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Question Number 124580 by Boucatchou last updated on 04/Dec/20

$$$$$$Letf\left(x\right)=x^3+2x-6,f^{-1}\left(x\right)=?$$

Answered by MJS_new last updated on 04/Dec/20

$$x=y^3+2y-6$$$$y^3+2y-(x+6)=0$$$$D=\frac{p^3}{27}+\frac{q^2}{4}=\frac{8}{27}+\frac{{(x+6)}^2}{4}>0\mathrm{\forall }x\in \mathbb{R}\Rightarrow \mathrm{we}\mathrm{need}$$$${\mathrm{Cardano}}^{\prime }\mathrm{s}\mathrm{solution}$$$$f^{-1}\left(x\right)=\sqrt[3]{\frac{x}{2}+3+\sqrt{\frac{x^2}{2}+6x+\frac{494}{27}}}-\sqrt[3]{-\frac{x}{2}-3+\sqrt{\frac{x^2}{2}+6x+\frac{494}{27}}}$$

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