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Higher one super hard function The function defined in satisfies that f (x) is equal to y (x) plus f (y)
The function (x) defined on the set of real numbers satisfies that f (x) is equal to f (x) plus f (y). It is proved that f (x) is an odd function
Prove: let x = 0, y = 0, f (0) = f (0) + F (0)
So f (0) = 0
For any real number x, Let f (0) = f (x-x) = f (x) + F (- x)
So f (- x) = - f (x)
So f (x) is an odd function
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