Given that the function f (x) satisfies f (x + y) + F (X-Y) = 2F (x) · f (y) & nbsp; (x ∈ R, y ∈ R), and f (0) ≠ 0, we try to prove that f (x) is an even function

It is proved that: let x = y = 0 ∵ f (x + y) + F (X-Y) = 2F (x) · f (y) ∵ f (0) ∵ f (0) ≠ 0, ∵ f (0) = 1, let x = 0 ∵ f (x + y) + F (X-Y) = 2F (x) · f (y) ∵ f (y) + F (- y) = 2F (0) · f (y) ∵ f (- y) = f (y), that is, f (x) is even function

Given that the function f (x) satisfies f (x + y) + F (X-Y) = 2F (x) · f (y) & nbsp; (x ∈ R, y ∈ R), and f (0) ≠ 0, we try to prove that f (x) is an even function

Given that the function f (x) satisfies f (x + y) + F (X-Y) = 2F (x) · f (y) & nbsp; (x ∈ R, y ∈ R), and f (0) ≠ 0, we try to prove that f (x) is an even function

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