Answer
$(f^{-1})'(3)=\frac{6}{7}$
Work Step by Step
$f$ is one-to-one because $f'(x)=\frac{1}{2}(x^3+4x+4)^{-\frac{1}{2}}(3x^2+4) >0$ and so $f$ is increasing.
To use theorem 7, we need to know $f^{-1}(3)$ and we can find this by inspection:
$f(1)=3$ so $f^{-1}(3)=1$ thus by theorem 7, $(f^{-1})'(3)=\frac{1}{f'(f^{-1}(3))}=\frac{1}{f'(1)}=\frac{1}{\frac{7}{6}}=\frac{6}{7}$.
