Respuesta :
[tex]\bf f(x)=y=\sqrt[3]{\cfrac{x}{8}}-4\impliedby \textit{first, let's switch the variables about}
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\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}
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\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