Find the absolute maximum and minimum values of the function below. -s*s 12 [-, 12], we can use the Closed Interval Method: f(x) = x³ - 9x? + 2 SOLUTION Since f is continuous on (x) = x3 - 9x² + 2 f'(x) = Since f(x) exists for all x, the only critical numbers of f occur when f'(x) = [ numbers lies in (-, 12). The values of f at these critical numbers are that is, x = 0 or x = ( Notice that each of these critical r(0) = and f(6) = C The values of f at the endpoints of the interval are (-글)-□ and (12) =| Comparing these four numbers, we see that the absolute maximum value is f(12) = O and the absolute minimum value is f(6) = 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

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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EXAMPLE 8
Find the absolute maximum and minimum values of the function below.
400
f(x) = x3 - 9x2 + 2
<x < 12
-
[- 3,12,
300
SOLUTION
Since f is continuous on
we can use the Closed Interval Method:
200
f(x)
x3 – 9x2 + 2
100
f'(x)
5
10
Since f'(x) exists for all x, the only critical numbers of f occur when f'(x)
that is, x = 0 or x =
Notice that each of these critical
-100|
numbers lies in
2, 12). The values of f at these critical numbers are
f(0)
and f(6) =
%3D
%3D
The values of f at the endpoints of the interval are
(-) -O
and
f(12)
%3D
Comparing these four numbers, we see that the absolute maximum value is f(12)
and the absolute minimum value is f(6) =
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:y EXAMPLE 8 Find the absolute maximum and minimum values of the function below. 400 f(x) = x3 - 9x2 + 2 <x < 12 - [- 3,12, 300 SOLUTION Since f is continuous on we can use the Closed Interval Method: 200 f(x) x3 – 9x2 + 2 100 f'(x) 5 10 Since f'(x) exists for all x, the only critical numbers of f occur when f'(x) that is, x = 0 or x = Notice that each of these critical -100| numbers lies in 2, 12). The values of f at these critical numbers are f(0) and f(6) = %3D %3D The values of f at the endpoints of the interval are (-) -O and f(12) %3D Comparing these four numbers, we see that the absolute maximum value is f(12) and the absolute minimum value is f(6) = 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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