Considere as funções $f$ e $g$

\[\begin{array}{*{20}{c}}
{f\left( x \right) = \frac{2}{{x – 1}}}&{\text{e}}&{g\left( x \right) = \frac{1}{{x + 3}}}
\end{array}\]

1. ${D_f} = \left\{ {x \in \mathbb{R}:x – 1 \ne 0} \right\} = \mathbb{R}\backslash \left\{ 1 \right\}$.

   ${D_g} = \left\{ {x \in \mathbb{R}:x + 3 \ne 0} \right\} = \mathbb{R}\backslash \left\{ { – 3} \right\}$.

2. \[{D_{f \circ g}} = \left\{ {x \in \mathbb{R}:x \in {D_g} \wedge g\left( x \right) \in {D_f}} \right\} = \left\{ {x \in \mathbb{R}:x \in \mathbb{R}\backslash \left\{ { – 3} \right\} \wedge \left( {\frac{1}{{x + 3}}} \right) \in \mathbb{R}\backslash \left\{ { – 3, – 2} \right\}} \right\} = \mathbb{R}\backslash \left\{ { – 3, – 2} \right\}\]
   \[{D_{f \circ g}} = \left\{ {x \in \mathbb{R}:x \in {D_g} \wedge g\left( x \right) \in {D_f}} \right\} = \left\{ {x \in \mathbb{R}:x \in \mathbb{R}\backslash \left\{ { – 3} \right\} \wedge \left( {\frac{1}{{x + 3}}} \right) \in \mathbb{R}\backslash \left\{ 1 \right\}} \right\} = \mathbb{R}\backslash \left\{ { – 3, – 2} \right\}\]
   \[\begin{array}{*{20}{c}}
   {f \circ g:}&{\mathbb{R}\backslash \left\{ { – 3, – 2} \right\} \to \mathbb{R}} \\
   {}&{x \to  – \frac{{2x + 6}}{{x + 2}}}
   \end{array}\]

   \[{D_{g \circ f}} = \left\{ {x \in \mathbb{R}:x \in {D_f} \wedge f\left( x \right) \in {D_g}} \right\} = \left\{ {x \in \mathbb{R}:x \in \mathbb{R}\backslash \left\{ 1 \right\} \wedge \left( {\frac{2}{{x – 1}}} \right) \in \mathbb{R}\backslash \left\{ { – 3} \right\}} \right\} = \mathbb{R}\backslash \left\{ {\frac{1}{3},1} \right\}\]
   \[\left( {g \circ f} \right)\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {\frac{2}{{x – 1}}} \right) = \frac{1}{{\frac{2}{{x – 1}} + 3}} = \frac{1}{{\frac{{2 + 3x – 3}}{{x – 1}}}} = \frac{{x – 1}}{{3x – 1}},\,\forall x \in {D_{g \circ f}}\]
   \[\begin{array}{*{20}{c}}
   {g \circ f:}&{\mathbb{R}\backslash \left\{ {\frac{1}{3},1} \right\} \to \mathbb{R}} \\
   {}&{x \to \frac{{x – 1}}{{3x – 1}}}
   \end{array}\]

   \[{D_{f \circ f}} = \left\{ {x \in \mathbb{R}:x \in {D_f} \wedge f\left( x \right) \in {D_f}} \right\} = \left\{ {x \in \mathbb{R}:x \in \mathbb{R}\backslash \left\{ 1 \right\} \wedge \left( {\frac{2}{{x – 1}}} \right) \in \mathbb{R}\backslash \left\{ 1 \right\}} \right\} = \mathbb{R}\backslash \left\{ {1,3} \right\}\]
   \[\left( {f \circ f} \right)\left( x \right) = f\left( {f\left( x \right)} \right) = f\left( {\frac{2}{{x – 1}}} \right) = \frac{2}{{\frac{2}{{x – 1}} – 1}} = \frac{2}{{\frac{{2 – x + 1}}{{x – 1}}}} = \frac{{2x – 2}}{{3 – x}},\,\forall x \in {D_{f \circ f}}\]
   \[\begin{array}{*{20}{c}}
   {f \circ f:}&{\mathbb{R}\backslash \left\{ {1,3} \right\} \to \mathbb{R}} \\
   {}&{x \to \frac{{2x – 2}}{{3 – x}}}
   \end{array}\]