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Question Number 164176 by mathlove last updated on 15/Jan/22

Answered by mr W last updated on 15/Jan/22

f(x)+f((1/(1−x)))=x    ...(i)  replace x with 1−(1/x)  ⇒f(((x−1)/x))+f(x)=1−(1/x)   ...(ii)  replace x with (1/(1−x))  ⇒ f((1/(1−x)))+f(((x−1)/x))=(1/(1−x))    ...(iii)    (i)+(ii)−(iii):  2f(x)=x+1−(1/x)−(1/(1−x))  ⇒f(x)=(1/2)(x+1−(1/x)−(1/(1−x)))

$${f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)={x}\:\:\:\:...\left({i}\right) \\ $$$${replace}\:{x}\:{with}\:\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$$$\Rightarrow{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)+{f}\left({x}\right)=\mathrm{1}−\frac{\mathrm{1}}{{x}}\:\:\:...\left({ii}\right) \\ $$$${replace}\:{x}\:{with}\:\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\Rightarrow\:{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)+{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:\:\:...\left({iii}\right) \\ $$$$ \\ $$$$\left({i}\right)+\left({ii}\right)−\left({iii}\right): \\ $$$$\mathrm{2}{f}\left({x}\right)={x}+\mathrm{1}−\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{1}−\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{1}−{x}}\right) \\ $$

Commented by Tawa11 last updated on 16/Jan/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by qaz last updated on 15/Jan/22

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

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

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