Answer
$(f\circ g)(x)=\dfrac{x^4}{1+x^4}$; $D_{f\circ g}=(-\infty,0)\cup(0,\infty)$
$(g\circ f)(x)=(x^2+1)^2$; $D_{g\circ f}=(-\infty,\infty)$
Work Step by Step
We are given the functions:
$f(x)=\dfrac{1}{x^2+1}$
$g(x)=x^{-2}=\dfrac{1}{x^2}$
Determine the domains of $f$ and $g$:
$D_f=(-\infty,\infty)$
$D_g=(-\infty,0)\cup(0.\infty)$
Determine $(f\circ g)$:
$(f\circ g)(x)=f(g(x))=f\left(\dfrac{1}{x^2}\right)=\dfrac{1}{\left(\dfrac{1}{x^2}\right)^2+1}=\dfrac{1}{\dfrac{1}{x^4}+1}=\dfrac{x^4}{1+x^4}$
The domain of $(f\circ g)$ is:
$D_{f\circ g}=(-\infty,0)\cup(0,\infty)$
Determine $(g\circ f)$:
$(g\circ f)(x)=g(f(x))=g\left(\dfrac{1}{x^2+1}\right)=\dfrac{1}{\left(\dfrac{1}{x^2+1}\right)^2}=(x^2+1)^2$
The domain of $(g\circ f)$ is:
$D_{g\circ f}=(-\infty,\infty)$
