Answer
For $f+g, f-g, fg$: $D=(-\infty,-4)\cup(-4,0)\cup(0,\infty)$,
For $\frac{f}{g}$: $(-\infty,-4)\cup(-4,0)\cup(0,\infty)$
Work Step by Step
$f(x)=\frac{2}{x}$,
$D_f=\{x|x \in \mathbb{R}-\{0\}\}$,
$g(x)=\frac{4}{x+4}$,
$D_g=\{x|x \in \mathbb{R}-\{-4\}\}$,
thus,
$f+g=\frac{2}{x}+\frac{4}{x+4}$,
$D_{f+g}=(-\infty,-4)\cup(-4,0)\cup(0,\infty)$,
$f-g=\frac{2}{x}-\frac{4}{x+4}$,
$D_{f-g}=(-\infty,-4)\cup(-4,0)\cup(0,\infty)$,
$f \times g=\frac{2}{x} \times \frac{4}{x+4}= \frac{8}{x^2+4x}$
$D_{f\times g}=(-\infty,-4)\cup(-4,0)\cup(0,\infty)$
$\frac{f}{g}=\frac{\frac{2}{x}}{\frac{4}{x+4}}=\frac{2x+8}{4x}$
$D_{\frac{f}{g}}=(-\infty,-4)\cup(-4,0)\cup(0,\infty)$
