Resposta :
- Tome f(x)=y
- "Inverta" as variáveis isolando o novo "y"
- Tome esse novo "y" como f^{-1}
[tex]f(x)=2x+1\\\\y=2x+1\\\\x=2y+1\\\\2y=x-1\\\\y=\frac{x-1}{2}\\\\f^{-1}(x)=\frac{x-1}{2}[/tex]
[tex]f(x)=x^3-8\\\\y=x^3-8\\\\x=y^3-8\\\\y^3=x+8\\\\y=\sqrt[3]{x+8}\\\\f^{-1}(x)=\sqrt[3]{x+8}[/tex]