Answer
$f\displaystyle \circ g(x)=\frac{1}{2x+4};\qquad $ domain: $(-\infty,-2)\cup(-2,\infty)$
$ g\displaystyle \circ f(x)=\frac{2}{x}+4; \quad$ domain: $(-\infty,0)\cup(0,\infty)$
$ f\circ f(x)=x; \quad$ domain: $(-\infty,0)\cup(0,\infty)$
$ g\circ g(x)=4\mathrm{x}+12; \quad$ domain: all reals, $(-\infty,\infty)$
Work Step by Step
f(x) is defined for all x, except 0.
g(x) is defined for all x
$f\displaystyle \circ g(x)=f[g(x)]=\frac{1}{g(x)},\qquad g(x)\neq 0$, on the domain of $g$(x)
$=\displaystyle \frac{1}{2x+4},\qquad 2x+4\neq 0\Rightarrow x\neq-2 $
$=\displaystyle \frac{1}{2x+4};\qquad $ domain: $(-\infty,-2)\cup(-2,\infty)$
$g\circ f(x)=g[f(x)]=2f(x)+4$
$=2(\displaystyle \frac{1}{x})+4,\quad x\neq 0 $, on the domain of f(x)
$=\displaystyle \frac{2}{x}+4; \quad$ domain: $(-\infty,0)\cup(0,\infty)$
$f\displaystyle \circ f(x)=f[f(x)]=\frac{1}{f(x)},\qquad f(x)\neq 0$
$=\displaystyle \frac{\frac{1}{x}}{x},\quad\frac{1}{x}\neq 0$ on the domain of f(x)
$=x; \quad$ domain: $(-\infty,0)\cup(0,\infty)$
$g\circ g(x)=g[g(x)]= 2g(x)+4$
$=2(2x+4)+4,\qquad$ no restrictions
$=4\mathrm{x}+12; \quad$ domain: all reals, $(-\infty,\infty)$
