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Question Number 115997 by bemath last updated on 30/Sep/20

Answered by bobhans last updated on 30/Sep/20

(1)f((1/x))+f(1−x) = x→f(1−x)=x−f((1/x)) replace 1−x by x (2) f((1/(1−x))) + f(x) = 1−x replace (1/x) by 1−x (3) f(1−x)+f(1−(1/(1−x)))=1−(1/(1−x)) f(1−x)+f((x/(x−1)))=(x/(x−1)) ⇔−f((1/x))+f((x/(x−1)))=(x/(x−1))−((x^2 −x)/(x−1)) ⇔ f((x/(x−1)))−f((1/x))=((−x^2 +2x)/(x−1)) continue

$$\left(\mathrm{1}\right){f}\left(\frac{\mathrm{1}}{{x}}\right)+{f}\left(\mathrm{1}−{x}\right)\:=\:{x}\rightarrow{f}\left(\mathrm{1}−{x}\right)={x}−{f}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$${replace}\:\mathrm{1}−{x}\:{by}\:{x} \\ $$$$\left(\mathrm{2}\right)\:{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\:+\:{f}\left({x}\right)\:=\:\mathrm{1}−{x} \\ $$$${replace}\:\frac{\mathrm{1}}{{x}}\:{by}\:\mathrm{1}−{x}\: \\ $$$$\left(\mathrm{3}\right)\:{f}\left(\mathrm{1}−{x}\right)+{f}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\:{f}\left(\mathrm{1}−{x}\right)+{f}\left(\frac{{x}}{{x}−\mathrm{1}}\right)=\frac{{x}}{{x}−\mathrm{1}} \\ $$$$\Leftrightarrow−{f}\left(\frac{\mathrm{1}}{{x}}\right)+{f}\left(\frac{{x}}{{x}−\mathrm{1}}\right)=\frac{{x}}{{x}−\mathrm{1}}−\frac{{x}^{\mathrm{2}} −{x}}{{x}−\mathrm{1}} \\ $$$$\Leftrightarrow\:{f}\left(\frac{{x}}{{x}−\mathrm{1}}\right)−{f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{−{x}^{\mathrm{2}} +\mathrm{2}{x}}{{x}−\mathrm{1}} \\ $$$${continue} \\ $$

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