f-R-R-g-R-R-f-x-g-x-f-x-g-x-f-x-g-x-f-x-g-x-f-x-g-x-f-x-f-x-g-x-g-x-f-x-g-x-f-x-g-x-

Question Number 874 by 123456 last updated on 08/Apr/15

$${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)={g}\left({x}\right)−{f}\left(−{x}\right) \\ $$$${g}\left({x}\right)={f}\left({x}\right)+{g}\left(−{x}\right) \\ $$$${f}\left({x}\right)=? \\ $$$${g}\left({x}\right)=? \\ $$$${f}\left({x}\right)+{g}\left({x}\right)=? \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\overset{?} {=}\boldsymbol{{f}}\left(−\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)\overset{?} {=}\boldsymbol{{g}}\left(−\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)\overset{?} {=}\boldsymbol{{f}}\left(−\boldsymbol{{x}}\right)\boldsymbol{{g}}\left(−\boldsymbol{{x}}\right) \\ $$

Commented by prakash jain last updated on 08/Apr/15

$${f}\left({x}\right)={g}\left({x}\right)−{f}\left(−{x}\right) \\ $$$${f}\left(−{x}\right)={g}\left(−{x}\right)−{f}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:={g}\left(−{x}\right)−\left[{g}\left({x}\right)−{f}\left(−{x}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:={g}\left(−{x}\right)−{g}\left({x}\right)+{f}\left(−{x}\right) \\ $$$$\Rightarrow{g}\left(−{x}\right)={g}\left({x}\right) \\ $$

Answered by prakash jain last updated on 08/Apr/15

$${g}\left({x}\right)={f}\left({x}\right)+{g}\left(−{x}\right) \\ $$$$\because{g}\left(−{x}\right)={g}\left({x}\right) \\ $$$$\therefore{f}\left({x}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)={g}\left(−{x}\right)−{f}\left(−{x}\right)\Rightarrow{g}\left(−{x}\right)={g}\left({x}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)=\mathrm{0} \\ $$$${g}\left({x}\right)=\mathrm{0} \\ $$

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