Answer
For $f+g, f-g, fg$: $D=\{x|x \in \mathbb{R}-\{-1\}\}$
For $\frac{f}{g}$: $D_{\frac{f}{g}}=\{x|x \in \mathbb{R}-\{-1,0\}\}$,
Work Step by Step
$f(x)=\frac{2}{x+1}$,
$D_f=D=\{x|x \in \mathbb{R}-\{-1\}\}$,
$g(x)=\frac{x}{x+1}$,
$D_g=D=\{x|x \in \mathbb{R}-\{-1\}\}$,
thus,
$f+g=\frac{2}{x+1}+\frac{x}{x+1}$,
$D_{f+g}=D=\{x|x \in \mathbb{R}-\{-1\}\}$,
$f-g=\frac{2}{x+1}-\frac{x}{x+1}$,
$D_{f-g}=D=\{x|x \in \mathbb{R}-\{-1\}\}$,
$f \times g=\frac{2}{x+1} \times \frac{x}{x+1}= \frac{2x}{x^2+2x+1}$
$D_{f\times g}=D=\{x|x \in \mathbb{R}-\{-1\}\}$
$\frac{f}{g}=\frac{\frac{2}{x+1}}{\frac{x}{x+1}}=\frac{2x+2}{x^2+x}$
$D_{\frac{f}{g}}=\{x|x \in \mathbb{R}-\{-1,0\}\}$,
