Answer
$(f\circ g)(t)=\sqrt{1-t^3}$; $D_{f\circ g}=(-\infty,1]$
$(g\circ f)(t)=1-t\sqrt t$; $D_{g\circ f}=[0,\infty)$
Work Step by Step
We are given the functions:
$f(t)=\sqrt t$
$g(x)=1-t^3$
Determine the domains of $f$ and $g$:
$D_f=(0,\infty)$
$D_g=(-\infty,\infty)$
Determine $(f\circ g)$:
$(f\circ g)(t)=f(g(t))=f(1-t^3)=\sqrt{1-t^3}$
The function $(f\circ g)$ is defined for the values of $t$ for which $1-t^3\geq 0$:
$1-t^3\geq 0$
$(1-t)(1+t+t^2)\geq 0$
$1-t\geq 0$
$t\leq 1$
$D_{f\circ g}=(-\infty,1]$
Determine $(g\circ f)$:
$(g\circ f)(x)=g(f(x))=g\left(\sqrt t\right)=1-(\sqrt t)^3=1-t\sqrt t$
The domain of $(g\circ f)$ is:
$D_{g\circ f}=[0,\infty)$
