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The function y = f (x) defined on R, when x ＞ 0, f (x) ＞ 1, and for any a, B belongs to R, f (a + b) = f (a) f (b), (1) Find f (0) = 1; (2) Proof: for any x belongs to R, f (x) > 0 (3) It is proved that f (x) is an increasing function on R; (4) If f (x) · f (2x-x2) > 1, find the value range of X
Let a = b = 0, f (0) = f (0) ^ 2, so f (0) = 0 or 1. Let B = 0, a > 0, f (a) = f (0) f (a) > 0, so f (0) = 1.2, any x > 0, then f (0) = f (x) f (- x), f (x) > 0, so f (- x) > 0, for any x belongs to R, there is always f (x) > 0.3, any X11, so f (x2) > F (x1), f (x) is an increasing function on r.4
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