Answer
$f^{-1}(x)=x^2+2$
$f(x)$: $\{x|x\ge2 \}$ and $\{y|y\ge0\}$.
$f^{-1}(x)$: $\{x|x\ge0\}$ and $\{y|y\ge2\}$.
Work Step by Step
1. $f(x)=\sqrt {x-2} \Longrightarrow y=\sqrt {x-2} \Longrightarrow x=\sqrt {y-2}\Longrightarrow y=x^2+2 \Longrightarrow f^{-1}(x)=x^2+2$
2. check $f(f^{-1}(x))=\sqrt {(x^2+2)-2}=x, x\ge0$. $f^{-1}(f(x))=(\sqrt {x-2})^2+2=x$.
3. The domain and the range of $f(x)$: $\{x|x\ge2 \}$ and $\{y|y\ge0\}$.
The domain and the range of $f^{-1}(x)$: $\{x|x\ge0\}$ and $\{y|y\ge2\}$.
