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Question Number 175866 by Linton last updated on 08/Sep/22

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

Answered by mr W last updated on 08/Sep/22

$$y=x^3+ax$$$$x^3+ax-y=0$$$$x=\sqrt[3]{\sqrt{\frac{y^2}{4}+\frac{a^3}{27}}+\frac{y}{2}}-\sqrt[3]{\sqrt{\frac{y^2}{4}+\frac{a^3}{27}}-\frac{y}{2}}$$$$\Rightarrow f^{-1}\left(x\right)=\sqrt[3]{\sqrt{\frac{x^2}{4}+\frac{a^3}{27}}+\frac{x}{2}}-\sqrt[3]{\sqrt{\frac{x^2}{4}+\frac{a^3}{27}}-\frac{x}{2}}$$

Commented by peter frank last updated on 08/Sep/22

$$\mathrm{please}\mathrm{explain}{3}^{\mathrm{rd}}\mathrm{line}$$

Commented by mr W last updated on 09/Sep/22

$${cardano}^{\prime }sformula$$$$seeQ89687$$

Commented by Tawa11 last updated on 15/Sep/22

$$\mathrm{Great}\mathrm{sir}$$

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