For the function f (x), f (x) is not zero, and X1 and X2 belong to any real number on R, there is f (x1 + x2) = 2 * f (x1) * f (x2). It is proved that f (x) is an even function

f(x)=1/2(e)^x
Then the condition of the title is satisfied
But f (x) is not even function
There is something wrong with the title

RELATED INFORMATIONS

1. Trigonometric function symmetry If f (x) = sin (π / 4 x + π / 6), G (x) is symmetric with respect to x = 1, then G (x) is obtained There is also an analytic expression of y = 3sin (π / 2x + π / 6) with respect to x = 2 π symmetry
2. Trigonometric function symmetry If the image of y = sin (x + a) + radical three cos (x-a) is symmetric about the Y axis, then the value of a is Why is it wrong to calculate with x = 0, y = plus or minus 2? The answer is a = k π - π / 6
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12. Let y = f (x) (x belongs to R) satisfy f (x1) + F (x2) = f (x1 * x2) for any real number x1, X2, and prove that (1) f (1) = f (- 1) = 0 (2) f (x) is even function
13. Function f (x), X belongs to R, if for any real number x1, X2 has f (x1 + x2) + F (x1-x2) = 2F (x1) f (x2), prove that f (x) is even function
14. Let f (x) (x ∈ R) satisfy f (x1 * x2) = f (x1) + F (x2) for any real number x1, X2, and prove that f (x) is an even function
15. It is known that the function f (x) is an even function on the set of real numbers R, and is a decreasing function on (0, + ∞). If x 10, then there is A.f（-x1）>f(-x2) B.f(-x1)=f(-x2) c.f(-x1)
16. Given the function y = F & nbsp; (x) (x ∈ R, X ≠ 0), for any non-zero real number x1, X2, f (x1x2) = f (x1) + F (x2), try to judge the parity of F (x)
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