Answer
(a) $ -\frac{x+1}{x-1}$, domain $\{x|x\ne0,1 \}$.
(b) $ \frac{x-1}{x+1}$, domain $\{x|x\ne\pm1 \}$.
(c) $ x$, domain $\{x|x\ne1 \}$.
(d) $ x$, domain $\{x|x\ne0 \}$.
Work Step by Step
Given $f(x)=\frac{x+1}{x-1}$ and $g(x)=\frac{1}{x}$, we have:
(a) $(f\circ g)(x)=\frac{\frac{1}{x}+1}{\frac{1}{x}-1}=-\frac{x+1}{x-1}$, domain $\{x|x\ne0,1 \}$.
(b) $(g\circ f)(x)=\frac{1}{\frac{x+1}{x-1}}=\frac{x-1}{x+1}$, domain $\{x|x\ne\pm1 \}$.
(c) $(f\circ f)(x)=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}=x$, domain $\{x|x\ne1 \}$.
(d) $(g\circ g)(x)=\frac{1}{\frac{1}{x}}=x$, domain $\{x|x\ne0 \}$.
