## Calculus 8th Edition

(a) $7$ (b)$\frac{1}{8}$
(a) Since $f(7)=3$ , then $f^{-1}(3)=7$ (b) since \begin{align*} (f^{-1})'(3)&=\frac{1}{f'(f^-1(3))}\\ &=\frac{1}{f'(7)}\\ &=\frac{1}{8} \end{align*}