EXAMPLE 8 Find the absolute maximum and minimum values of the function below. F(x) = x³ - 3x² + 2 (- 4), we can use the Closed Interval Method: SOLUTION Since fis continuous on f(x) = x³ - 3x2 + 2 F'(x) = |, that is, x = 0 or x =| Since f'(x) exists for all x, the only critical numbers of f occur when f'(x) = [ numbers lies in (- 4). The values of f at these critical numbers are Notice that each of these critical (0) = | and f(2) = | The values of f at the endpoints of the interval are and f(4) = Comparing these four numbere we see that the ahsolute mavimum value is ff4) and the ahsolute minimum value is f2)

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
Chapter5: A Survey Of Other Common Functions
Section5.4: Combining And Decomposing Functions
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EXAMPLE 8
Find the absolute maximum and minimum values of the function below.
f(x) = x3 - 3x2 + 2
15
SOLUTION Since f is continuous on
we can use the Closed Interval Method:
10
f(x) = x3 - 3x² + 2
f'(x) =
5
that is, x = o or x =
3
4
6
Since f'(x) exists for all x, the only critical numbers off occur when f'(x) =
Notice that each of these critical
numbers lies in
(- 4). 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
(-) -C
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 fis 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 15 SOLUTION Since f is continuous on we can use the Closed Interval Method: 10 f(x) = x3 - 3x² + 2 f'(x) = 5 that is, x = o or x = 3 4 6 Since f'(x) exists for all x, the only critical numbers off occur when f'(x) = Notice that each of these critical numbers lies in (- 4). 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 (-) -C 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 fis sketched in the figure.
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