Answer
$f\displaystyle \circ g(x)==\frac{2x-1}{2x} \qquad $ domain: $(-\infty,0)\cup(0,\infty)$
$ g\displaystyle \circ f(x)=\frac{x-1}{x+1}; \quad$ domain: $(-\infty,-1)\cup(-1,\infty)$
$ f\displaystyle \circ f(x)=\frac{x}{2x+1}; \quad$ domain: $(-\displaystyle \infty,-1)\cup(-1,-\frac{1}{2})\cup(-\frac{1}{2},\infty)$
$ g\circ g(x)=4x-3; \quad$ domain: $(-\infty,\infty)$
Work Step by Step
f(x) is defined for all x except $-1$
g(x) is defined for all x
$ f\displaystyle \circ g(x)=f[g(x)]= \frac{g(x)}{g(x)+1}\quad$on the domain of $g$(x), and $g(x)+1\neq 0$
$=\displaystyle \frac{2x-1}{2x-1+1},\quad$for$ \left\{\begin{array}{l}
2x-1+1\neq 0\\
x\neq 0
\end{array}\right\}$
$=\displaystyle \frac{2x-1}{2x} \qquad $ domain: $(-\infty,0)\cup(0,\infty)$
$g\circ f(x)=g[f(x)]=2f(x)-1 \quad$on the domain of f(x),
$=2(\displaystyle \frac{x}{x+1})-1,\quad$for $x\neq-1$
$=\displaystyle \frac{2x-(x+1)}{x+1}$
$=\displaystyle \frac{x-1}{x+1}; \quad$ domain: $(-\infty,-1)\cup(-1,\infty)$
$ f\displaystyle \circ f(x)=f[f(x)]=\frac{f(x)}{f(x)+1},\qquad$on the domain of f(x), and $f(x)+1\neq 0$
$=\displaystyle \frac{\frac{x}{x+1}}{\frac{x}{x+1}+1},\quad $ for $\left\{\begin{array}{ll}
x\neq-1 & \frac{x}{x+1}+1\neq 0\\
& \frac{x+x+1}{x+1}\neq 0\\
& 2x+1\neq 0\\
& x\neq-1/2
\end{array}\right\}$
$=\displaystyle \frac{\frac{x}{x+1}}{\frac{x}{x+1}+1}\times\frac{x+1}{x+1}$
$=\displaystyle \frac{x}{x+x+1}$
$=\displaystyle \frac{x}{2x+1}; \quad$ domain: $(-\displaystyle \infty,-1)\cup(-1,-\frac{1}{2})\cup(-\frac{1}{2},\infty)$
$ g\circ g(x)=g[g(x)]=2g(x)-1 \quad$on the domain of $g$(x)
$=2(2x-1)-1 \quad$on $(-\infty,\infty)$
$=4x-3; \quad$ domain: $(-\infty,\infty)$