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Question Number 57011 by 121194 last updated on 28/Mar/19
![f(((x+y)/2))f(((x−y)/2))=g(x) g(x+y)g(x−y)=[f(x)]^2 −[f(y)]^2 f(x),g(x)=?](Q57011.png)
$${f}\left(\frac{{x}+{y}}{\mathrm{2}}\right){f}\left(\frac{{x}−{y}}{\mathrm{2}}\right)={g}\left({x}\right) \\ $$$${g}\left({x}+{y}\right){g}\left({x}−{y}\right)=\left[{f}\left({x}\right)\right]^{\mathrm{2}} −\left[{f}\left({y}\right)\right]^{\mathrm{2}} \\ $$$${f}\left({x}\right),{g}\left({x}\right)=? \\ $$
Answered by kaivan.ahmadi last updated on 28/Mar/19
![g(x) ★x=y=0⇒f(0)f(0)=g(0) g(0)g(0)=0→g(0)=0→f(0)=0 ★x=y⇒ f(x)f(0)=g(x)⇒g(x)=0 ★y=0⇒ g(x)g(x)=[f(x)]^2 ⇒[f(x)]^2 =0⇒f(x)=0](Q57017.png)
$${g}\left({x}\right) \\ $$$$\bigstar{x}={y}=\mathrm{0}\Rightarrow{f}\left(\mathrm{0}\right){f}\left(\mathrm{0}\right)={g}\left(\mathrm{0}\right) \\ $$$${g}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)=\mathrm{0}\rightarrow{g}\left(\mathrm{0}\right)=\mathrm{0}\rightarrow{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\bigstar{x}={y}\Rightarrow \\ $$$${f}\left({x}\right){f}\left(\mathrm{0}\right)={g}\left({x}\right)\Rightarrow{g}\left({x}\right)=\mathrm{0} \\ $$$$\bigstar{y}=\mathrm{0}\Rightarrow \\ $$$${g}\left({x}\right){g}\left({x}\right)=\left[{f}\left({x}\right)\right]^{\mathrm{2}} \Rightarrow\left[{f}\left({x}\right)\right]^{\mathrm{2}} =\mathrm{0}\Rightarrow{f}\left({x}\right)=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$