2. The definition field of function f (x) = LG1 / (a ^ 2x-a ^ x + m) is r, and the value range of M is obtained 3. The range of function f (x) = LG (a ^ 2x-a ^ x + m) is r, and the range of M is obtained

2. The definition field of function f (x) = LG1 / (a ^ 2x-a ^ x + m) is r, and the value range of M is obtained 3. The range of function f (x) = LG (a ^ 2x-a ^ x + m) is r, and the range of M is obtained

2. According to the meaning of the title
The value of M must satisfy that for any real number x,
a^2x-a^x+m>0 ==> (a^x-1/2)^2+m-1/4>0
For any x, only M-1 / 4 > 0 is needed
That is, M > 1 / 4
3. According to the meaning of the title
The value of M must satisfy that for any real number x,
a^2x-a^x+m>0
It is easy to know that M > 1 / 4

Given that the definition field of even function f (x) is r, when x ≥ 0, f (x) = x2 + 3x-1, find the analytic expression of F (x)

Set x0
When x ≥ 0, f (x) = x2 + 3x-1,
Then f (- x) = x2-3x-1
And because f (- x) = f (x)
So when x

The function f (x) is an even function defined on R, and for any x belonging to R, f (2 + x) = - f (x). When x belongs to [0,2], f (x) = 3x + 2. Then the analytic expression of the function f (x) in the interval [- 4,0] is?

The problem is that f (x + 4) = - f (x + 2) = f (x), so it is a periodic function with period 4
We know that [0,2], f (x) = 3x + 2. By using even function, we can get f (x) = - 3x + 2 on [- 2,0]
Using the period of 4, we know that in [- 4, - 2] f (x) = f (x + 4) = 3 (x + 4) + 2 = 3x + 14