The function f (x) is an even function over the domain R, and it decreases monotonically on X ∈ (0, + ∞), Then the solution set of the inequality f (x) ≥ f (- 2) is?

-2

If the domain of function y = f (x) is [0,2], then what is the domain of function g (x) = X-1 / F (2x)

Because the definition field of F (x) is:
[0,2]
By substituting x with 2x, we can get the following results
0≤2x≤2
Then the definition domain of F (2x) is obtained
[0,1]
Since g (x) = f (2x) / (x-1)
Therefore:
x≠1
In conclusion, the definition domain of G (x) is as follows:
[0,1)

Let y = f (x) be defined in R, and satisfy f (x + y) = f (x) + F (y), f (1 / 3) = 1, and when x > 0, f (x)

1. Let x = y = 0, f (0 + 0) = f (0) + F (0), i.e. f (0) = 2F (0) get f (0) = 02, let x = - y, i.e. y = - x substitute into f (x-x) = f (x) + F (- x), i.e. f (x) = f (x) + F (- x) shift term to get f (- x) = - f (x) is odd function 3, because f (1 / 3) = 1, let x = y = 1 / 3 substitute into f (1 / 3 + 1 / 3) = f (1 / 3) + F (1 / 3) = 2, i.e. f (2 / 3) = 2 (

The domain of the function f (2x + 3) is [- 4,5]. Find the domain of F (2x-3)
I don't understand why 2x-3 can replace 2x + 3?

The domain of F (2x + 3) is [- 4,5), that is - 4 ＜ = x ＜ 5, then - 5 ＜ = 2x + 3 ＜ 13
Because both 2x-3 and 2x + 3 are two elements of function f (x), 2x-3 can replace 2x + 3
So - 5 ＜ = 2x-3 ＜ 13, so - 1 ＜ = x ＜ 8

Given that the domain of F (x + 3) is [- 4,5], the domain of F (2x-3) is___

The domain of F (x + 3) is [- 4,5], which is derived from the domain of F (x)
For example, if f (x) domain is [- 1,8], then f (x + 3) domain is [- 4 + 3,5 + 3] = [- 1 + 8]
Since the domain of F (x) is [- 1,8], then the domain of F (2x-3) is defined by
－1《2x-3《8 1《x《11/2
Then the domain of F (2x-3) is [1,11 / 2]

Given the function FX = ln (x) - ax (a ∈ R) 1. When a = 2, find the monotone interval of FX. 2. When a > 0, find the minimum value of FX on [1,2]

The function FX = ln (x) - ax (a ∈ R) 1 is known. When a = 2, find the monotone interval of FX. 2. When a > 0, find the minimum value of FX on [1,2] (1) analytic: ∵ if the function FX = ln (x) - ax (a ∈ R) makes a = 2, then the function FX = ln (x) - 2x makes f '(x) = 1 / X-2 = 0 = = > x = 1 / 2F' (x) = - 1 / x ^ 2x = 1 / a ∵ the function f (x) is in X

It is known that the square of the function f (x) = ax + B / 1 + X is an odd function over (- 1,1) and f (1) = 1 / 2

(1) ∵ the function f (x) = (AX + b) / (1 + x ^ 2) is an odd function over the domain (- 1,1); ∵ f (0) = b = 0, f (1 / 2) = A / 2 = 1 / 2, ∵ a = 1, f (x) = x / (1 + x ^ 2); [x0d (2) let 0 & lt; X1 & lt; x2 & lt; 1, (x0df (x1) - f (x2) = (x1-x2) (1-x1x2) / [(1 + X1 ^ 2) (1 + x2 ^ 2)] & lt; 0

Given that the function f (x) = ax + B / 1 + x square is an odd function over (- 1,1) and f (1 / 2) = 2 / 5, the analytic expression of F (x) is obtained
Quick solution, it's due tomorrow. I have to sleep

Given that f (x) is an odd function, f (x) = f (- x), ax + B / 1 + x ^ 2 = - ax + B / 1 + x ^ 2 deduces B = 0, then f (1 / 2) = 2 / 5 is brought into f (x) = ax to get a = 4 / 5, so f (x) = 4 / 5x

We know that the domain of F (x) is a closed interval from zero to one. We can find the domain of F (x + a) + F (x-a)
come on.

The domain of F (x) is a closed interval from zero to one
0≤x+a≤1,-a≤x≤1-a
0≤x-a≤1,a≤x≤1+a
If: 1-a1 / 2
Or, 1 + A

If the domain of F (x-1) is a closed interval 1 to 2, then the domain of F (x) is

∵1≤x≤2;
∴0≤x-1≤1;
The definition field of F (x) is [0,1]
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
If you don't understand this question, you can ask,

The function f (x) is an even function over the domain R, and it decreases monotonically on X ∈ (0, + ∞), Then the solution set of the inequality f (x) ≥ f (- 2) is?