Answer
$(f+g)(x) =\displaystyle \frac{x+2}{x+1}$, domain: $(-\infty,-1)\cup(-1,+\infty).$
$(f-g)(x) =\displaystyle \frac{2-x}{x+1}$, domain: $(-\infty,-1)\cup(-1,+\infty).$
$(fg)(x) =\displaystyle \frac{2x}{(x+1)^{2}}$, domain: $(-\infty,-1)\cup(-1,+\infty).$
$(\displaystyle \frac{f}{g})(x) =\frac{2}{x}$ , domain: $(-\infty,-1)\cup(-1,0)\cup(0,+\infty)$
Work Step by Step
The domain of f is $\{x|x\neq-1\}$, because
f is defined only for x that yield a nonzero number in the denominator.
The domain of g is $\{x|x\neq-1\}$, for the same reason.
Their intersection is $\{x|x\neq-1\}= (-\infty,-1)\cup(-1,+\infty)$
$(f+g)(x)=\displaystyle \frac{2}{x+1}+\frac{x}{x+1} =\displaystyle \frac{x+2}{x+1}$, domain: $(-\infty,-1)\cup(-1,+\infty).$
$(f-g)(x)=\displaystyle \frac{2}{x+1}-\frac{x}{x+1} =\displaystyle \frac{2-x}{x+1}$, domain: $(-\infty,-1)\cup(-1,+\infty).$
$(fg)(x)=\displaystyle \frac{2}{x+1}\cdot\frac{x}{x+1} =\displaystyle \frac{2x}{(x+1)^{2}}$, domain: $(-\infty,-1)\cup(-1,+\infty).$
$(\displaystyle \frac{f}{g})(x)=\frac{\frac{2}{x+1}}{\frac{x}{x+1}}=\frac{2(x+1)}{x(x+1)}=\frac{2}{x}$ , domain: $\{x|x\neq-1,x\neq 0\}=(-\infty,-1)\cup(-1,0)\cup(0,+\infty)$
