For any function f (x) defined on (0, positive infinity), n belongs to (0, positive infinity), f (MN) = f (m) + F (n) holds. When x > 1, f(x)＜0. (1) Calculation f (1) (2) It is proved that f (x) is a subtractive function on (0, + infinity) (3) When f (2) = - 1 / 2, find f (x) ²- Value range of variable X of 3x) > - 1

For any function f (x) defined on (0, positive infinity), n belongs to (0, positive infinity), f (MN) = f (m) + F (n) holds. When x > 1, f(x)＜0. (1) Calculation f (1) (2) It is proved that f (x) is a subtractive function on (0, + infinity) (3) When f (2) = - 1 / 2, find f (x) ²- Value range of variable X of 3x) > - 1

m=n=1
f(1*1) = f(1)+f(1)
f(1) = 0
Let X11, f (a) < 0
There is f (x2) = f (a) + F (x1) < f (x1)
So f (x) is a subtractive function
f(2)=-1/2
f(4) = f(2)+f(2) = -1
f(x^2 -3x) > -1
x^2 - 3x < 4
(x+1)(x-4) < 0
-1 < x < 4

The function f (x) defined on R satisfies f (x + 4) = f (x). When 2 ≤ x ≤ 6, f (x)= ½^ I x-m I + N, f (4) = 31, find the value of M, n

∵ f(x+4)=f(x) ∴ f(6)=f(2)
When 2 ≤ x ≤ 6, f (x) =^ Ⅰx-mⅠ+n
∴ ^Ⅰ6-mⅠ+n=?^ Ⅰ2-mⅠ+n ,∴ m=4
∵ f (4) = 31, ^ Ⅰ 4-4 Ⅰ + n = 31, ∵ n = 30

The function f (x) defined on R satisfies f (x + 4) = f (x). When 2 ≤ x ≤ 6, f (x) = (1 / 2) ^|x-m| + N, f (4) = 31 Find the value of M, n

Meaning of the question:
f(6)=f(2+4)=f(2)
therefore
[(1/2)^|6-m|]+n=[(1/2)^|2-m|]+n
|6-m|=|2-m|
It is discussed that M = 4
Substitute f (4) = 31 into f (4) = (1 / 2) ^|0-0| + n = 1 + n
n=30
So m = 4, n = 30

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