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Question Number 165848 by leonhard77 last updated on 09/Feb/22

f((1/x))+f(1−x)=x f(x)=?

$$\:{f}\left(\frac{\mathrm{1}}{{x}}\right)+{f}\left(\mathrm{1}−{x}\right)={x} \\ $$$$\:\:{f}\left({x}\right)=? \\ $$

Answered by qaz last updated on 10/Feb/22

f(x)+f(1−(1/x))=(1/x) ⇒f(x)=(1/x)−f(1−(1/x)) =(1/x)−[(1/(1−(1/x)))−f(1−(1/(1−(1/x))))] =(1/x)−(x/(x−1))+f((1/(1−x))) =(1/x)−(x/(x−1))+1−x−f(x) ⇒f(x)=(1/2)((1/x)−(x/(x−1))+1−x)

$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}\right)=\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}}−\mathrm{f}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{x}}−\left[\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}}−\mathrm{f}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}+\mathrm{1}−\mathrm{x}−\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}+\mathrm{1}−\mathrm{x}\right) \\ $$

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