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Answer:
[tex]f^{-1}(x)=\dfrac{\sqrt[3]{x}+1}{2}}[/tex]
Step-by-step explanation:
Interchange x and y in the function definition and solve for y.
[tex]x=(2y-1)^3\\\\\sqrt[3]{x}=2y-1\qquad\text{take cube root}\\\\\sqrt[3]{x}+1=2y\qquad\text{add 1}\\\\y=\dfrac{\sqrt[3]{x}+1}{2}\qquad\text{divide by 2}\\\\\boxed{f^{-1}(x)=\dfrac{\sqrt[3]{x}+1}{2}}[/tex]
