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: 33

Answer

Over the specified interval, the function has an absolute maximum equal to $1$ and an absolute minimum equal to $-\dfrac{1}{2}.$

Work Step by Step

$f'(x)=\cos{x}.$ $f'(x)=0\to x=\dfrac{\pi}{2}.$ The interval's endpoints and $x=\dfrac{\pi}{2}$ are possible points at which the function can attain absolute extrema. $f(\frac{5}{6}\pi)=\dfrac{1}{2}.$ $f(\frac{1}{2}\pi)=1.$ $f(\frac{11}{6}\pi)=-\dfrac{1}{2}.$ Over the specified interval, the function has an absolute maximum equal to $1$ and an absolute minimum equal to $-\dfrac{1}{2}.$
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