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Question Number 192080 by sciencestudentW last updated on 07/May/23

$$f^{-1}\left(\frac{x+1}{x}\right)=x^3$$$$f^{-1}\left(x\right)+f\left(8\right)=?$$

Answered by deleteduser1 last updated on 07/May/23

$$f^{-1}\left(\frac{x+1}{x}\right)=x^3\Rightarrow f\left(x^3\right)=\frac{x+1}{x}=1+\frac{1}{x}$$$$\Rightarrow f\left(x\right)=1+\frac{1}{\sqrt[3]{x}}$$$$f\left(8\right)=\frac{3}{2}$$$$Lety=1+\frac{1}{\sqrt[3]{x}}$$$$\sqrt[3]{x}=\frac{1}{y-1}\Rightarrow x=\frac{1}{{(y-1)}^3}\Rightarrow f^{-1}\left(x\right)=\frac{1}{{(x-1)}^3}$$$$\Rightarrow f^{-1}\left(x\right)+f\left(8\right)=\frac{2+3{(x-1)}^3}{2{(x-1)}^3}$$$$$$

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