Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.1 Inverse Functions - 6.1 Exercises - Page 407: 46



Work Step by Step

We are given $G(x)=\frac{1}{f^{-1}(x)}$, so $G'(x)=\frac{-(f^{-1})'(x)}{f^{-1}(x)^2}$. This implies that:$G'(2)=\frac{-(f^{-1})'(2)}{f^{-1}(2)^2}$. We are given $f(3)=2$, so $f^{-1}(2)=3$. By theorem 7, $(f^{-1})'(2)=\frac{1}{f'(f^{-1}(2))}=\frac{1}{f'(3)}=9$. Now substitute: $G'(2)=\frac{-9}{3^2}=-1$.
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