College Algebra, Chapter 3, 3.6, Section 3.6, Problem 44

Determine the functions $f \circ g, \quad g \circ f, \quad f \circ f$ and $g \circ g$ and their domains if $\displaystyle f(x) = \frac{2}{x}$ and $\displaystyle g(x) = \frac{x}{x+2}$
For $f \circ g$,

$
\begin{equation}
\begin{aligned}
(f \circ g)(x) &= f(g(x)) && \text{Definition of } f \circ g\\
\\
(f \circ g)(x) &= f \left( \frac{x}{x + 2} \right) && \text{Definition of } g\\
\\
(f \circ g)(x) &= \frac{2}{\frac{x}{x+2}} && \text{Simplify}\\
\\
(f \circ g)(x) &= \frac{2(x+2)}{x} && \text{Definition of } f
\end{aligned}
\end{equation}
$

The function can't have a denominator equal to zero.
So the domain of $f \circ g$ is $(-\infty, 0) \bigcup (0,\infty)$

For $g \circ f$

$
\begin{equation}
\begin{aligned}
(g \circ f)(x) &= g(f(x)) && \text{Definition of } g \circ f\\
\\
(g \circ f)(x) &= g \left( \frac{2}{x} \right) && \text{Definition of } f\\
\\
(g \circ f)(x) &= \frac{\frac{2}{x}}{\frac{2}{x}+2} && \text{Simplify}\\
\\
(g \circ f)(x) &= \frac{2x}{2x(1 + x)} && \text{Simplify}\\
\\
(g \circ f)(x) &= \frac{1}{1+x} && \text{Definition of } g
\end{aligned}
\end{equation}
$

The denominator is not defined when $x = y$. So the domain of $g \circ f$ is $(-\infty, -1) \bigcup (-1, \infty)$

For $f \circ f$,

$
\begin{equation}
\begin{aligned}
(f \circ f)(x) &= f (f (x) ) && \text{Definition of } f \circ f\\
\\
(f \circ f)(x) &= f \left( \frac{2}{x} \right) && \text{Definition of } f\\
\\
(f \circ f)(x) &= \frac{\frac{2}{2}}{x} && \text{Simplify}\\
\\
(f \circ f)(x) &= x && \text{Definition of } f
\end{aligned}
\end{equation}
$

The function is define for all values of $x$, so the domain of $f \circ f$ is $(-\infty, \infty)$

For $g \circ g$,

$
\begin{equation}
\begin{aligned}
(g \circ g)(x) &= g(g(x)) && \text{Definition of } g \circ g\\
\\
(g \circ g)(x) &= g \left( \frac{x}{x+2} \right) && \text{Definition of } g\\
\\
(g \circ g)(x) &= \frac{\frac{x}{x+2}}{\frac{x}{x+2}+2} && \text{Simplify}\\
\\
(g \circ g)(x) &= \frac{\frac{x}{\cancel{x+2}}}{\frac{x+2x+4}{\cancel{x+2}}} && \text{Simplify}\\
\\
(g \circ g)(x) &= \frac{x}{3x+4} && \text{Definition of } g
\end{aligned}
\end{equation}
$

The denominator is not define when $\displaystyle x = -\frac{4}{3}$. So the domain of $g \circ g$ is $\displaystyle \left( -\infty, \frac{-4}{3} \right) \bigcup \left( \frac{-4}{3}, \infty \right)$