Question Number 132135 by benjo_mathlover last updated on 11/Feb/21
Answered by Olaf last updated on 11/Feb/21
![f(x)+f(((x−1)/x)) = 1+x (1) Let x = ((u−1)/u) (1) : f(((u−1)/u))+f(((((u−1)/u)−1)/((u−1)/u))) = 1+((u−1)/u) f(((u−1)/u))+f((1/(1−u))) = 1+((u−1)/u) (2) (1)−(2) : f(x)−f((1/(1−x))) = x−((x−1)/x) (3) Let v = (1/(1−x)) (1) : f((1/(1−v)))+f((((1/(1−v))−1)/(1/(1−v)))) = 1+(1/(1−v)) f((1/(1−v)))+f(v) = 1+(1/(1−v)) (4) (3)+(4) : 2f(x) = x−((x−1)/x)+1+(1/(1−x)) f(x) = (1/2)[x+(1/x)+(1/(1−x))]](https://www.tinkutara.com/question/Q132152.png)
$${f}\left({x}\right)+{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)\:=\:\mathrm{1}+{x}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{x}\:=\:\frac{{u}−\mathrm{1}}{{u}} \\ $$$$\left(\mathrm{1}\right)\::\:{f}\left(\frac{{u}−\mathrm{1}}{{u}}\right)+{f}\left(\frac{\frac{{u}−\mathrm{1}}{{u}}−\mathrm{1}}{\frac{{u}−\mathrm{1}}{{u}}}\right)\:=\:\mathrm{1}+\frac{{u}−\mathrm{1}}{{u}} \\ $$$${f}\left(\frac{{u}−\mathrm{1}}{{u}}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{u}}\right)\:=\:\mathrm{1}+\frac{{u}−\mathrm{1}}{{u}}\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\:: \\ $$$${f}\left({x}\right)−{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\:=\:{x}−\frac{{x}−\mathrm{1}}{{x}}\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\mathrm{Let}\:{v}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\left(\mathrm{1}\right)\::\:{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{v}}\right)+{f}\left(\frac{\frac{\mathrm{1}}{\mathrm{1}−{v}}−\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{1}−{v}}}\right)\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−{v}} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{v}}\right)+{f}\left({v}\right)\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−{v}}\:\:\:\:\:\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{3}\right)+\left(\mathrm{4}\right)\:: \\ $$$$\mathrm{2}{f}\left({x}\right)\:=\:{x}−\frac{{x}−\mathrm{1}}{{x}}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[{x}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\right] \\ $$
Answered by benjo_mathlover last updated on 11/Feb/21
