If the nonzero function f (x) has f (a + b) = f (a) &; f (b) for any real number A.B, and if x1, (1) prove that f (x) > 0 (2) prove that f (x) is a decreasing function (3) solve the inequality f (x-3) &; f (6-2x) ≤ 1 / 4 when f (4) = 1 / 16

If the nonzero function f (x) has f (a + b) = f (a) &; f (b) for any real number A.B, and if x1, (1) prove that f (x) > 0 (2) prove that f (x) is a decreasing function (3) solve the inequality f (x-3) &; f (6-2x) ≤ 1 / 4 when f (4) = 1 / 16

(1) Let a = 0, then f (0 + b) = f (0) &; f (b), that is, f (b) = f (0) &; f (b), because f (x) is a non-zero function, f (b) ≠ 0,
So f (0) = 1 > 0;
Let x < 0, then f (x) > 1 > 0, and - x > 0,
1=f(0)=f(x+(-x))=f(x)•f(-x),
So f (- x) = 1 / F (x) > 0
In conclusion, for X ∈ R, f (x) > 0
(2) Let a < B, then A-B < 0, f (a-b) > 1, and from (1), f (b) > 0,
So f (a) - f (b) = f ((a-b) + b) - f (b) = f (a-b) · f (b) - f (b) = f (b) [f (a-b) - 1] > 0
So f (x) is a decreasing function of R
(3) F (4) = f (2 + 2) = f (2) · f (2) = 1 / 16, because f (x) > 0, f (2) = 1 / 4
F (x-3) &; f (6-2x) ≤ 1 / 4 is equivalent to
f[(x-3)+(6-2x)]≤f(2)
It is known from (2) that f (x) is a decreasing function of R, so
(x-3)+(6-2x)≥2
The solution is x ≤ 1