Answer
a. $\frac{2x}{3x+1}$
b. $(-\infty,-\frac{1}{3})\cup(-\frac{1}{3},0)\cup(0,\infty)$
Work Step by Step
Given $f(x)=\frac{2}{x+3}, x\ne-3$ and $g(x)=\frac{1}{x}, x\ne0$, we have:
a. $(f\circ g)(x)=\frac{2}{1/x+3}=\frac{2x}{3x+1}$
b. The domain of the above function can be found as $\{x|x\ne-\frac{1}{3},0\}$ or $(-\infty,-\frac{1}{3})\cup(-\frac{1}{3},0)\cup(0,\infty)$ (note that $x=-3$ is in the domain)
