## Precalculus (6th Edition) Blitzer

The above statement can be verified by applying the identity $\underbrace{{{f}^{-1}}\left( f\left( x \right) \right)}_{\text{LHS}}=\underbrace{x}_{\text{RHS}}$ By solving the left hand side: ${f}^{-1}(f(x))=$ \begin{align} & \left( \frac{f\left( x \right)+4}{5} \right) \\ & \left( \frac{5x-4+4}{5} \right) \\ & \left( \frac{5x}{5} \right) \\ \end{align} $= (x)$ Thus, the left-hand side of the expression is equal to the right hand side of the expression $=x$