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: 44


absolute minimum value =0 absolute maximum value =27

Work Step by Step

Given $f(\theta)=3\theta^{\frac{2}{3}}$ Critical point occurs where first derivative is zero: so${\frac{df(theta)}{d\theta}=\frac{3\theta^{\frac{2}{3}}}{d\theta}}$ ${\frac{df(\theta)}{d\theta}}=\frac{2\times3}{3}\theta^{\frac{-1}{3}}=0$ $\theta=0$ so the critical point is: $f(0)=0$ test end points: $f(-1)=3(-27)^{\frac{2}{3}}=27$ $f(8)=3\times8^{\frac{2}{3}}=12$ absolute minimum value =0 absolute maximum value =27
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