Question Number 68260 by aliesam last updated on 08/Sep/19
$${find}\:{f}\left({x}\right)\:{if}\: \\ $$$${f}\left(\frac{\mathrm{1}}{{x}}\right)+{f}\left(\mathrm{1}−{x}\right)={x} \\ $$
Commented by Prithwish sen last updated on 08/Sep/19
![Put x= (1/x) ⇒f(x)+f(1−(1/x))=(1/x)......(i) Putx= (x/(x−1))⇒f(1−(1/x))+f((1/(1−x)))=(x/(x−1)).....(ii) Putx = 1−x⇒f((1/(1−x)))+f(x)=1−x.....(iii) (i)−(ii)+(iii) we get f(x)=(1/2)[(1/x)−(x/(x−1))+1−x] please check.](Q68268.png)
$$\mathrm{Put}\:\mathrm{x}=\:\frac{\mathrm{1}}{\mathrm{x}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}\right)=\frac{\mathrm{1}}{\mathrm{x}}......\left(\mathrm{i}\right) \\ $$$$\mathrm{Putx}=\:\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\Rightarrow\mathrm{f}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\right)=\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}.....\left(\mathrm{ii}\right) \\ $$$$\mathrm{Putx}\:=\:\mathrm{1}−\mathrm{x}\Rightarrow\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\right)+\mathrm{f}\left(\mathrm{x}\right)=\mathrm{1}−\mathrm{x}.....\left(\mathrm{iii}\right) \\ $$$$\left(\boldsymbol{\mathrm{i}}\right)−\left(\boldsymbol{\mathrm{ii}}\right)+\left(\boldsymbol{\mathrm{iii}}\right)\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}} \\ $$$$\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}−\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{x}}−\mathrm{1}}+\mathrm{1}−\boldsymbol{\mathrm{x}}\right]\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$