## Calculus 8th Edition

$G'(2)=-1$.
We are given $G(x)=\frac{1}{f^{-1}(x)}$, so $G'(x)=\frac{-(f^{-1})'(x)}{f^{-1}(x)^2}$. This implies that:$G'(2)=\frac{-(f^{-1})'(2)}{f^{-1}(2)^2}$. We are given $f(3)=2$, so $f^{-1}(2)=3$. By theorem 7, $(f^{-1})'(2)=\frac{1}{f'(f^{-1}(2))}=\frac{1}{f'(3)}=9$. Now substitute: $G'(2)=\frac{-9}{3^2}=-1$.