## Calculus 8th Edition

$(f^{-1})'(5)=\frac{1}{6}$
$f$ is one-to-one because $f'(x)=9x^2+8x+6 >0$ and so $f$ is increasing. To use theorem 7, we need to know $f^{-1}(5)$ and we can find this by inspection: $f(0)=5$ so $f^{-1}(5)=0$ thus by theorem 7, $(f^{-1})'(5)=\frac{1}{f'(f^{-1}(5))}=\frac{1}{f'(0)}=\frac{1}{6}$.