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Question Number 154133 by qaz last updated on 14/Sep/21

$$\mathrm{Given}\mathrm{f}\left(\mathrm{x}\right)=\sqrt[3]{\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}}+\sqrt[3]{\mathrm{x}-\sqrt{1+{\mathrm{x}}^2}}$$$$\mathrm{Find}{\mathrm{f}}^{-1}\left(\mathrm{x}\right)=?$$

Answered by EDWIN88 last updated on 14/Sep/21

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

Answered by mr W last updated on 14/Sep/21

$$\mathrm{f}\left(\mathrm{x}\right)=y=\sqrt[3]{\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}}+\sqrt[3]{\mathrm{x}-\sqrt{1+{\mathrm{x}}^2}}$$$$y=\sqrt[3]{-(-\mathrm{x})+\sqrt{1^3+{(-x)}^2}}-\sqrt[3]{(-\mathrm{x})+\sqrt{1^3+{(-x)}^2}}$$$$thatmeansyistherootofcubiceqn.$$$$withrespecttot:$$$$t^3+3t-2x=0$$$$i.e.$$$$y^3+3y-2x=0$$$$\Rightarrow x=\frac{y(y^2+3)}{2}$$$$\Rightarrow f^{-1}\left(x\right)=\frac{x(x^2+3)}{2}$$

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