Given that the function f (x) has: F (x1x2) = f (x1) + F (x2) for any x1, X2 on R, it is proved that f (x) is an even function

f(1)+f(1)=f(1*1)=f(1)
So: F (1) = 0
f(-1)+f(-1)=f((-1)*(-1))=f(1)=0
f(-1)=0
So: F (1) = f (- 1) = 0
f(-x)=f(-1)+f(x)=0+f(x)=f(x)
F (x) is an even function

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