Answer
$ f\circ g(x)=3^{x^{2}+1},\qquad(-\infty,+\infty$)
$ g\circ f(x)=3^{2x}+1,\qquad(-\infty,+\infty$)
Work Step by Step
$f\circ g(x)=f(g(x))=3^{g(x)}$
$=3^{x^{2}+1},$
no restriction,defined for all reals
domain=$\mathbb{R}=(-\infty,+\infty$)
$g\circ f(x)=g(f(x)=[f(x)]^{2}+1=(3^{x})^{2}+1$
$=3^{2x}+1$
no restriction,defined for all reals
domain=$\mathbb{R}=(-\infty,+\infty$)
