Single Variable Calculus, Chapter 7, 7.1, Section 7.1, Problem 28

Find a formula for the inverse of the function $f(x) = 2x^2 - 8x \quad x \geq 2$
If $f(x) = 2x^2 - 8x$, then

$
\begin{equation}
\begin{aligned}
f^{-1}(x) \quad \Longrightarrow \quad x &= 2y^2 - 8y\\
\\
\frac{x}{2} &= y^2 - 4y\\
\\
\text{by completing the square,}\\
\\
\frac{x}{2} + 4 &= y^2 - 4y + y\\
\\
\frac{x}{2} + 4 &= (y - 2)^2\\
\\
\pm \sqrt{\frac{x}{2}+4} &= y - 2\\
\\
y &= 2 \pm \sqrt{\frac{x}{2} + 4}
\end{aligned}
\end{equation}
$

We got two values of $y$, however, the function is defined only for $x \geq 2$. The domain of $f$ is $x \geq 2$ and its range $y \geq 2(2)^2 -8(2) \quad \Longleftarrow \quad -8$. Thus the domain of $f^{-1}(x)$ is $x \geq -8$ and range $y \geq 2$. If we check both $y$,

when $x = -6$,

$
\begin{equation}
\begin{aligned}
y &= 2 + \sqrt{\frac{-6}{2} + 4}\\
\\
y &= 3
\end{aligned}
\end{equation}
$

when $x = -6$,

$
\begin{equation}
\begin{aligned}
y &= 2 - \sqrt{\frac{-6}{2} + 4}\\
\\
y &= 1
\end{aligned}
\end{equation}
$


Hence, $f^{-1}(x) = 2 + \sqrt{\frac{x}{2}+4}$