It is known that the definition domain of function f (x) is (0, + ∞) and satisfies that f (XY) = f (x) + F (y), f (1 / 2) = 1. If 0 < x < y, there is f (x) > F (Y) Solving inequality f (x-3) + F (x + 3) ≥ - 2

It is known that the definition domain of function f (x) is (0, + ∞) and satisfies that f (XY) = f (x) + F (y), f (1 / 2) = 1. If 0 < x < y, there is f (x) > F (Y) Solving inequality f (x-3) + F (x + 3) ≥ - 2

According to f (XY) = f (x) + F (y), f (x-3) + F (x + 3) ≥ - 2 is reduced to f ((x-3) (x + 3)) > = - 2 and (x-3) > 0, (x + 3) > 0
So f (x ^ 2-9) > = - 2, that is, f (x ^ 2-9) + 2 > = 0
Because f (1 / 4) = f (1 / 2 * 1 / 2) = f (1 / 2) + F (1 / 2) = 1 + 1 = 2
Therefore, f (x ^ 2-9) + F (1 / 4) > = 0 is obtained from F (x ^ 2-9) + 2 > = 0, and then f ((x ^ 2-9) / 4) > = 0 is obtained from F (XY) = f (x) + F (Y)
Since f (1) = f (1 * 1) = f (1) + F (1), f (1) = 0, and from the meaning of the question, it is known that when x is greater than 0, the function is a subtractive function
So if f ((x ^ 2-9) / 4) > = 0, then f ((x ^ 2-9) / 4) > = f (1), so ((x ^ 2-9) / 4) 0, solve this inequality and add (x-3) > 0, (x + 3) > 0

It is known that the definition domain of function f (x) is (0, + ∞), and satisfies that f (XY) = f (x) + F (y), f (1 / 2) = 1. If for 0 < x < y, there is f (x) > F (y) (1) Find f (1) (2) Solving inequality f (- x) + F (3-x) > - 2

Take x = y = 1 and substitute it into the relationship f (1) = f (1) + F (1) to get f (1) = 0. Because f (1 / 2) = 1, f (1 / 4) = f (1 / 2) + F (1 / 2) = 20 < x < y, all have f (x) > F (y), so f (x) monotonically decreases f (- x) + F (3-x) > - 2 on (0, + 00) to get f (- x) + F (3-x) + 2 > 0f (- x (3-x) / 4) > 0 = f (1)

It is known that the definition domain of function f (x) is (0, positive infinity) and satisfies that f (XY) = f (x) + F (y), f (1 / 2) = 1. If for 0 < x < y, there is f (x) > F (y). Solve the inequality f (- x) + F (3-x) ≥ - 2

In F (XY) = f (x) + F (y), let x = y = 1
f(1)=0,
In F (XY) = f (x) + F (y), let y = 1 / X
Get f (1) = f (x) + F (1 / x) = 0
F (1 / 2) = 1, so f (2) = - 1, f (4) = f (2) + F (2) = - 2
F (- x) + F (3-x) ≥ - 2, there must be x first

It is known that f (x) is an increasing function on its definition domain (0, positive infinity), and satisfies f (XY) = f (x) + F (y), f (2) = 1, (1) find f (8) = 3 (2) if x satisfies It is known that f (x) is an increasing function on its definition domain (0, positive infinity), and satisfies f (XY) = f (x) + F (y), f (2) = 1, (1) Find f (8) = 3 (2) If x satisfies f (x) - f (X-2) > 3, find the value range of X

(1)f(8)=f(2 × 4）=f(2)+f(4)=1+f(2 × 2）=1+f(2)+f(2)=1+1+1=3
(2)
f(x)-f(x-2)>3
f(x)-f(x-2)>f(8)
f(x)>f(x-2)+f(8)=f(8x-16)
∵ f (x) is an increasing function on its definition domain (0, positive infinity)
∴x>0,x-2>0,x>8x-16
Solution 2

It is known that the domain of the increasing function y = f (x) is (0, + ∞), and satisfies that f (2) = 1, f (XY) = f (x) + F (y) Find f (1) f (4) Find the range of X satisfying f (x) + F (x-3) ≤ 2

f(2)=f(2 × 1) = f (2) + F (1) = 1. Then f (1) = 0
f(4)=f(2 × 2)=f(2)+f(2)=2
f(x)+f(x-3)=f(x^2-3x)<=2
That is, x ^ 2-3x < = 4
So - 1 < = x < = 4
Because the domain of y = f (x) is (0, + ∞)
So 0

Known function f (x) = 2x-1 2x+1． (1) Find the value range of the function; (2) Judge and prove the monotonicity of function

（1）∵2x=1+y
1-y，
And 2x > 0, i.e. 1 + y
1-y＞0，
The solution is - 1 < y < 1
The value range of function f (x) is (- 1, 1)
(2) The function f (x) is a monotonically increasing function on X ∈ R
Proof: F (x) = 2x-1
2x+1=1-2
2x+1
Take any two real numbers x1, X2 in the definition field, and X1 < x2
f(x1)-f(x2)=2(2x1-2x2)
(2x1+1)(2x2+1)
x1＜x2
∴2x1＜2x2
Thus, f (x1) - f (x2) < 0
Therefore, the function f (x) is a monotonically increasing function on X ∈ R