Answer
$(f\circ g)(t)=\dfrac{1}{\sqrt{-t^2}}$; $D_{f\circ g}=\phi$
$(g\circ f)(t)=-\dfrac{1}{t}$; $D_{g\circ f}=(0,\infty)$
Work Step by Step
We are given the functions:
$f(t)=\dfrac{1}{\sqrt t}$
$g(x)=-t^2$
Determine the domains of $f$ and $g$:
$D_f=(0,\infty)$
$D_g=(-\infty,\infty)$
Determine $(f\circ g)$:
$(f\circ g)(t)=f(g(t))=f(-t^2)=\dfrac{1}{\sqrt{-t^2}}$
As $-t^2\leq 0$ for any real $t$ and for $t=0$ the denominator $\sqrt{-t^2}$ is not defined, it means that $(f\circ g)$ is not defined at all.
$D_{f\circ g}=\phi$
Determine $(g\circ f)$:
$(g\circ f)(x)=g(f(x))=g\left(\dfrac{1}{\sqrt t}\right)=-\left(\dfrac{1}{\sqrt t}\right)^2=-\dfrac{1}{t}$
The domain of $(g\circ f)$ is:
$D_{g\circ f}=(0,\infty)$
