Suppose that $f( x) =\dfrac{1}{x+2}$ and $g( x) = \dfrac{4}{x-1}$. Find $f\circ g$ and its domain.

Solution $$( f\circ g) ( x) =f( g( x) ) =\dfrac{1}{g( x) +2}=\dfrac{1}{\dfrac{4}{x-1}+2}= \dfrac{x-1}{4+2( x-1) }=\dfrac{x-1}{2x+2}$$

To find the domain of $f\circ g,$ first note that the domain of $g$ is $\{ x|x\neq 1\} ,$ so we exclude 1 from the domain of $f\circ g.$ Next note that the domain of $f$ is $\{ x|x\neq -2\}$, which means $g( x)$ cannot equal $-2.$ To determine what additional values of $x$ to exclude, we solve the equation $g( x) =-2$: $$\begin{array}{l} \frac{4}{{x - 1}} = - 2 \quad \quad {\color{#0066A7}{g(x) = - 2}} \\ \,\, \,\,\,\,\,\,\,4 = - 2(x - 1) \\ \,\,\, \,\,\,\,\,\,4 = - 2x + 2 \\ \,\,\,\,\,\,2x = - 2 \\ \,\,\,\, \,\,\,\,\,x = - 1 \\ \end{array}$$

We also exclude $-1$ from the domain of $f\,{\circ}\, g.$

The domain of $f\circ g$ is $\{ x|x\neq -1$, $x\neq 1\}$.

We could also find the domain of $f\circ g$ by first finding the domain of $g$: $\{ x|x\neq 1\}$. So, exclude $1$ from the domain of $f\circ g$. Then looking at $( f\circ g) ( x) =\dfrac{ x-1}{2x+2}=\dfrac{x-1}{2( x+1) }$, notice that $x\neq -1$, so we exclude $-1$ from the domain of $f\circ g.$ Therefore, the domain of $f\circ g$ is $\{ x|x\neq -1$, $x\neq 1\}$.