problem: If $f(x) = \frac{3x+2}{5}$ what is the value of $\left[f^{-1}(4)\right]^{-1}$?
solution: Substituting $f^{-1}(x)$ into our expression for $f$, we find \[f(f^{-1}(x))=\frac{3f^{-1}(x)+2}{5}.\]Since $f(f^{-1}(x))=x$ for all $x$ in the domain of $f^{-1}$, we have \[x=\frac{3f^{-1}(x)+2}{5}.\]Solving for $f^{-1}(x)$, we obtain $f^{-1}(x) = \frac{5x-2}{3}$.  In particular, $f^{-1}(4) = \frac{5 \cdot 4 - 2}{3} = 18/3=6$, so $[f^{-1}(4)]^{-1} = \boxed{\frac16}$.