Answer
(a) $ \frac{2}{3x+2}$, domain $\{x|x\ne0,-\frac{2}{3} \}$.
(b) $ \frac{2x+6}{x}$, domain $\{x|x\ne-3,0 \}$.
(c) $ \frac{x}{4x+9}$, domain $\{x|x\ne-3,-\frac{9}{4} \}$.
(d) $ x$, domain $\{x|x\ne0 \}$.
Work Step by Step
Given $f(x)=\frac{x}{x+3}$ and $g(x)=\frac{2}{x}$, we have:
(a) $f\circ g=\frac{\frac{2}{x}}{\frac{2}{x}+3}=\frac{2}{3x+2}$, domain $\{x|x\ne0,-\frac{2}{3} \}$.
(b) $g\circ f=\frac{2}{\frac{x}{x+3}}=\frac{2x+6}{x}$, domain $\{x|x\ne-3,0 \}$.
(c) $f\circ f=\frac{\frac{x}{x+3}}{\frac{x}{x+3}+3}=\frac{x}{4x+9}$, domain $\{x|x\ne-3,-\frac{9}{4} \}$.
(d) $g\circ g=\frac{2}{\frac{2}{x}}=x$, domain $\{x|x\ne0 \}$.
