Answer
a) $\dfrac{3}{x(x+2)}$; domain $(-\infty,-2)\cup(-2,0)(0,\infty)$
b) $\dfrac{x^2+2}{x^2-1}$; domain: $(-\infty,-1)\cup(-1,1)\cup(1,\infty)$
Work Step by Step
We are given the functions:
$f(x)=\dfrac{3}{x^2-1}$
$g(x)=x+1$
Determine the domains $D_f$ and $D_g$ of the two functions:
$x^2-1=0\Rightarrow x=\pm 1$
$D_f=(-\infty,-1)\cup(-1,1)\cup(1,\infty)$
$D_g=(-\infty,\infty)$
a) Find $f\circ g$ and its domain $D_{f\circ g}$:
$(f\circ g)(x)=f(g(x))=f\left(x+1\right)=\dfrac{3}{(x+1)^2-1}=\dfrac{3}{x^2+2x+1-1}=\dfrac{3}{x^2+2x}=\dfrac{3}{x(x+2)}$
$x(x+2)=0\Rightarrow x=-2,x=0$
$D_{f\circ g}=(-\infty,-2)\cup(-2,0)\cup(0,\infty)$
b) Find $g\circ f$ and its domain $D_{g\circ f}$:
$(g\circ f)(x)=g(f(x))=g\left(\dfrac{3}{x^2-1}\right)=\dfrac{3}{x^2-1}+1=\dfrac{3+x^2-1}{x^2-1}=\dfrac{x^2+2}{x^2-1}$
$x^2-1=0\Rightarrow x=\pm 1$
$D_{g\circ f}=(-\infty,-1)\cup(-1,1)\cup(1,\infty)$
