Answer
a) $\dfrac{3x}{2-x}$; $D=(-\infty,0)\cup(0,2)\cup(2,\infty)$
b) $\dfrac{2x-2}{3}$; $D=(-\infty,1)\cup(1,\infty)$
c) $\dfrac{3x-3}{4-x}$; $D=(-\infty,1)\cup(1,4)\cup(4,\infty)$
d) $x$; $D=(-\infty,0)\cup(0,\infty)$
Work Step by Step
We are given the functions:
$f(x)=\dfrac{3}{x-1}$
$g(x)=\dfrac{2}{x}$
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 $D_{f\circ g}$:
$(f\circ g)(x)=f(g(x))=f\left(\dfrac{2}{x}\right)=\dfrac{3}{\dfrac{2}{x}-1}=\dfrac{3}{\dfrac{2-x}{x}}=\dfrac{3x}{2-x}$
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:
$g(x)=1$
$\dfrac{2}{x}=1$
$x=2$
So we also exclude 2 from the domain of $f\circ g$.
We got:
$D_{f\circ g}=(-\infty,0)\cup(0,2)\cup(2,\infty)$
b) Determine $g\circ f$ and its domain $D_{g\circ f}$:
$(g\circ f)(x)=g(f(x))=g\left(\dfrac{3}{x-1}\right)=\dfrac{2}{\dfrac{3}{x-1}}=\dfrac{2(x-1)}{3}=\dfrac{2x-2}{3}$
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:
$f(x)=0$
$\dfrac{3}{x-1}=0$
No solution
We got:
$D_{g\circ f}=(-\infty,1)\cup(1,\infty)$
c) Determine $f\circ f$ and its domain $D_{f\circ f}$:
$(f\circ f)(x)=f(f(x))=f\left(\dfrac{3}{x-1}\right)=\dfrac{3}{\dfrac{3}{x-1}-1}=\dfrac{3}{\dfrac{4-x}{x-1}}=\dfrac{3(x-1)}{4-x}=\dfrac{3x-3}{4-x}$
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:
$f(x)=1$
$\dfrac{3}{x-1}=1$
$x-1=3$
$x=4$
So we also exclude 4 from the domain of $f\circ f$.
We got:
$D_{f\circ f}=(-\infty,1)\cup(1,4)\cup(4,\infty)$
d) Determine $g\circ g$ and its domain $D_{g\circ g}$:
$(g\circ g)(x)=g(g(x))=g\left(\dfrac{2}{x}\right)=\dfrac{2}{\dfrac{2}{x}}=x$
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:
$g(x)=0$
$\dfrac{2}{x}=0$
No solution
We got:
$D_{g\circ g}=(-\infty,0)\cup(0,\infty)$
