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Question Number 139032 by bramlexs22 last updated on 21/Apr/21

Answered by EDWIN88 last updated on 21/Apr/21

$$\ell \mathrm{et}\mathrm{y}=\frac{1-\mathrm{x}}{1+\mathrm{x}}\Rightarrow \mathrm{xy}+\mathrm{y}=1-\mathrm{x}$$$$\mathrm{xy}+\mathrm{x}=1-\mathrm{y}\Rightarrow \mathrm{x}=\frac{1-\mathrm{y}}{1+\mathrm{x}}$$$${\left(\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)\right)}^2.\mathrm{f}\left(\mathrm{x}\right)=\frac{1-\mathrm{x}}{1+\mathrm{x}}\dots \left(1\right)$$$$\left(\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)\right).{\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2=\mathrm{x}\dots \left(2\right)$$$$\mathrm{squaring}\mathrm{equation}\left(2\right)\mathrm{give}{\left(\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)\right)}^2.{\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^4={\mathrm{x}}^2$$$$\iff \frac{{\left(\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)\right)}^2.{\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^4}{{\left(\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)\right)}^2.\mathrm{f}\left(\mathrm{x}\right)}=\frac{{\mathrm{x}}^2(1+\mathrm{x})}{1-\mathrm{x}}$$$$\iff {\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^3=\frac{{\mathrm{x}}^2(1+\mathrm{x})}{1-\mathrm{x}}$$$$\therefore \mathrm{f}\left(\mathrm{x}\right)=\sqrt[3]{\frac{{\mathrm{x}}^2(1+\mathrm{x})}{1-\mathrm{x}}}.$$

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