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

Answered by EDWIN88 last updated on 21/Apr/21

ℓet y=((1−x)/(1+x)) ⇒xy+y=1−x xy+x=1−y ⇒ x=((1−y)/(1+x)) (f(((1−x)/(1+x))))^2 .f(x) = ((1−x)/(1+x)) …(1) (f(((1−x)/(1+x)))).(f(x))^2 = x …(2) squaring equation (2) give (f(((1−x)/(1+x))))^2 .(f(x))^4 =x^2 ⇔ (((f(((1−x)/(1+x))))^2 .(f(x))^4 )/((f(((1−x)/(1+x))))^2 .f(x))) = ((x^2 (1+x))/(1−x)) ⇔ (f(x))^3 = ((x^2 (1+x))/(1−x)) ∴ f(x) = (((x^2 (1+x))/(1−x)))^(1/3) .

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

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