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Question Number 57010 by 121194 last updated on 28/Mar/19

$$f\left(\frac{x+y}{2}\right)=\frac{f\left(x\right)f\left(y\right)}{f\left(2\right)}$$$$f\left(x\right)=?$$

Commented by maxmathsup by imad last updated on 29/Mar/19

$$x=y\Rightarrow f\left(x\right)=\frac{f^2\left(x\right)}{f\left(2\right)}withf\left(2\right)\ne 0\Rightarrow f^2\left(x\right)-f\left(2\right)\left(x\right)=0\Rightarrow f\left(x\right)\{f\left(x\right)-f\left(2\right)\}=0$$$$\Rightarrow f\left(x\right)=0off\left(x\right)=f\left(2\right)forallx.$$

Answered by kaivan.ahmadi last updated on 28/Mar/19

$$★x=y=0\Rightarrow f\left(0\right)=\frac{f^2\left(0\right)}{2}\Rightarrow f\left(0\right)=0orf\left(0\right)=2$$$$★x=0,y=2\Rightarrow f\left(1\right)=\frac{f\left(0\right)f\left(2\right)}{f\left(2\right)}=f\left(0\right)$$$$★x=1,y=1\Rightarrow f\left(1\right)=\frac{f\left(1\right)f\left(1\right)}{f\left(2\right)}\Rightarrow f\left(2\right)=f\left(1\right)=f\left(0\right)$$$$★x=y\Rightarrow f\left(x\right)=\frac{f^2\left(x\right)}{f\left(2\right)}\Rightarrow f^2\left(x\right)-2f\left(x\right)=0\Rightarrow$$$$f\left(x\right)=2$$

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