Chapter 5
Graphing and
Optimization
Section 5
Absolute Maxima and
Minima
Objectives for Section 5.5
Absolute Maxima and Minima
■ The student will be able to
identify absolute maxima
and minima.
■ The student will be able to
use the second derivative
test to classify extrema.
Barnett/Ziegler/Byleen Business Calculus 12e
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Extreme Value Theorem
Theorem 1. (Extreme Value Theorem)
A function f that is continuous on a closed interval [a, b] has
both an absolute maximum value and an absolute minimum
value on that interval.
Barnett/Ziegler/Byleen Business Calculus 12e
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Absolute Extrema in a
Closed Interval
Barnett/Ziegler/Byleen Business Calculus 12e
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Absolute Extrema in a Closed
Interval
Theorem 2. Absolute extrema (if they exist) must always
occur at critical values of the derivative, or at end points.
a. Find 𝑓′(𝑥)
b. Find the critical values in the interval [a, b].
c. Plug in the critical values and the end points a and b into 𝑓(𝑥).
d. The absolute maximum on [a, b] is the largest of the values
found in step c. The absolute minimum on [a, b] is the smallest
of the values found in step c. (Note: there can be more than one
absolute maximum/minimum)
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 1
Find the absolute maximum and absolute minimum value of
𝑓(𝑥) on [1, 7].
f ( x)  x 3  6 x 2
𝑓 ′ x = 3𝑥 2 − 12𝑥
0 = 3𝑥(𝑥 − 4)
x = 0, 4
𝑓 4 = −32
0 is not in the interval [1, 7]
𝑓 1 = −5
𝑓 7 = 49
Abs. max. of 49 occurs at x=7.
Abs. min. of -32 occurs at x=4.
Barnett/Ziegler/Byleen Business Calculus 12e
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Absolute Maxima and Minima
in an Open Interval
No absolute min./max.
Local max. at −2, 16
3
−16
Local min. at 2, 3
Absolute max. at (0,4)
Absolute min. at (0,0)
The type of graph will often indicate whether
It has an absolute max or min.
Barnett/Ziegler/Byleen Business Calculus 12e
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Second Derivative Test
Theorem 3. Let f be continuous on an “open” interval I with
only one critical value c.
Examples of “open” intervals: (-, ) or (-, 0) or (a, )
If f (c) = 0 and f (c) > 0, then f (c) is the absolute minimum
of f on I.
If f (c) = 0 and f (c) < 0, then f (c) is the absolute maximum
of f on I.
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 2
 Find the absolute max./min. value of 𝑓 𝑥 = 4𝑥 − 𝑥 2 in
the interval (-∞, ∞).
𝑓 ′ 𝑥 = 4 − 2𝑥
0 = 4 − 2𝑥
𝑥=2
Since there is only 1 critical value, we
can use the 2nd derivative test.
𝑓 ′′ 𝑥 = −2
𝑓 ′′ 2 = −2
𝑓 2 =4
𝑓 ′′ 2 < 0
There is an absolute maximum of 4
when x = 2.
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 3
Find the absolute max/min of 𝑓 𝑥 = 𝑥
4
′
𝑓 𝑥 =1− 2
𝑥
4
0=1− 2
𝑥
4
−1 = − 2
𝑥
2
𝑥 =4
𝑥 = −2, 2
4
+
𝑥
on (0, )
𝑓 ′′ 𝑥 = 8𝑥 −3
𝑓 ′′ 2 = 1
𝑓 2 =4
f (2) > 0
There is an absolute minimum value
of 4 when x = 2.
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 4
1
 Find the absolute max/min 𝑓 𝑥 = 2 in the interval (-∞, ∞)
𝑥 +1
−2𝑥
𝑓′(𝑥) = 2
(𝑥 + 1)2
−2𝑥
0= 2
(𝑥 + 1)2
0 = −2x
x=0
𝑓 ′′
2(3𝑥 2 − 1)
𝑥 =
(𝑥 2 + 1)3
𝑓 ′′ 0 = −2
𝑓 0 =1
Barnett/Ziegler/Byleen Business Calculus 12e
f (0) < 0
There is an absolute maximum
value of 1 when x = 0.
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Homework
Barnett/Ziegler/Byleen Business Calculus 12e
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