Example

The "inside" function $g,$ in $f( g ( x ) ) ,$ is evaluated first.

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

$( f\circ g ) ( 0 ) =f ( g (0 ) ) =f ( -4 ) =\dfrac{1}{-4+2}=-\dfrac{1}{2} \quad \quad {\color{#0066A7}{{g} ( {x} )}} {\color{#0066A7}{=}} {\color{#0066A7}{\dfrac{4}{x-1}}}{\color{#0066A7}{;\;}} {\color{#0066A7}{{g}}} {\color{#0066A7}{( {0} )}} {\color{#0066A7}{=}}{\color{#0066A7}{-4}}$

$\begin{array}{l} (g \circ f)(1) = g(f(1))\mathop = \limits_ {\color{#0066A7}{\uparrow}} g\left( {\frac{1}{3}} \right) = \frac{4}{{\frac{1}{3} - 1}} = \frac{4}{{ - \frac{2}{3}}} = - 6 \\ {\color{#0066A7}{ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f(x) = \frac{1}{{x + 2}};\,f(1) = \frac{1}{3}}} \\ \end{array}$

$( f\circ f ) ( 1 ) =f ( f ( 1 ) ) =f \left( \dfrac{1}{3} \right) =\dfrac{1}{\dfrac{1}{3}+2}= \dfrac{1}{\dfrac{7}{3}}=\dfrac{3}{7}$

$\begin{array}{l} (g \circ g)( - 3) = g(g(-3))\mathop = \limits_ {\color{#0066A7}{\uparrow}} g\left( { - 1} \right) = \frac{4}{{ - 1 - 1}} = - 2 \\ {\color{#0066A7}{ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,g(x) = \frac{4}{{x - 1}};\,g( - 3) = \frac{4}{{ - 3 - 1}} = - 1}} \\ \end{array}$