The domain of definition of function f (x) is d. if f (x1) ≤ f (x2) exists for any x1, X2 ∈ d when X1 < X2, then function f (x) is called non decreasing function on D, Let f (x) be a non decreasing function on [0,1] and satisfy the following three conditions: ① f (0) = 0, ② f (1-x) + F (x) = 1, ③ f (x / 3) = 1 / 2F (x), then the value of F (1 / 3) + F (5 / 12) is___________ .

The domain of definition of function f (x) is d. if f (x1) ≤ f (x2) exists for any x1, X2 ∈ d when X1 < X2, then function f (x) is called non decreasing function on D, Let f (x) be a non decreasing function on [0,1] and satisfy the following three conditions: ① f (0) = 0, ② f (1-x) + F (x) = 1, ③ f (x / 3) = 1 / 2F (x), then the value of F (1 / 3) + F (5 / 12) is___________ .

It is easy to get that f (1) = 1, f (1 / 3) = 1 / 2, f (1 / 2) = 1 / 2; because f (x) is not decreasing on [0,1], then 1 / 3

The domain of definition of function f (x) is d. if for any x 1, x 2 ∈ D, f (x 1) ≤ f (x 2) exists when x 1 ＜ x 2, then function f (x) is called non decreasing function on D
Let f (x) be a non decreasing function on [0,1] and satisfy: ① f (0) = 0; ② f (1-x) + F (x) = 1; ③ f (x / 3) = 0.5f (x), then what is the value of F (1 / 3) + f (1 / 8)?

According to f (x / 3) = 0.5f (x), if x = 1, f (1 / 3) = 0.5f (1). If f (1-x) + F (x) = 1, if x = 1, f (1-1) + F (1) = 1, namely f (0) + F (1) = 1, if f (0) = 0, f (1) = 1, if f (1) = 1 and f (1 / 3) = 0.5f (1), f (1 / 3) = 1 / 2, if x = 1 / 3, f (1 / 9) = 1 / 4, if f (1-x) + F (x) = 1

The definition field of function f (x) is D, if for any x1, X2 belongs to D, if x1

F (1-x) = 1-f (x). When x = 0, f (1) = 1-f (0) = 1
F (1-x) = 1-f (x), when x = 1 / 2, we can get f (1 / 2) = 1-f (1 / 2), we can get f (1 / 2) = 1 / 2
F (x / 3) = 1 / 2F (x). When x = 1, f (1 / 3) = 1 / 2F (1) and f (1 / 3) = 1 / 2 can be obtained
F (x / 3) = 1 / 2F (x), when x = 1 / 2, we can get f (1 / 6) = 1 / 2F (1 / 2), = 1 / 4
F (x / 3) = 1 / 2F (x), when x = 1 / 3, we can get f (1 / 9) = 1 / 2F (1 / 3), = 1 / 4
By being X1 on [0,1]

The definition field of function f (x) is d = {x | x ≠ 0}. For any x 1, x 2 ∈ D, f (x 1 · x 2) = f (x 1) + F (x 2)
If x > 1, f (x) > 0, it is proved that f (x) is an increasing function in the interval (0, + ∞)

Let x1x2 = x, from the equation: F (x) = f (x1) + F (x / x1), that is, f (x) - f (x1) = f (x / x1)
Let x2 > X1 > 0, then x2 / X1 > 1, f (x2 / x1) > 0
So f (x2) - f (x1) = f (x2 / x1) > 0
So f (x2) > F (x1)
So it's an increasing function