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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

$$f\left(x\right)+f\left(\frac{1}{1-x}\right)=x...\left(i\right)$$$$replacexwith1-\frac{1}{x}$$$$\Rightarrow f\left(\frac{x-1}{x}\right)+f\left(x\right)=1-\frac{1}{x}...\left(ii\right)$$$$replacexwith\frac{1}{1-x}$$$$\Rightarrow f\left(\frac{1}{1-x}\right)+f\left(\frac{x-1}{x}\right)=\frac{1}{1-x}...\left(iii\right)$$$$$$$$\left(i\right)+\left(ii\right)-\left(iii\right):$$$$2f\left(x\right)=x+1-\frac{1}{x}-\frac{1}{1-x}$$$$\Rightarrow f\left(x\right)=\frac{1}{2}(x+1-\frac{1}{x}-\frac{1}{1-x})$$

Commented by Tawa11 last updated on 16/Jan/22

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

Answered by qaz last updated on 15/Jan/22

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

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