![f(y)=x\\\\64y^3-1=x\\\\64y^3=x+1 \qquad\text{add1}\\\\4y=\sqrt[3]{x+1} \qquad\text{cube root}\\\\y=\dfrac{\sqrt[3]{x+1}}{4} \qquad\text{divide by 4}\\\\\boxed{f^{-1}(x)=\dfrac{\sqrt[3]{x+1}}{4}}](/tpl/images/1024/4233/dd3ba.png)
What is the inverse of function f?
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