Question Number 169842 by mathlove last updated on 10/May/22
Answered by Rasheed.Sindhi last updated on 10/May/22
$${f}^{\:\mathrm{2}} \left({x}\right).{f}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)={x}^{\mathrm{2}} ………….\left(\mathrm{A}\right) \\ $$$$\left(\mathrm{A}\right)^{\mathrm{2}} :\:\:{f}^{\:\:\mathrm{4}} \left({x}\right).{f}^{\:\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)={x}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{f}^{\:\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)=\frac{{x}^{\mathrm{4}} }{{f}\:^{\mathrm{4}} \left({x}\right)} \\ $$$$\bullet{x}\rightarrow\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\: \\ $$$$\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\rightarrow\frac{\mathrm{1}−\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{\mathrm{1}+\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}=\frac{\mathrm{1}+{x}−\mathrm{1}+{x}}{\mathrm{1}+{x}+\mathrm{1}−{x}}=\frac{\mathrm{2}{x}}{\mathrm{2}}={x} \\ $$$$\left(\mathrm{A}\right)\Rightarrow \\ $$$${f}^{\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right).{f}\left({x}\right)=\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{4}} }{{f}\:^{\mathrm{4}} \left({x}\right)}.{f}\left({x}\right)=\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{4}} }{{f}\:^{\mathrm{3}} \left({x}\right)}=\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{2}} \\ $$$${f}\:^{\mathrm{3}} \left({x}\right)={x}^{\mathrm{4}} .\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)^{\mathrm{2}} \\ $$
Commented by mathlove last updated on 11/May/22
$${thanks}\:{sir} \\ $$