Caracterize a função inversa

1. Ora, ${{D}_{f}}=\mathbb{R}$ e ${{D}_{f}}’=\mathbb{R}$.
   \[y=3x+2\Leftrightarrow 3x=y-2\Leftrightarrow x=\frac{1}{3}y-\frac{2}{3}\]
   Logo, \[\begin{array}{*{35}{l}}
   {{f}^{-1}}: & \mathbb{R}\to \mathbb{R}  \\
   {} & x\to \frac{1}{3}x-\frac{2}{3}  \\
   \end{array}\]
2. Ora, ${{D}_{g}}=\mathbb{R}\backslash \left\{ 0 \right\}$ e ${{D}_{g}}’=\mathbb{R}\backslash \left\{ -1 \right\}$ (note que $g(x)=\frac{2-x}{x}=-1+\frac{2}{x}$).
   \[y=\frac{2-x}{x}\Leftrightarrow yx=2-x\Leftrightarrow x(y+1)=2\Leftrightarrow x=\frac{2}{y+1}\]
   Logo, \[\begin{array}{*{35}{l}}
   {{g}^{-1}}: & \mathbb{R}\backslash \left\{ -1 \right\}\to \mathbb{R}\backslash \left\{ 0 \right\}  \\
   {} & x\to \frac{2}{x+1}  \\
   \end{array}\]
3. Ora, ${{D}_{h}}=\mathbb{R}\backslash \left\{ -2 \right\}$ e ${{D}_{h}}’=\mathbb{R}\backslash \left\{ 1 \right\}$ (note que $h(x)=\frac{x-5}{x+2}=\frac{x+2-7}{x+2}=1-\frac{7}{x+2}$).
   \[y=\frac{x-5}{x+2}\Leftrightarrow yx+2y=x-5\Leftrightarrow x(y-1)=-2y-5\Leftrightarrow x=\frac{2y+5}{1-y}\]
   Logo, \[\begin{array}{*{35}{l}}
   {{h}^{-1}}: & \mathbb{R}\backslash \left\{ 1 \right\}\to \mathbb{R}\backslash \left\{ -2 \right\}  \\
   {} & x\to \frac{2x+5}{1-x}  \\
   \end{array}\]
4. Ora, ${{D}_{i}}=\mathbb{R}$ e ${{D}_{i}}’=\mathbb{R}$.
   \[y={{x}^{3}}-3\Leftrightarrow {{x}^{3}}=y+3\Leftrightarrow x=\sqrt[3]{y+3}\]
   Logo, \[\begin{array}{*{35}{l}}
   {{i}^{-1}}: & \mathbb{R}\to \mathbb{R}  \\
   {} & x\to \sqrt[3]{x+3}  \\
   \end{array}\]

Qual é a relação entre as coordenadas de pontos simétricos relativamente ao eixo de simetria de equação $y=x$?