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Question Number 213000 by golsendro last updated on 28/Oct/24

$$\mathrm{f}\left(\frac{\mathrm{x}-3}{\mathrm{x}+1}\right)+\mathrm{f}\left(\frac{\mathrm{x}+3}{1-\mathrm{x}}\right)=\mathrm{x},\mathrm{x}\ne \pm 1$$$$\mathrm{f}\left(\mathrm{x}\right)=?$$

Answered by Ghisom last updated on 28/Oct/24

$$\frac{u-3}{u+1}=v\iff u=\frac{v+3}{1-v}$$$$2f\left(x\right)=\frac{x-3}{x+1}+\frac{x+3}{1-x}-x$$$$f\left(x\right)=\frac{x(x^2+7)}{2(1-x^2)}$$

Answered by ajfour last updated on 28/Oct/24

$$f\left(u\right)+f\left(v\right)=x$$$$u=h\left(x\right)=\frac{x-3}{1+x}$$$$\Rightarrow x=\frac{u+3}{1-u}$$$$v=-h(-x)=\frac{x+3}{1-x}$$$$x=\frac{v-3}{1+v}=\frac{u+3}{1-u}$$$$v-3-uv+3u=u+3+uv+3v$$$$(v-u)+uv+3=0$$$$orv=\frac{u-3}{1+u}\mathrm{\&}u=\frac{v+3}{1-v}$$$$f\left(x\right)+f\left(\frac{x-3}{1+x}\right)=\frac{x+3}{1-x}$$$$\Rightarrow f\left(x\right)+x-f\left(\frac{x+3}{1-x}\right)=\frac{x+3}{1-x}$$$$\Rightarrow f\left(x\right)-f\left(\frac{x+3}{1-x}\right)=\frac{x^2+3}{x(1-x)}..\left(i\right)$$$$\mathrm{\&}f\left(x\right)+f\left(\frac{x+3}{1-x}\right)=\frac{x-3}{1-x}..\left(ii\right)$$$$adding\left(i\right)\mathrm{\&}\left(ii\right)$$$$2f\left(x\right)=\frac{x^2+3+x(x-3)}{x(1-x)}$$$$\Rightarrow f\left(x\right)=\frac{2x^2-3x+3}{2x(1-x)}$$$$$$

Commented by Ghisom last updated on 30/Oct/24

$$\mathrm{test}\mathrm{your}\mathrm{solution}!$$

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