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Question Number 49389 by behi83417@gmail.com last updated on 06/Dec/18

$$forx\ne 0,y\ne 0,\mathrm{xy}\ne -1,f\left(1\right)=\frac{1}{2}$$$$\mathbf{f}\left(\mathbf{x}\right)+\mathbf{f}\left(\mathbf{y}\right)=\mathbf{f}(\mathbf{x}+\mathbf{y})+\frac{\mathbf{x}+\mathbf{y}}{1+\mathbf{xy}}$$$$1.find:f\left(x\right),\left[ifpossible\right]$$$$2.find:f^{-1}\left(1\right),\left[ifpossible\right].$$

Answered by kaivan.ahmadi last updated on 07/Dec/18

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

Commented by Abdo msup. last updated on 08/Dec/18

$$f\left(x\right)=y\iff x=f^{-1}\left(y\right)$$$$f\left(x\right)=y\iff \frac{x}{1+x^2}=y\iff x=(1+x^2)y\iff y+{yx}^2-x=0$$$$\iff {yx}^2-x+y=0$$$$\mathrm{\Delta }=1-4y^2\mathrm{\Delta }⩾0\iff 4y^2⩽1\iff \mid y\mid ⩽\frac{1}{2}$$$$x_1=\frac{1+\sqrt{1-4y^2}}{2y}and{x}_2=\frac{1-\sqrt{1-4y^2}}{2y}\Rightarrow$$$$f^{-1}\left(y\right)=\frac{1\stackrel{-}{+}\sqrt{1-4y^2}}{2y}andweseethat{f}^{-1}\left(1\right)dontexist$$$$(1\notin [-\frac{1}{2},\frac{1}{2}])$$

Answered by kaivan.ahmadi last updated on 07/Dec/18

$${\mathrm{f}}^{-1}\left(1\right)\mathrm{is}\mathrm{not}\mathrm{exist},\mathrm{since}\mathrm{if}(\mathrm{x},1)\in \mathrm{f}\mathrm{then}$$$${\mathrm{x}}^2-\mathrm{x}+1=0\mathrm{that}\mathrm{has}\mathrm{no}\mathrm{root}$$

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