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Question Number 125968 by bramlexs22 last updated on 16/Dec/20

$$Findf\left(x\right)suchthat$$$$f\left(x\right)+f\left(\frac{1}{1-x}\right)=x$$

Answered by liberty last updated on 16/Dec/20

$$\left(1\right)f\left(x\right)+f\left(\frac{1}{1-x}\right)=x$$$$replacex\to \frac{1}{1-x}$$$$\left(2\right)f\left(\frac{1}{1-x}\right)+f\left(\frac{x-1}{x}\right)=\frac{1}{1-x}$$$$replace\frac{1}{1-x}\to x$$$$\left(3\right)f\left(\frac{x-1}{x}\right)+f\left(x\right)=\frac{x-1}{x}$$$$\iff \left(1\right)+\left(3\right)\Rightarrow 2f\left(x\right)+f\left(\frac{x-1}{x}\right)+f\left(\frac{1}{1-x}\right)=\frac{x^2+x-1}{x}$$$$\Rightarrow 2f\left(x\right)+\frac{1}{1-x}=\frac{x^2+x-1}{x}$$$$\Rightarrow 2f\left(x\right)=\frac{x^2+x-1}{x}-\frac{1}{1-x}$$$$\Rightarrow 2f\left(x\right)=\frac{x^2+x-1}{x}+\frac{1}{x-1}=\frac{x^3-x^2+x^2-x-x+1+x}{x(x-1)}$$$$f\left(x\right)=\frac{x^3-x+1}{2x(x-1)}$$$$$$

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