Answer
(a) $\sqrt {\frac{4x-2}{x}}$, $(-\infty,0)U[\frac{1}{2},\infty)$
(b) $ -\frac{2}{\sqrt {x+4}}$, $(-4,\infty)$
Work Step by Step
(a) Given $f(x)=\sqrt {x+4}, x\ge-4$ and $g(x)=-\frac{2}{x}, x\ne0$, we have $(f\circ g)(x)=f(g(x))=\sqrt {-\frac{2}{x}+4}=\sqrt {\frac{4x-2}{x}}=\sqrt {\frac{2(2x-1)}{x}}$. From $\frac{2x-1}{x}\ge0$, we have $x\lt0$ or $x\ge\frac{1}{2}$. we can find its domain: $(-\infty,0)U[\frac{1}{2},\infty)$
(b) $(g\circ f)(x)=g(f(x))=-\frac{2}{\sqrt {x+4}}$. From $x+4\gt0$, we have $x\gt-4$, and its domain: $(-4,\infty)$
