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 belongs to R, f (a + b) = f (a) f (b) (1) , prove that f (0) = 1; (2), prove that for any x belonging to R, f (x) > 0; (3), prove that f (x) is an increasing function on R; (4), if f (x) * f (2x-x Square) > 1, find X

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 belongs to R, f (a + b) = f (a) f (b) (1) , prove that f (0) = 1; (2), prove that for any x belonging to R, f (x) > 0; (3), prove that f (x) is an increasing function on R; (4), if f (x) * f (2x-x Square) > 1, find X

1. Because f (a + b) = f (a) f (b), let a = b = 0 in the formula: F (0) = f (0) * f (0), because f (0) is not equal to 0, the two equations eliminate f (0) at the same time, and get: 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