EXAMPLE 8 Find the absolute maximum and minimum values of the function below. f(x) = x³ - 3x² + 2 15 SOLUTION Since fis continuous on we can use the Closed Interval Method: 10아 f(x) = x³ - 3x² + 2 f'(x) = 3 4 Since f'(x) exists for all x, the only critical numbers of f occur when f'(x) =| 5 6 that is, x = 0 or Notice that each of these critical numbers lies in The values of f at these critical numbers are f(0) = and f(2) = The values of f at the endpoints of the interval are and f(4) = | Comparing these four numbers, we see that the absolute maximum value is f(4) = and the absolute minimum value is f(2) = Note that in this example the absolute maximum occurs at an endpoint, whereas the absolute minimum occurs at a critical number. The graph of f is sketched in the figure.

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter1: Functions
Section1.2: Functions Given By Tables
Problem 32SBE: Does a Limiting Value Occur? A rocket ship is flying away from Earth at a constant velocity, and it...
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EXAMPLE 8
Find the absolute maximum and minimum values of the function below.
f(x) = x3 - 3x2 + 2
< x< 4
15
SOLUTION
Since fis continuous on
we can use the Closed Interval Method:
10
f(x) = x3 - 3x2 + 2
5
f'(x) =
X
3
5
Since f'(x) exists for all x, the only critical numbers off occur when f'(x) =
-1
1
4
6
that is, x = 0 or
(-글).
Notice that each of these critical numbers lies in
The values of f at these critical
numbers are
f(0) =
and
f(2) =
The values of f at the endpoints of the interval are
and f(4) =
Comparing these four numbers, we see that the absolute maximum value is f(4) =
and the absolute
minimum value is f(2) =
Note that in this example the absolute maximum occurs at an endpoint, whereas the absolute minimum
occurs at a critical number. The graph of f is sketched in the figure.
Transcribed Image Text:EXAMPLE 8 Find the absolute maximum and minimum values of the function below. f(x) = x3 - 3x2 + 2 < x< 4 15 SOLUTION Since fis continuous on we can use the Closed Interval Method: 10 f(x) = x3 - 3x2 + 2 5 f'(x) = X 3 5 Since f'(x) exists for all x, the only critical numbers off occur when f'(x) = -1 1 4 6 that is, x = 0 or (-글). Notice that each of these critical numbers lies in The values of f at these critical numbers are f(0) = and f(2) = The values of f at the endpoints of the interval are and f(4) = Comparing these four numbers, we see that the absolute maximum value is f(4) = and the absolute minimum value is f(2) = Note that in this example the absolute maximum occurs at an endpoint, whereas the absolute minimum occurs at a critical number. The graph of f is sketched in the figure.
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