Question Number 192084 by sciencestudentW last updated on 07/May/23
$${f}\left({x}\right)+{x}\centerdot{f}\left(−{x}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${f}\left(\sqrt{\mathrm{2}}\right)=? \\ $$
Answered by AST last updated on 07/May/23
$${x}^{\mathrm{2}} +\mathrm{1}=−\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)+\mathrm{2}=−\left(\mathrm{1}+{x}−{x}\left(\mathrm{1}+{x}\right)\right)+\mathrm{2} \\ $$$$=\mathrm{1}−{x}+{x}\left(\mathrm{1}+{x}\right)={f}\left({x}\right)+{x}\centerdot{f}\left(−{x}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{1}−{x}\Rightarrow{f}\left(\sqrt{\mathrm{2}}\right)=\mathrm{1}−\sqrt{\mathrm{2}} \\ $$
Answered by York12 last updated on 07/May/23
$${f}\left({x}\right)+{xf}\left(−{x}\right)={x}^{\mathrm{2}} +\mathrm{1}\:\rightarrow\:{xf}\left({x}\right)+{x}^{\mathrm{2}} {f}\left(−{x}\right)={x}^{\mathrm{3}} +{x}\:\rightarrow\left({I}\right) \\ $$$${f}\left(−{x}\right)−{xf}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{1}\:\rightarrow\left({II}\right) \\ $$$${summing}\:{I}\:{and}\:{II} \\ $$$${we}\:{get}\:: \\ $$$${f}\left(−{x}\right)=\frac{\left({x}^{\mathrm{4}} −\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}={x}+\mathrm{1}\:\rightarrow\:{f}\left({x}\right)=\mathrm{1}−{x} \\ $$$$\therefore\:{f}\left(\sqrt{\mathrm{2}}\right)=\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\:\rightarrow\:\left({That}'{s}\:{it}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{\mathscr{BY}\:\:{YORK}\right\} \\ $$$$ \\ $$
Answered by mehdee42 last updated on 07/May/23
$${x}=\sqrt{\mathrm{2}}\rightarrow{f}\left(\sqrt{\mathrm{2}}\right)+\sqrt{\mathrm{2}}{f}\left(−\sqrt{\mathrm{2}}\right)=\mathrm{3}\:\:\:\left({i}\right) \\ $$$${x}=−\sqrt{\mathrm{2}}\rightarrow{f}\left(−\sqrt{\mathrm{2}}\right)−\sqrt{\mathrm{2}}{f}\left(\sqrt{\left.\mathrm{2}\right)}=\mathrm{3}\:\:\:\overset{×−\sqrt{\mathrm{2}}} {\Rightarrow}−\:\sqrt{\mathrm{2}}{f}\left(−\sqrt{\mathrm{2}}\right)+\mathrm{2}{f}\left(\sqrt{\mathrm{2}}\right)=−\mathrm{3}\sqrt{\mathrm{2}}\:\:\left({ii}\right)\right. \\ $$$$\left({i}\right)+\left({ii}\right)\rightarrow\mathrm{3}{f}\left(\sqrt{\mathrm{2}}\right)=\mathrm{3}−\mathrm{3}\sqrt{\mathrm{2}}\Rightarrow{f}\left(\sqrt{\mathrm{2}}\right)=\mathrm{1}−\sqrt{\mathrm{2}}\:\:\checkmark \\ $$
Answered by gatocomcirrose last updated on 08/May/23
$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}+\mathrm{b} \\ $$$$\mathrm{ax}+\mathrm{b}−\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}=\mathrm{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\Rightarrow\mathrm{a}=−\mathrm{1},\:\mathrm{b}=\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=−\mathrm{x}+\mathrm{1}\Rightarrow\mathrm{f}\left(\sqrt{\mathrm{2}}\right)=\mathrm{1}−\sqrt{\mathrm{2}} \\ $$