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: 42



Work Step by Step

$f$ is one-to-one because $f'(x)=\frac{1}{2}(x^3+4x+4)^{-\frac{1}{2}}(3x^2+4) >0$ and so $f$ is increasing. To use theorem 7, we need to know $f^{-1}(3)$ and we can find this by inspection: $f(1)=3$ so $f^{-1}(3)=1$ thus by theorem 7, $(f^{-1})'(3)=\frac{1}{f'(f^{-1}(3))}=\frac{1}{f'(1)}=\frac{1}{\frac{7}{6}}=\frac{6}{7}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.