## Calculus 8th Edition

The inverse function is $$f^{-1}(x)=2+\frac{\sqrt{16+2x}}{2}.$$
The function $f(x)=2x^2-8x=2x(x-4)$ is a quadratic function with zeroes $x_1=0$ and $x_2=4$. Knowing that the quadratic function achieves its minimum in the midpoint between its zeroes which is here $x_m=2$ (the value of the minimum is $y_m=2\cdot2^2-8\cdot2=-8$), we see that in its domain it is an increasing function (because it 'passed' its only minimum). Now denote $y=2x^2-8x.$ To find the inverse function we have to express $x$ in terms of $y$ and then $x=f^{-1}(y)$. $$y=2x^2-8x\Rightarrow 2x^2-8x-y=0.$$ This is a quadratic equation whose solutions are given by $$x=\frac{8\pm\sqrt{(-8)^2-4\cdot2\cdot(-y)}}{2\cdot2}=\frac{8\pm\sqrt{64+8y}}{4}=2\pm\frac{\sqrt{16+2y}}{2}.$$ Because we need that $x\geq2$ then we have to take the '+' sign i.e. $$x=2+\frac{\sqrt{16+2y}}{2}$$ giving $$f^{-1}(y)=2+\frac{\sqrt{16+2y}}{2}.$$ Renaming $y$ back to $x$ we get $$f^{-1}(x)=2+\frac{\sqrt{16+2x}}{2}$$