Function f (x) is an even function defined as function set R. it is an increasing function in the interval [0, + ∞). If f (m) ≥ f (- 2), then the value range of real number m? Mainly write the analysis process, which the teacher asked to write,

Function f (x) is an even function defined as function set R. it is an increasing function in the interval [0, + ∞). If f (m) ≥ f (- 2), then the value range of real number m? Mainly write the analysis process, which the teacher asked to write,

（-∞,-2]U[2,+∞）

If f (A-1) > F (2-A), then what is the value range of real number a

Even function f (x) is a decreasing function on (a ∞, 0]
When (f) is + 0, the increasing function is infinite
[（a-1)+(2-a)]/2=1/2＞0
∵ f (A-1) > F (2-A),
∴a-1＞2-a
∴a＞3/2
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If the definition domain of even function f (x) is [- 10,10], and if [0,10] is a minus function, and f (m-1) - f (2m-1) > 0, then the value range of real number m is

F (x) is an even function and a minus function in [0,10]
/ / F (x) is an increasing function on [- 10,0]
The larger the absolute value of the independent variable x, the smaller the value of F (x)
∴f(m-1)-f(2m-1)>0
f(m-1)>f(2m-1)
{-10≤m-1≤10
｛-10≤2m-1≤10
{ |m-1|

Let f (x) be monotonically decreasing on the interval [0,2] if f (1-m) is an even function on [- 2,2] Math homework help users 2017-10-19 report Use this app to check the operation efficiently and accurately!

Classified discussion
When 1-m > = 0 m > = 0, 0 is obtained=

The even function f (x) of the definition field on [- 2,2] is an increasing function on the interval [0,2], and f (1-m) < f (m), so as to find the value range of M

The even function f (x) defined on the domain [- 2,2] is an increasing function on the interval [0,2]
Then it is a minus function on [- 20]
1-mm may fall on the same side or the opposite side
(1) On the left side of the y-axis
M

It is known that even function f (x) defined on real number set R is monotonically decreasing function on interval (0, + infinity), if f (1) Math homework help users 2017-09-24 report Use this app to check the operation efficiently and accurately!

Even function f (x) defined on the set of real numbers R, then there is f (x) = f (- x) = f (| x |)
If f (1) is on the interval (0, + infinity), the decreasing function is monotone,
So, 1 > | 3x-1|
That is - 1 < 3x-1 < 1,
The solution is: 0
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If the definition domain of the function f (x) = (√ (ax-x ^ 2)) / (LG (2x-1)) is (1 / 2,1) U (1,2], then the value of real number a is

True number 2x-1 > 0
x>1/2
Denominator is not equal to 0
So 2x-1 ≠ 1
x≠1
Therefore, the radical ax-x 2 > = 0
x²-ax<=0
x(x-a)<=0
Here the solution set should get x < = 2
That is, 0 < = x < = 2
So a = 2

The known function f (x) = 1 lg(5x+4 If the definition domain of 5x + m) is r, then the value range of real number m is () A. （-3，+∞） B. （-∞，-3） C. （-4，+∞） D. （-∞，-2）

∵5x＞0，∴5x+4
If and only if 5x = 4
5x, that is, when x = log52, the equal sign is taken,
According to the fact that there is no logarithm between negative number and 0, we get: 5x + 4
When 5x + m ≥ 4 + m > 0, m ＞ - 4 is obtained,
According to the denominator is not 0, we get: 5x + 4
If 5x + m ≠ 1, let 5x = t > 0, then t + 4
t+m≠1，
∵ T > 0, ᙽ when T2 + (m-1) t + 4 = 0 has no solution or the solution is negative, T2 + (m-1) t + 4 ≠ 0,
If △ = (m-1) 2-16 ＜ 0, the solution is: - 3 ＜ m ＜ 5, and the equation has no solution, which satisfies the meaning of the question;
If T2 + (m-1) t + 4 = 0 has no positive number solution, according to the product of two is 4 > 0, two are the same sign,
Therefore, to ensure that the two are negative, the sum of the two must be 1-m < 0, and the solution m > 1,
To sum up, the range of real number m is m ＞ - 3,
Then the value range of real number m is (- 3, + ∞)
So choose a

Given that the definition domain of function f (x) = LG (x ^ 2-ax + A / 2 + 2) is all real numbers, find the value range of A

x^2-ax+a/2+2=(x-a/2)^2+a/2+2-a^2/4
Because the definition field of lgx must be x > 0, it holds for all real numbers when a / 2 + 2-A ^ 2 / 4 > 0
So we get a ^ 2-2a-8 (A-4) (a + 2) - 2

The function f (x) = LG [(a ^ 2-1) x ^ 2 + (a + 1) x + 1], if the definition field of F (x) is r, then the value range of real number a is

(a ^ 2-1) x ^ 2 + (a + 1) x + 1 > 0``
It solves the parametric inequality```
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