We actually don't need to do any computation. By definition, the inverse function [tex] f^{-1} [/tex] changes the role of input and output. So, if a function f maps x onto y, the inverse function maps y onto x.
You have to think like this: if the function makes a step further, the inverse function makes that same step back.
This means that the composition [tex] f \circ f^{-1} [/tex] is always the identity function [tex] f(x) = x [/tex]. In fact,
[tex] (f\circ f^{-1})(x) = f(f^{-1}(x)) = x [/tex]
So, for every function, you have
[tex] (f\circ f^{-1})(4) = f(f^{-1}(4)) = 4 [/tex]
