Answer
For $x\in(-\infty,4]$ the inverse and its domain are:
$f^{-1}(x)=-\sqrt x+4$
$D_{f^{-1}}=[0,\infty)$
For $x\in[4,\infty)$ the inverse and its domain are:
$f^{-1}(x)=\sqrt x+4$
$D_{f^{-1}}=[0,\infty)$
Work Step by Step
We are given the function:
$f(x)=(x-4)^2$
The domain of and range of $f$ are:
$D_f=(-\infty,\infty)$
$R_f=[0,\infty)$
Determine the inverse $f^{-1}$:
$y=(x-4)^2$
$x=(y-4)^2$ (we switched $x$ and $y$)
$\sqrt x=\pm (y-4)$
$\sqrt x=-(y-4)\Rightarrow \sqrt x=-y+4\Rightarrow y=-\sqrt x+4$
$\sqrt x=y-4\Rightarrow y=\sqrt x+4$
There are two inverses:
For $x\in(-\infty,4]$ the inverse and its domain are:
$f^{-1}(x)=-\sqrt x+4$
$D_{f^{-1}}=[0,\infty)$
For $x\in[4,\infty)$ the inverse and its domain are:
$f^{-1}(x)=\sqrt x+4$
$D_{f^{-1}}=[0,\infty)$
