problem: If $f(x)=4x^3+1$, find $f^{-1}(33)$.
solution: If we substitute $f^{-1}(x)$ into our expression for $f$ we get  \[f(f^{-1}(x))=4(f^{-1}(x))^3+1.\]This reduces to  \[x=4(f^{-1}(x))^3+1.\]If we solve for $f^{-1}(x)$, we find that $f^{-1}(x)=\sqrt[3]{\frac{x-1}{4}}$. Therefore, $f^{-1}(33)=\sqrt[3]{\frac{33-1}{4}}=\sqrt[3]8=\boxed{2}$.