Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - 4.2 Extreme Values - Exercises - Page 181: 8


1, 0 and $\frac{1}{2}$

Work Step by Step

$g'(z)=\frac{-1}{(z-1)^{2}}-\frac{-1}{z^{2}}$ $=\frac{1}{z^{2}}-\frac{1}{(z-1)^{2}}=\frac{1-2z}{(z-1)^{2}z^{2}}$ $g'(z)=0\implies 1-2z=0\implies 1=2z\implies z=\frac{1}{2}$ $g'(z)$ does not exist when $z=1$ or when $z=0$ as the denominator vanishes in both of the cases. The critical points of $g$ are those $z$ values for which $g'(z)=0$ or $g'(z)$ does not exist. Therefore, the critical points are 1, 0 and $\frac{1}{2}$.
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