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Question Number 95150 by bobhans last updated on 23/May/20

$$\mathrm{how}\mathrm{do}\mathrm{you}\mathrm{solve}\mathrm{f}\left(\mathrm{x}\right)+2\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)=\mathrm{x}$$$$\mathrm{for}\mathrm{f}?$$

Answered by mr W last updated on 23/May/20

$$f\left(x\right)+2f\left(\frac{1}{1-x}\right)=x...\left(i\right)$$$$f\left(\frac{1}{1-x}\right)+2f\left(\frac{x-1}{x}\right)=\frac{1}{1-x}...\left(ii\right)$$$$f\left(\frac{x-1}{x}\right)+2f\left(x\right)=\frac{x-1}{x}...\left(iii\right)$$$$\left(i\right)-2\left(ii\right):$$$$f\left(x\right)-4f\left(\frac{x-1}{x}\right)=x-\frac{2}{1-x}...\left(iv\right)$$$$4\left(iii\right)+\left(iv\right):$$$$9f\left(x\right)=\frac{4(x-1)}{x}+x-\frac{2}{1-x}$$$$\Rightarrow f\left(x\right)=\frac{x^3+3x^2-6x+4}{9x(x-1)}$$

Commented by bobhans last updated on 23/May/20

$$\mathrm{ok}.\mathrm{i}\mathrm{compare}\mathrm{to}\mathrm{my}\mathrm{way}$$$$f\left(x\right)+2f\left(\frac{1}{1-x}\right)=x$$$$f\left(\frac{1}{1-x}\right)+2f\left(\frac{1}{1-\frac{1}{1-x}}\right)=\frac{1}{1-x}$$$$f\left(\frac{1}{1-x}\right)+2f\left(\frac{x-1}{x}\right)=\frac{1}{1-x}...\left(2\right)$$$$f\left(\frac{1}{1-\frac{1}{1-x}}\right)+2f\left(\frac{\frac{1}{1-x}-1}{\frac{1}{1-x}}\right)=\frac{1}{1-\frac{1}{1-x}}$$$$f\left(\frac{x-1}{x}\right)+2f\left(x\right)=\frac{x-1}{x}...\left(3\right)$$$$\left(1\right)-2\times \left(2\right)+4\times \left(3\right)$$$$9f\left(x\right)=\frac{x^3+3x^2-6x+4}{x(x-1)}\Rightarrow f\left(x\right)=\frac{x^3+3x^2-6x+4}{9x(x-1)}$$

Commented by I want to learn more last updated on 23/May/20

$$\mathrm{Weldone}\mathrm{sir}$$

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