The function defined on R is y = f (x), and f (0) is not equal to 0. When x > 0, f (x) > 1, and for any a and B, f (0) = 1 is proved by F (a + b) = f (a) * f (b) (1) The function defined on R is y = f (x), f (0) is not equal to 0, when x > 0, f (x) > 1, and for any a, B, f (a + b) = f (a) * f (b) (1) Prove that f (0) = 1 (2) prove that for any x belongs to R, f (x) > 0 (3) prove that f (x) is an increasing function on R. 4) if f (x) * f (2x-x2) > 1, find the value range of X

The function defined on R is y = f (x), and f (0) is not equal to 0. When x > 0, f (x) > 1, and for any a and B, f (0) = 1 is proved by F (a + b) = f (a) * f (b) (1) The function defined on R is y = f (x), f (0) is not equal to 0, when x > 0, f (x) > 1, and for any a, B, f (a + b) = f (a) * f (b) (1) Prove that f (0) = 1 (2) prove that for any x belongs to R, f (x) > 0 (3) prove that f (x) is an increasing function on R. 4) if f (x) * f (2x-x2) > 1, find the value range of X

1. Because f (a + b) = f (a) f (b), let a = b = 0 get: F (0) = f (0) * f (0), because f (0) is not equal to 0, so f (0) is deleted from both equations, and f (0) = 1
2. Let f (a + b) = f (a) f (b) where a = b = x / 2, then f (x) = f (0.5x) * f (0.5x) = (f (0.5x)) ^ 2 > = 0. If a = x, B = - x, then f (0) = f (x) * f (- x), because f (0) is not equal to 0, for any f (x) and f (- x), f (x) is not equal to 0, so f (x) > 0
3. Let X1 > X2, for any x belongs to R, always f (x) > 0, so f (x1) / F (x2) = f (x1 + x2-x2) / F (x2) = (f (x1-x2) * f (x2)) / F (x2), and the numerator denominator is reduced to f (x2), so f (x1) / F (x2) = f (x1-x2), because X1 > X2, so x1-x2 > 0, so f (x1-x2) > 1, so f (x1) / F (x2) > 1, so f (x1) > F (x2), so f (x) is an increasing function on R
4. F (x) * f (2x-x Square) = f (3x-x ^ 2) > 1, because when x > 0, f (x) > 1, f (x) is an increasing function on R, so only when 3x-x ^ 2 > 0, there will be f (x) * f (2x-x Square) > 1, at this time, 0