Answer
We see that $f[g(x)]=g[f(x)]=x$. This means that $f(x)$ and $g(x)$ are inverses of each other.
Since for $f(x)$, $x\geq2$, then $x\lt2$ are values to be excluded from the domain of $f(x)$.
For $g(x)$, $x\ge 0$, so $x\lt 0$ are values to be excluded from the domain of $g(x)$.
Work Step by Step
We wish to plug $g(x)$ into $f(x)$ to obtain:
$$f(g(x))=f(\sqrt{ x}+2) \\ =((\sqrt{ x}+2)-2)^2 \\ =(\sqrt{ x})^2 \\=x$$
We wish to plug $f(x)$ into $g(x)$ to obtain:
$$ g[f(x)]=g((x-2)^2) \\ =\sqrt{(x-2)^2}+2\\ =x-2+2\\ =x$$
We see that $f[g(x)]=g[f(x)]=x$. This means that $f(x)$ and $g(x)$ are inverses of each other.
Since for $f(x)$, $x\geq2$, then $x\lt2$ are values to be excluded from the domain of $f(x)$.
For $g(x)$, $x\ge 0$, so $x\lt 0$ are values to be excluded from the domain of $g(x)$.
