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Question Number 196464 by mokys last updated on 25/Aug/23

$$iff\left(x\right)=x^3+x^2,g\left(x\right)=f^{-1}\left(x\right),find{g}^{{}^{\prime }}\left(3\right)?$$

Answered by Frix last updated on 25/Aug/23

$$y=x^3+x^2$$$$dy=x(3x+2)dx$$$$g^{\prime }\left(x\right)=\frac{dx}{dy}=\frac{1}{x(3x+2)}$$$$x^3+x^2=3\Rightarrow x=-\frac{1}{3}+\sqrt[3]{\frac{79}{54}+\frac{\sqrt{77}}{6}}+\sqrt[3]{\frac{79}{54}-\frac{\sqrt{77}}{6}}$$$$x\approx 1.17455941$$$$g^{\prime }\left(x\right)\approx .154133358$$$$[\mathrm{This}\mathrm{is}\mathrm{the}\mathrm{real}\mathrm{solution}\mathrm{of}{r}^3+\frac{r}{231}-\frac{1}{231}=0]$$

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