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

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