Answer
$(f+g)(x)=\displaystyle \frac{2}{x}+\frac{4}{x+4}$ , domain: $\{x|x\neq-4,x\neq 0\}.$
$(f-g)(x)=\displaystyle \frac{2}{x}-\frac{4}{x+4}$ , domain: $\{x|x\neq-4,x\neq 0\}.$
$(fg)(x)=\displaystyle \frac{2}{x}\cdot\frac{4}{x+4}$ , domain: $\{x|x\neq-4,x\neq 0\}.$
$(\displaystyle \frac{f}{g})(x)=\frac{x+4}{2x}$ , domain: $\{x|x\neq-4,x\neq 0\}.$
In interval notation, all domains are
$(-\infty,-4)\cup(-4,0)\cup(0,+\infty)$
Work Step by Step
The domain of f is $\{x|x\neq 0\}$, because
f is defined only for x that yield a nonzero numberin the denominator.
The domain of g is $\{x|x\neq-4\}$, for the same reason.
Their intersection is $\{x|x\neq-4,x\neq 0\}.$
$(f+g)(x)=\displaystyle \frac{2}{x}+\frac{4}{x+4}$ , domain: $\{x|x\neq-4,x\neq 0\}.$
$(f-g)(x)=\displaystyle \frac{2}{x}-\frac{4}{x+4}$ , domain: $\{x|x\neq-4,x\neq 0\}.$
$(fg)(x)=\displaystyle \frac{2}{x}\cdot\frac{4}{x+4}$ , domain: $\{x|x\neq-4,x\neq 0\}.$
$(\displaystyle \frac{f}{g})(x)=\frac{\frac{2}{x}}{\frac{4}{x+4}}=\frac{2(x+4)}{4x}=\frac{x+4}{2x}$ , domain: $\{x|x\neq-4,x\neq 0\}.$
In interval notation, all domains are
$(-\infty,-4)\cup(-4,0)\cup(0,+\infty)$