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Question Number 133168 by bemath last updated on 19/Feb/21

Answered by floor(10²Eta[1]) last updated on 19/Feb/21

$$\mathrm{x}\to 1-\frac{1}{\mathrm{x}}$$$$\left(\mathrm{I}\right):\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}-1}\right)+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\frac{\mathrm{x}-1}{\mathrm{x}}$$$$\mathrm{but}\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{x}-\mathrm{f}(1-\mathrm{x}),\mathrm{so}$$$$\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}-1}\right)+\mathrm{x}-\mathrm{f}(1-\mathrm{x})=\frac{\mathrm{x}-1}{\mathrm{x}}$$$$\left(\mathrm{II}\right):\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}-1}\right)-\mathrm{f}(1-\mathrm{x})=\frac{-{\mathrm{x}}^2+\mathrm{x}-1}{\mathrm{x}}$$$$\mathrm{x}\to \frac{1}{1-\mathrm{x}}$$$$\left(\mathrm{III}\right):\mathrm{f}(1-\mathrm{x})+\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}-1}\right)=\frac{1}{1-\mathrm{x}}$$$$\left(\mathrm{III}\right)-\left(\mathrm{II}\right)=2\mathrm{f}(1-\mathrm{x})=\frac{1}{1-\mathrm{x}}-\left(\frac{-{\mathrm{x}}^2+\mathrm{x}-1}{\mathrm{x}}\right)$$$$\mathrm{x}\to 1-\mathrm{x}$$$$2\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}-\left(\frac{-{(1-\mathrm{x})}^2+(1-\mathrm{x})-1}{(1-\mathrm{x})}\right)$$$$=\frac{1}{\mathrm{x}}+\frac{{\mathrm{x}}^2-\mathrm{x}+1}{1-\mathrm{x}}=\frac{1-\mathrm{x}+{\mathrm{x}}^3-{\mathrm{x}}^2+\mathrm{x}}{\mathrm{x}-{\mathrm{x}}^2}$$$$\Rightarrow 2\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2-1-{\mathrm{x}}^3}{{\mathrm{x}}^2-\mathrm{x}}$$

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