Answer
$f\circ g(x)=x-3;\qquad $ domain: $[3,\infty)$
$ g\circ f(x)=\sqrt{x^{2}-3}; \quad$ domain: $(-\infty,-\sqrt{3})\cup(\sqrt{3},\infty)$
$ f\circ f(x)=x^{4}; \quad$ domain: $(-\infty,\infty)$
$ g\circ g(x)==\sqrt{\sqrt{x-3}-3}; \quad$ domain: all reals, $[12,\infty)$
Work Step by Step
f(x) is defined for all x,
g(x) is defined on $[3,\infty)$ , because the radicand can not be negative.
$ f\circ g(x)=f[g(x)]= [g(x)]^{2}\quad$on the domain of $g$(x)
$=[\sqrt{x-3}]^{2}\quad$ on $[3,\infty)$
$=x-3;\qquad $ domain: $[3,\infty)$
$g\circ f(x)=g[f(x)]=\sqrt{f(x)-3} $on the domain of f(x),\ and $f(x)\geq 3$
$=\sqrt{x^{2}-3},\quad x^{2}\geq 3\Rightarrow x\leq-\sqrt{3}$ or $x\geq\sqrt{3}$
$=\sqrt{x^{2}-3}; \quad$ domain: $(-\infty,-\sqrt{3})\cup(\sqrt{3},\infty)$
$ f\circ f(x)=f[f(x)]=[f(x)]^{2},\qquad$on the domain of f(x)
$=(x^{2})^{2},\quad $ on $(-\infty,\infty)$
$=x^{4}; \quad$ domain: $(-\infty,\infty)$
$g\circ g(x)=g[g(x)]=\sqrt{g(x)-3} \quad$on the domain of $g$(x), and $g(x)\geq 3$
$=\sqrt{\sqrt{x-3}-3},\ \qquad\left\{\begin{array}{ll}
\sqrt{x-3}\geq 3 & /(..)^{2}\\
x-3\geq 9 & \\
x\geq 12 &
\end{array}\right\}$ so the domain is $[12,\infty)$