Answer
a) $\dfrac{x}{3x-2}$; $D=(-\infty,0)\cup\left(0,\dfrac{2}{3}\right)\cup\left(\dfrac{2}{3},\infty\right)$
b) $-2x-6$; $D=(-\infty,-3)\cup(-3,\infty)$
c) $\dfrac{x+3}{3x+10}$; $D=\left(-\infty,-\dfrac{10}{3}\right)\cup\left(-\dfrac{10}{3},-3\right)\cup(-3,\infty)$
d) $x$; $D=(-\infty,0)\cup(0,\infty)$
Work Step by Step
We are given the functions:
$f(x)=\dfrac{1}{x+3}$
$g(x)=-\dfrac{2}{x}$
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 $D_{f\circ g}$:
$(f\circ g)(x)=f(g(x))=f\left(-\dfrac{2}{x}\right)=\dfrac{1}{-\dfrac{2}{x}+3}=\dfrac{1}{\dfrac{3x-2}{x}}=\dfrac{x}{3x-2}$
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:
$D_{f\circ g}=(-\infty,0)\cup\left(0,\dfrac{2}{3}\right)\cup\left(\dfrac{2}{3},\infty\right)$
b) Determine $g\circ f$ and its domain $D_{g\circ f}$:
$(g\circ f)(x)=g(f(x))=g\left(\dfrac{1}{x+3}\right)=-\dfrac{2}{\dfrac{1}{x+3}}=-2(x+3)=-2x-6$
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:
$f(x)=0$
$\dfrac{1}{x+3}=0$
No solution
We got:
$D_{g\circ f}=(-\infty,-3)\cup(-3,\infty)$
c) Determine $f\circ f$ and its domain $D_{f\circ f}$:
$(f\circ f)(x)=f(f(x))=f\left(\dfrac{1}{x+3}\right)=\dfrac{1}{\dfrac{1}{x+3}+3}=\dfrac{1}{\dfrac{3x+10}{x+3}}=\dfrac{x+3}{3x+10}$
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:
$f(x)=-3$
$\dfrac{1}{x+3}=-3$
$-3x-9=1$
$-3x=10$
$x=-\dfrac{10}{3}$
So we also exclude $-\dfrac{10}{3}$ from the domain of $f\circ f$.
We got:
$D_{f\circ f}=\left(-\infty,-\dfrac{10}{3}\right)\cup\left(-\dfrac{10}{3},-3\right)\cup(-3,\infty)$
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:
$D_{g\circ g}=(-\infty,0)\cup(0,\infty)$
