Respuesta :

[tex]\bf f(x)=y=6x^3-8\qquad inverse\implies \begin{array}{lll} x=&6y^3-8\\ &\uparrow\\ &\textit{switched variables} \end{array} \\\\\\ x+8=6y^3\implies \cfrac{x+8}{6}=y^3\impliedby taking\ \sqrt[3]{\qquad } \\\\\\ \sqrt[3]{\cfrac{x+8}{6}}=\sqrt[3]{y^3}\implies \sqrt[3]{\cfrac{x+8}{6}}=y\impliedby f^{-1}(x)[/tex]

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