Question Number 213000 by golsendro last updated on 28/Oct/24
$$\:\:\:\mathrm{f}\left(\frac{\mathrm{x}−\mathrm{3}}{\mathrm{x}+\mathrm{1}}\right)\:+\:\mathrm{f}\left(\frac{\mathrm{x}+\mathrm{3}}{\mathrm{1}−\mathrm{x}}\right)\:=\:\mathrm{x}\:,\:\mathrm{x}\neq\:\pm\:\mathrm{1} \\ $$$$\:\:\:\mathrm{f}\left(\mathrm{x}\right)=? \\ $$
Answered by Ghisom last updated on 28/Oct/24
$$\frac{{u}−\mathrm{3}}{{u}+\mathrm{1}}={v}\:\Leftrightarrow\:{u}=\frac{{v}+\mathrm{3}}{\mathrm{1}−{v}} \\ $$$$\mathrm{2}{f}\left({x}\right)=\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}+\frac{{x}+\mathrm{3}}{\mathrm{1}−{x}}−{x} \\ $$$${f}\left({x}\right)=\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{7}\right)}{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)} \\ $$
Answered by ajfour last updated on 28/Oct/24
$${f}\left({u}\right)+{f}\left({v}\right)={x} \\ $$$${u}={h}\left({x}\right)=\frac{{x}−\mathrm{3}}{\mathrm{1}+{x}} \\ $$$$\Rightarrow\:{x}=\frac{{u}+\mathrm{3}}{\mathrm{1}−{u}} \\ $$$${v}=−{h}\left(−{x}\right)=\frac{{x}+\mathrm{3}}{\mathrm{1}−{x}} \\ $$$${x}=\frac{{v}−\mathrm{3}}{\mathrm{1}+{v}}=\frac{{u}+\mathrm{3}}{\mathrm{1}−{u}} \\ $$$${v}−\mathrm{3}−{uv}+\mathrm{3}{u}={u}+\mathrm{3}+{uv}+\mathrm{3}{v} \\ $$$$\left({v}−{u}\right)+{uv}+\mathrm{3}=\mathrm{0} \\ $$$${or}\:\:{v}=\frac{{u}−\mathrm{3}}{\mathrm{1}+{u}}\:\:\:\:\:\:\:\&\:\:\:{u}=\frac{{v}+\mathrm{3}}{\mathrm{1}−{v}} \\ $$$${f}\left({x}\right)+{f}\left(\frac{{x}−\mathrm{3}}{\mathrm{1}+{x}}\right)=\frac{{x}+\mathrm{3}}{\mathrm{1}−{x}} \\ $$$$\Rightarrow\:\:\:{f}\left({x}\right)+{x}−{f}\left(\frac{{x}+\mathrm{3}}{\mathrm{1}−{x}}\right)=\frac{{x}+\mathrm{3}}{\mathrm{1}−{x}} \\ $$$$\Rightarrow{f}\left({x}\right)−{f}\left(\frac{{x}+\mathrm{3}}{\mathrm{1}−{x}}\right)=\frac{{x}^{\mathrm{2}} +\mathrm{3}}{{x}\left(\mathrm{1}−{x}\right)}\:\:\:\:..\left({i}\right) \\ $$$$\&\:\:{f}\left({x}\right)+{f}\left(\frac{{x}+\mathrm{3}}{\mathrm{1}−{x}}\right)=\frac{{x}−\mathrm{3}}{\mathrm{1}−{x}}\:\:\:\:\:\:\:..\left({ii}\right) \\ $$$${adding}\:\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$$\mathrm{2}{f}\left({x}\right)=\frac{{x}^{\mathrm{2}} +\mathrm{3}+{x}\left({x}−\mathrm{3}\right)}{{x}\left(\mathrm{1}−{x}\right)} \\ $$$$\Rightarrow\:\:\:{f}\left({x}\right)=\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{3}}{\mathrm{2}{x}\left(\mathrm{1}−{x}\right)} \\ $$$$ \\ $$
Commented by Ghisom last updated on 30/Oct/24
$$\mathrm{test}\:\mathrm{your}\:\mathrm{solution}! \\ $$