## Precalculus (6th Edition) Blitzer

The inverse function is, ${{f}^{-1}}\left( x \right)=\frac{1}{4}\left( {{x}^{2}}+7 \right)$
Consider the given function, $y=\sqrt{4x-7}$ Square both sides to get, \begin{align} & {{y}^{2}}=4x-7 \\ & x=\frac{1}{4}\left( {{y}^{2}}+7 \right) \end{align} Therefore, ${{f}^{-1}}\left( x \right)=\frac{1}{4}\left( {{x}^{2}}+7 \right)$