Answer
$$(f \circ g)(x)=1-x$$
$$(g \circ f)(x)=\sqrt{1-x^2}$$
$$D_{f \circ g}= (- \infty , 1]$$
$$D_{g \circ f}= [-1,1]$$
Work Step by Step
$$(f \circ g)(x)=f(g(x))=(\sqrt{1-x})^2=1-x$$
$$(g \circ f)(x)=g(f(x))=\sqrt{1-x^2}$$
The domain of $f \circ g$ consists of all $x$ in the domain of $g$ for which $g(x)$ is in the domain of $f$. So we have
$$D_{f \circ g}= \{ x \le 1 \mid \sqrt{1-x} \in \mathbb{R} \}=(- \infty , 1]$$ $$D_{g \circ f}= \{ x \in \mathbb{R} \mid x^2 \le 1 \}=[-1,1].$$
