The function f (x) defined on a positive real number belongs to a positive real number for any m and N, and f (MN) = f (m) + F (n) holds. When x > 1, f (x) < 0. It is proved that f (x) is a subtractive function on a positive real number

The function f (x) defined on a positive real number belongs to a positive real number for any m and N, and f (MN) = f (m) + F (n) holds. When x > 1, f (x) < 0. It is proved that f (x) is a subtractive function on a positive real number

Assuming X1 > x2 > 1, X1 / x2 > 1
Then f (x1) = f (x2 * X1 / x2) = f (x2) + F (x1 / x2)
Then f (x1) - f (x2) = f (x1 / x2) < 0, then f (x) is a subtractive function on (1, + ∞)
Assuming 1 > X1 > x2 > 0, X1 / x2 > 1
Then f (x1) = f (x2 * X1 / x2) = f (x2) + F (x1 / x2)
Then f (x1) - f (x2) = f (x1 / x2) < 0, then f (x) is the subtraction function on (0,1)
Because f (MN) = f (m) + F (n) makes n = 1
Then f (MN) = f (m) + F (n), that is, f (m) = f (m) + F (1), then f (1) = 0
To sum up, we can know that f (x) is a subtractive function on (0, + ∞)

Judging the parity of function f (x) = root sign (1-x ^ 2) / |x + 3| - 3

Definition field 1-x ^ 2 > = 0
x^2<=1
-1<=x<=1
So x + 3 > = 0
|x+3|=x+3
So denominator = x + 3-3 = X
f(x)=√(1-x^2)/x
f(-x)=-√(1-x^2)/x=-f(x)
Is an odd function

Find the monotonicity of F (x) with known function f (x) = log2 (root sign (x ^ 2 + 1) - x)

 
 

2 ^ x = 1 / 2 is known. Find the maximum and minimum values of function f (x) = log2 (x / 2) * log root 2 (root X / 2)

2^x≤256=2^8
x≤8
Log (2) x ≥ 1 / 2, get x ≥ √ 2
So the X range is [√ 2,8]
Function f (x) = log2's (x / 2) * log √ 2's (√ X / 2)
=Log2 (x / 2 * x / 4)
=Log2 (x ^ 2 / 8)
When x = √ 2, f (x) min = - 2
When x = 8, f (x) max = 3

The domain of function f (x) = log2 (x / 2) multiplied by log2 (x / 4) is the solution set of inequality 2 (Log1 / 2 (x)) * 2 + 7log1 / 2 (x) + 3 Find the maximum and minimum values of F (x),

Where are the inequalities? In addition, does finding the maximum and minimum of F (x) have anything to do with the inequality? Please confirm the title

Let f (x) and G (x) be singular functions defining the domain R, the solution set of inequality f (x) > 0 (m, n), and the solution set of inequality g (x) > 0 be (M / 2, N / 2), where 0

F (x) * g (x) > 0, then f (x) > 0 and G (x) > 0, case 1 or F (x) < 0 and G (x) < 0, case 2. When case 2, the solution set can be obtained as (m, N / 2) according to known conditions. When case 2, because f (x) and G (x) are odd functions defining domain R, f (- x) = - f (x), then the solution set of F (x) < 0 is (- N, - M) and the solution set of G (x) < 0 is (- n