Answer
a) $\dfrac{2}{3x+2}$; $D=\left(-\infty,-\dfrac{2}{3}\right)\cup\left(-\dfrac{2}{3},0\right)\cup\left(0,\infty\right)$
b) $\dfrac{2x+6}{x}$; $D=(-\infty,-3)\cup(-3,0)\cup(0,\infty)$
c) $\dfrac{x}{4x+9}$; $D=\left(-\infty,-3\right)\cup\left(-3-\dfrac{9}{4}-\dfrac{9}{4}\right)\cup\left(1,\infty\right)$
d) $x$; $D=(-\infty,0)\cup(0,\infty)$
Work Step by Step
We are given the functions:
$f(x)=\dfrac{x}{x+3}$
$g(x)=\dfrac{2}{x}$
Let's note:
$D_f$=the domain of $f$
$D_g$=the domain of $g$
We have:
$D_f=(-\infty,-3)\cup(-3,\infty)$
$D_g=(-\infty,0)\cup(0,\infty)$
a) Determine $f\circ g$ and its domain $D_{f\circ g}$:
$(f\circ g)(x)=f(g(x))=f\left(\dfrac{2}{x}\right)=\dfrac{\dfrac{2}{x}}{\dfrac{2}{x}+3}=\dfrac{2}{3x+2}$
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 -3, so $g(x)\not=-3$. Solve:
$g(x)=-3$
$\dfrac{2}{x}=-3$
$x=-\dfrac{2}{3}$
So we also exclude $-\dfrac{2}{3}$ from the domain of $f\circ g$.
We got:
$D_{f\circ g}=\left(-\infty,-\dfrac{2}{3}\right)\cup\left(-\dfrac{2}{3},0\right)\cup\left(0,\infty\right)$
b) Determine $g\circ f$ and its domain $D_{g\circ f}$:
$(g\circ f)(x)=g(f(x))=g\left(\dfrac{x}{x+3}\right)=\dfrac{2}{\dfrac{x}{x+3}}=\dfrac{2(x+3}{x}=\dfrac{2x+6}{x}$
As $D_f=(-\infty,-3)\cup(-3,\infty)$ we exclude -3 from the domain of $g\circ f$. Also the domain of $g$ doesn't include 0, so $f(x)\not=0$. Solve:
$f(x)=0$
$\dfrac{x}{x+3}=0$
$x=0$
We exclude 0 from the domain of $f\circ g$ also.
We got:
$D_{g\circ f}=(-\infty,-3)\cup(-3,0)\cup(0,\infty)$
c) Determine $f\circ f$ and its domain $D_{f\circ f}$:
$(f\circ f)(x)=f(f(x))=f\left(\dfrac{x}{x+3}\right)=\dfrac{\dfrac{x}{x+3}}{\dfrac{x}{x+3}+3}=\dfrac{x}{4x+9}$
As $D_f=(-\infty,-3)\cup(-3,\infty)$ we exclude -3 from the domain of $f\circ f$. Also the domain of $f$ doesn't include -3, so $f(x)\not=-3$. Solve:
$f(x)=-3$
$\dfrac{x}{x+3}=-3$
$-3x-9=x$
$4x=-9$
$x=-\dfrac{9}{4}$
We exclude $-\dfrac{9}{4}$ from the domain of $f\circ f$.
We got:
$D_{f\circ f}=\left(-\infty,-3\right)\cup\left(-3-\dfrac{9}{4}-\dfrac{9}{4}\right)\cup\left(1,\infty\right)$
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:
$D_{g\circ g}=(-\infty,0)\cup(0,\infty)$
