University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.1 - Extreme Values of Functions - Exercises - Page 216: 43


absolute minimum value =-8 absolute maximum value =1

Work Step by Step

Given $f(\theta)=\theta^{\frac{3}{5}}$ Critical point occurs where first derivative is zero: so${\frac{df(theta)}{dx}=\frac{\theta^{\frac{3}{5}}}{d\theta}}$ ${\frac{df(\theta)}{dx}}=\frac{3}{5}\theta^{\frac{-2}{5}}$ so the critical point is: $f(0)=0$ test end points: $f(-1)=(-32)^{\frac{3}{5}}=-8$ $f(8)=1^{\frac{3}{5}}=1$ absolute minimum value =-8 absolute maximum value =1
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