## Precalculus (10th Edition)

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) $x$; $D=(-\infty,0)\cup(0,\infty)$
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: $D_{g\circ g}=(-\infty,0)\cup(0,\infty)$