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Question Number 96829 by bobhans last updated on 05/Jun/20

$$\mathrm{If}2\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}(1-\mathrm{x})={\mathrm{x}}^2.\mathrm{determine}\mathrm{f}\left(\mathrm{x}\right)$$

Commented by john santu last updated on 05/Jun/20

$$\mathrm{replace}\mathrm{x}\mathrm{by}1-\mathrm{x}\mathrm{into}\mathrm{eq}\left(1\right)$$$$\left(2\right)2\mathrm{f}(1-\mathrm{x})+\mathrm{f}\left(\mathrm{x}\right)={(1-\mathrm{x})}^2$$$$2\mathrm{f}(1-\mathrm{x})+\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-2\mathrm{x}+1$$$$\mathrm{multiply}\mathrm{eq}\left(1\right)\mathrm{by}2$$$$\left(1\right)2\mathrm{f}(1-\mathrm{x})+4\mathrm{f}\left(\mathrm{x}\right)={2\mathrm{x}}^2$$$$\mathrm{subtract}\mathrm{eq}\left(2\right)\mathrm{by}\mathrm{eq}\left(1\right)$$$$3\mathrm{f}\left(\mathrm{x}\right)={2\mathrm{x}}^2-({\mathrm{x}}^2-2\mathrm{x}+1)$$$$3\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+2\mathrm{x}-1$$$$\therefore \mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2+2\mathrm{x}-1}{3}.$$

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