Answer
a) $\dfrac{4}{x+4}$; $D=(-\infty,-4)\cup(-4,0)\cup\left(0,\infty\right)$
b) $\dfrac{4-4x}{x}$; $D=(-\infty,0)\cup(0,1)\cup(1,\infty)$
c) $x$; $D=\left(-\infty,1\right)\cup\left(1,\infty\right)$
d) $x$; $D=(-\infty,0)\cup(0,\infty)$
Work Step by Step
We are given the functions:
$f(x)=\dfrac{x}{x-1}$
$g(x)=-\dfrac{4}{x}$
Let's note:
$D_f$=the domain of $f$
$D_g$=the domain of $g$
We have:
$D_f=(-\infty,1)\cup(1,\infty)$
$D_g=(-\infty,0)\cup(0,\infty)$
a) Determine $f\circ g$ and its domain $D_{f\circ g}$:
$(f\circ g)(x)=f(g(x))=f\left(-\dfrac{4}{x}\right)=\dfrac{-\dfrac{4}{x}}{-\dfrac{4}{x}-1}=\dfrac{-4}{-4-x}=\dfrac{4}{x+4}$
As $D_g=(-\infty,0)\cup(0,\infty)$ we exclude 0 from the domain of $f\circ g$. Also the domain of $f$ doesn't include 1, so $g(x)\not=1$. Solve:
$g(x)=1$
$-\dfrac{4}{x}=1$
$x=-4$
So we also exclude $-4$ from the domain of $f\circ g$.
We got:
$D_{f\circ g}=(-\infty,-4)\cup(-4,0)\cup\left(0,\infty\right)$
b) Determine $g\circ f$ and its domain $D_{g\circ f}$:
$(g\circ f)(x)=g(f(x))=g\left(\dfrac{x}{x-1}\right)=-\dfrac{4}{\dfrac{x}{x-1}}=-\dfrac{4(x-1)}{x}=\dfrac{4-4x}{x}$
As $D_f=(-\infty,1)\cup(1,\infty)$ we exclude 1 from the domain of $g\circ f$. Also the domain of $g$ doesn't include 0, so $f(x)\not=0$. Solve:
$f(x)=0$
$\dfrac{x}{x-1}=0$
$x=0$
We exclude 0 from the domain of $f\circ g$ also.
We got:
$D_{g\circ f}=(-\infty,0)\cup(0,1)\cup(1,\infty)$
c) Determine $f\circ f$ and its domain $D_{f\circ f}$:
$(f\circ f)(x)=f(f(x))=f\left(\dfrac{x}{x-1}\right)=\dfrac{\dfrac{x}{x-1}}{\dfrac{x}{x-1}-1}=\dfrac{x}{1}=x$
As $D_f=(-\infty,1)\cup(1,\infty)$ we exclude -1 from the domain of $f\circ f$. Also the domain of $f$ doesn't include 1, so $f(x)\not=1$. Solve:
$f(x)=1$
$\dfrac{x}{x-1}=1$
$x=x-1$
$0=-1$
No solution
We got:
$D_{f\circ f}=\left(-\infty,1\right)\cup(1,\infty)$
d) Determine $g\circ g$ and its domain $D_{g\circ g}$:
$(g\circ g)(x)=g(g(x))=g\left(-\dfrac{4}{x}\right)=-\dfrac{4}{-\dfrac{4}{x}}=x$
As $D_g=(-\infty,0)\cup(0,\infty)$ we exclude 0 from the domain of $g\circ g$. Also the domain of $g$ doesn't include 0, so $g(x)\not=0$. Solve:
$g(x)=0$
$-\dfrac{4}{x}=0$
No solution
We got:
$D_{g\circ g}=(-\infty,0)\cup(0,\infty)$