Question Number 206679 by mathlove last updated on 22/Apr/24
$${if}\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)+\mathrm{2}{g}\left(\mathrm{1}−{x}\right)={x}^{\mathrm{2}} \\ $$$${and}\:\:\:\:\:\:\:\:{f}\left(\mathrm{1}−{x}\right)−{g}\left({x}\right)={x}^{\mathrm{2}} \\ $$$${then}\:\:\:\:\:\:\:{f}\left({x}\right)=? \\ $$
Answered by A5T last updated on 22/Apr/24
$${f}\left({x}\right)+\mathrm{2}{g}\left(\mathrm{1}−{x}\right)={x}^{\mathrm{2}} …\left({i}\right) \\ $$$$\mathrm{2}{f}\left({x}\right)−\mathrm{2}{g}\left(\mathrm{1}−{x}\right)=\left(\mathrm{1}−{x}\right)^{\mathrm{2}} =\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)\Rightarrow\mathrm{3}{f}\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}\Rightarrow{f}\left({x}\right)={x}^{\mathrm{2}} −\frac{\mathrm{4}{x}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by mathlove last updated on 22/Apr/24
$${thanks} \\ $$