Respuesta :

[tex]\bf f(x)=y=\sqrt[3]{\cfrac{x}{8}}-4\impliedby \textit{first, let's switch the variables about} \\\\\\ \underline{x}=\sqrt[3]{\cfrac{\underline{y}}{8}}-4\implies x+4=\sqrt[3]{\cfrac{\underline{y}}{8}}\implies (x+4)^3=\cfrac{y}{8} \\\\\\ \boxed{8(x+4)^3=y}\impliedby f^{-1}(x)[/tex]

Answer:

[tex]f^{-1}(x)=8(x+4)^{3}[/tex]

Step-by-step explanation:

we have

[tex]f(x)=\sqrt[3]{\frac{x}{8}}-4[/tex]

Let

[tex]y=f(x)[/tex]

[tex]y=\sqrt[3]{\frac{x}{8}}-4[/tex]

Exchanges the variable x for y and y for x

[tex]x=\sqrt[3]{\frac{y}{8}}-4[/tex]

Isolate the variable y

[tex]x+4=\sqrt[3]{\frac{y}{8}}[/tex]

elevates to the cube both members

[tex](x+4)^{3}=\frac{y}{8} \\ \\y=8(x+4)^{3}[/tex]

Let

[tex]f^{-1}(x)=y[/tex]

[tex]f^{-1}(x)=8(x+4)^{3}[/tex] ------> inverse function

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