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Question Number 118150 by bobhans last updated on 15/Oct/20

$$\mathrm{Determine}\mathrm{all}\mathrm{function}\mathrm{f}:\mathbb{R}╲\{0,1\}\to \mathbb{R}$$$$\mathrm{satisfying}\mathrm{the}\mathrm{functional}\mathrm{relation}$$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)=\frac{2(1-2\mathrm{x})}{\mathrm{x}(1-\mathrm{x})};\mathrm{for}\mathrm{x}\ne 0\mathrm{and}\mathrm{x}\ne 1$$

Commented by bemath last updated on 15/Oct/20

$$goodquestion$$

Answered by bemath last updated on 15/Oct/20

$$lettingw=\frac{1}{1-x},thegivenfunctional$$$$equationcanbewrittenas$$$$f\left(x\right)+f\left(w\right)=2(\frac{1}{x}-w)...\left(i\right)$$$$Ifwesettingz=\frac{1}{1-w}thenx=\frac{1}{1-z}$$$$Hencewealsohavetherelation$$$$f\left(w\right)+f\left(z\right)=2(\frac{1}{w}-z)...\left(ii\right)$$$$andf\left(z\right)+f\left(x\right)=2(\frac{1}{z}-x)...\left(iii\right)$$$$adding\left(i\right)and\left(iii\right)weget$$$$2\{f\left(x\right)+f\left(w\right)+f\left(z\right)\}=2(\frac{1}{x}-x)-2w+\frac{2}{x}$$$$using\left(ii\right)relation,thisreduces$$$$to\Rightarrow 2f\left(x\right)=2(\frac{1}{x}-x)-2(w+\frac{1}{w})+2(z+\frac{1}{z})$$$$nowusingw+\frac{1}{w}=\frac{1}{1-x}+1-x$$$$z+\frac{1}{z}=\frac{x-1}{x}+\frac{x}{x-1}$$$$thuswegetf\left(x\right)=\frac{x+1}{x-1}$$

Answered by bobhans last updated on 15/Oct/20

$$\left(1\right)\mathrm{replacing}\frac{1}{1-\mathrm{x}}\mathrm{by}\mathrm{x},\mathrm{gives}$$$$\mathrm{f}\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)+\mathrm{f}\left(\mathrm{x}\right)=\frac{2(1-2\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right))}{\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)(1-\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right))}$$$$\mathrm{f}\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)+\mathrm{f}\left(\mathrm{x}\right)=\frac{2\mathrm{x}(2-\mathrm{x})}{(\mathrm{x}-1)}$$$$\left(2\right)\mathrm{replacing}\mathrm{x}\mathrm{by}\frac{1}{1-\mathrm{x}}$$$$\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)+\mathrm{f}\left(\frac{1}{1-\left(\frac{1}{1-\mathrm{x}}\right)}\right)=\frac{2(1-2\left(\frac{1}{1-\mathrm{x}}\right))}{\left(\frac{1}{1-\mathrm{x}}\right)(1-\left(\frac{1}{1-\mathrm{x}}\right))}$$$$\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)+\mathrm{f}\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)=\frac{2(1-\mathrm{x})(1+\mathrm{x})}{\mathrm{x}}$$$$\mathrm{now}\mathrm{we}\mathrm{have}$$$$\left(1\right)\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)=\frac{2-4\mathrm{x}}{\mathrm{x}-{\mathrm{x}}^2}$$$$\left(2\right)\mathrm{f}\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)+\mathrm{f}\left(\mathrm{x}\right)=\frac{4\mathrm{x}-{2\mathrm{x}}^2}{\mathrm{x}-1}$$$$\left(3\right)\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)+\mathrm{f}\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)=\frac{2-{2\mathrm{x}}^2}{\mathrm{x}}$$$$\mathrm{adding}\left(1\right),\left(2\right)\mathrm{and}\left(3\right)\mathrm{we}\mathrm{get}$$$$2[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)+\mathrm{f}\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)]=\frac{-{4\mathrm{x}}^3+{6\mathrm{x}}^2+6\mathrm{x}-4}{{\mathrm{x}}^2-\mathrm{x}}$$$$\mathrm{or}\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)+\mathrm{f}\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)=\frac{-{2\mathrm{x}}^3+{3\mathrm{x}}^2+3\mathrm{x}-2}{{\mathrm{x}}^2-\mathrm{x}}$$$$\mathrm{substitute}\mathrm{third}\mathrm{equation},\mathrm{gives}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{-{2\mathrm{x}}^3+{3\mathrm{x}}^2+3\mathrm{x}-2}{{\mathrm{x}}^2-\mathrm{x}}-\frac{2-{2\mathrm{x}}^2}{\mathrm{x}}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{-{2\mathrm{x}}^3+{3\mathrm{x}}^2+3\mathrm{x}-2-(2\mathrm{x}-2-{2\mathrm{x}}^3+{2\mathrm{x}}^2)}{{\mathrm{x}}^2-\mathrm{x}}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2+\mathrm{x}}{{\mathrm{x}}^2-\mathrm{x}}=\frac{\mathrm{x}+1}{\mathrm{x}-1}.$$

Commented by bemath last updated on 15/Oct/20

$$waw....great...$$

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