Answer
a. $(f\circ g)(x)=\frac{1+x}{1-2x}$
b. $(-\infty,0)\cup(0,\frac{1}{2})\cup(\frac{1}{2}, \infty)$
Work Step by Step
a. Given $f(x)=\frac{x+1}{x-2}, x\ne2$ and $g(x)=\frac{1}{x}, x\ne0$, we have $(f\circ g)(x)=\frac{1/x+1}{1/x-2}=\frac{1+x}{1-2x}$
b. The domain of $(f\circ g)(x)$ can be found as $(-\infty,0)\cup(0,\frac{1}{2})\cup(\frac{1}{2}, \infty)$
