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: 44



Work Step by Step

Since $f(2)=8$, this implies $g^{-1}(8)=2$ and so, by theorem 7, $(g^{-1})'(8)=\frac{1}{g'(g^{-1}(8))}=\frac{1}{g'(2)}=\frac{1}{5}.$
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