Answer
a) $2-x$; $[0,\infty)$
b) $\sqrt{2-x^2}$; $[-\sqrt 2,\sqrt 2]$
Work Step by Step
We are given the functions:
$f(x)=2-x^2$
$g(x)=\sqrt x$
a) Determine the function $f\circ g$:
$(f\circ g)(x)=f(g(x))=f(\sqrt x)=2-(\sqrt x)^2=2-x$
The domain of the function $f\circ g$ is:
$D=[0,\infty)$
b) Determine the function $g\circ f$:
$(g\circ f)(x)=g(f(x))=g(2-x^2)=\sqrt{2-x^2}$
Determine the domain of the function $g\circ f$:
$2-x^2\geq 0$
$x\in[-\sqrt 2,\sqrt 2]$
$D=[-\sqrt 2,\sqrt 2]$
