Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.2 Inverse Functions - Exercises - Page 335: 26



Work Step by Step

Since $f(x)=\sqrt{3-x}$, then $f'(x)=-\frac{1}{2\sqrt{3-x}}$. Hence by Theorem 2 and assuming that $g(x)=f^{-1}(x)$, we have $$y=\sqrt{3-x}\Longrightarrow y^2=3-x\Longrightarrow x=3-y^2,$$ Therefore, $$g(x)=3-x^2, \quad g'(x)=\frac{1}{f'(g(x))}\\=-\frac{1}{ (1/(2\sqrt{3-(3-x^2))}) }= -2x.$$
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.