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: 45


$(f^{-1})'(0)=\frac{1}{\sqrt 28}$

Work Step by Step

Let $x=3$, then $f(3)=\int\sqrt {1+t^3}dt$=0, so $f^{-1}(0)=3$. Furthermore, $f'(x)=\sqrt{1+t^3}$. Now, by theorem 7, $(f^{-1})'(0)=\frac{1}{f'(f^{-1}(0))}=\frac{1}{f'(3)}=\frac{1}{\sqrt {28}}.$
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