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

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)

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

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

f(x)=g(x)−f(−x) f(−x)=g(−x)−f(x) =g(−x)−[g(x)−f(−x)] =g(−x)−g(x)+f(−x) ⇒g(−x)=g(x)

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

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

g(x)=f(x)+g(−x) ∵g(−x)=g(x) ∴f(x)=0 f(x)=g(−x)−f(−x)⇒g(−x)=g(x)=0 f(x)=0 g(x)=0

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

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