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Question Number 82431 by Power last updated on 21/Feb/20

Commented by Power last updated on 21/Feb/20

$$\mathrm{solution}\mathrm{sir}\mathrm{please}\mathrm{prove}\mathrm{it}$$

Commented by mr W last updated on 21/Feb/20

$${what}^{\prime }syourresultsir?f\left(x\right)=?$$

Commented by mr W last updated on 21/Feb/20

$$igotf\left(x\right)=\frac{1}{2}(x-\frac{1}{1-x}+\frac{x-1}{x}),right?$$

Commented by Power last updated on 21/Feb/20

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^3-\mathrm{x}+1}{2\mathrm{x}(\mathrm{x}-1)}$$

Commented by mr W last updated on 21/Feb/20

$$misterpower:$$$$f\left(x\right)=\frac{1}{2}(x-\frac{1}{1-x}+\frac{x-1}{x})and$$$$f\left(x\right)=\frac{x^3-x+1}{2x(x-1)}$$$$arethesame!$$

Answered by mr W last updated on 21/Feb/20

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

Commented by Power last updated on 21/Feb/20

$$\mathrm{yes}\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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