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Question Number 1130 by 123456 last updated on 22/Jun/15
f:R→R  f(xy)+f(x+y)=f(x)f(y)+f(x)+f(y)  f(−1)=?  f(0)=?  f(+1)=?
$${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({xy}\right)+{f}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right)+{f}\left({x}\right)+{f}\left({y}\right) \\ $$$${f}\left(−\mathrm{1}\right)=? \\ $$$${f}\left(\mathrm{0}\right)=? \\ $$$${f}\left(+\mathrm{1}\right)=? \\ $$
Answered by prakash jain last updated on 23/Jun/15
f(x)=x  f(x+y)+f(xy)=x+y+xy=f(x)+f(y)+f(x)f(y)  f(0)=0  f(1)=1  f(−1)=−1
$${f}\left({x}\right)={x} \\ $$$${f}\left({x}+{y}\right)+{f}\left({xy}\right)={x}+{y}+{xy}={f}\left({x}\right)+{f}\left({y}\right)+{f}\left({x}\right){f}\left({y}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${f}\left(−\mathrm{1}\right)=−\mathrm{1} \\ $$
Commented by Rasheed Soomro last updated on 15/Jul/15
You have proved yourself as an intelligent mathematition!
$${You}\:{have}\:{proved}\:{yourself}\:{as}\:{an}\:{intelligent}\:{mathematition}! \\ $$$$ \\ $$

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