = 4 − x 2 on [−2, 1].
Solution
x
We have f (x) = − √ , which is zero when x = 0. (f is not
4 − x2
diﬀerentiable at ±2 as well.)
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
87.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = 4 − x 2 on [−2, 1].
Solution
x
We have f (x) = − √ , which is zero when x = 0. (f is not
4 − x2
diﬀerentiable at ±2 as well.) So our points to check are:
f (−2) =
f (0) =
f (1) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
88.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = 4 − x 2 on [−2, 1].
Solution
x
We have f (x) = − √ , which is zero when x = 0. (f is not
4 − x2
diﬀerentiable at ±2 as well.) So our points to check are:
f (−2) = 0
f (0) =
f (1) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
89.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = 4 − x 2 on [−2, 1].
Solution
x
We have f (x) = − √ , which is zero when x = 0. (f is not
4 − x2
diﬀerentiable at ±2 as well.) So our points to check are:
f (−2) = 0
f (0) = 2
f (1) =
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
90.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = 4 − x 2 on [−2, 1].
Solution
x
We have f (x) = − √ , which is zero when x = 0. (f is not
4 − x2
diﬀerentiable at ±2 as well.) So our points to check are:
f (−2) = 0
f (0) = 2
√
f (1) = 3
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
91.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = 4 − x 2 on [−2, 1].
Solution
x
We have f (x) = − √ , which is zero when x = 0. (f is not
4 − x2
diﬀerentiable at ±2 as well.) So our points to check are:
f (−2) = 0 (absolute min)
f (0) = 2
√
f (1) = 3
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
92.
Extreme values of an algebraic function
Example
Find the extreme values of f (x) = 4 − x 2 on [−2, 1].
Solution
x
We have f (x) = − √ , which is zero when x = 0. (f is not
4 − x2
diﬀerentiable at ±2 as well.) So our points to check are:
f (−2) = 0 (absolute min)
f (0) = 2 (absolute max)
√
f (1) = 3
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 31 / 32
93.
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
Show your work unless you want to end up like Fermat!
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 32 / 32
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