Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 3 - Applications of Differentiation - 3.1 Exercises: 21

Answer

Over the specified interval, the function has an absolute maximum equal to $2$ and an absolute minimum equal to $-\frac{5}{2}.$

Work Step by Step

$f'(x)=3x^2-3x=3x(x-1).$ $f'(x)$ is defined for all x in the interval. $f'(x)=0\to x=0$ or $x=1.$ Since both are in the specified interval, they, along with the endpoints of the interval, are possible candidates for absolute extrema. $f(-1)=-\frac{5}{2}.$ $f(0)=0.$ $f(1)=-\frac{1}{2}.$ $f(2)=2.$ Over the specified interval, the function has an absolute maximum equal to $2$ and an absolute minimum equal to $-\frac{5}{2}.$
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