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: 40



Work Step by Step

$f$ is one-to-one because $f'(x)=3x^2+3cosx-2sinx >0$ and so $f$ is increasing. To use theorem 7, we need to know $f^{-1}(2)$ and we can find this by inspection: $f(0)=2$ so $f^{-1}(2)=0$ thus by theorem 7, $(f^{-1})'(2)=\frac{1}{f'(f^{-1}(2))}=\frac{1}{f'(0)}=\frac{1}{3}$.
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