It is known that the odd function f (x) is a subtractive function on the definition field [- 2,2]. If f (2a + 1) + F (4a-3) > 0, find the value range of the real number a

It is known that the odd function f (x) is a subtractive function on the definition field [- 2,2]. If f (2a + 1) + F (4a-3) > 0, find the value range of the real number a

Because f (x) is an odd function,
Therefore, f (2a + 1) + F (4a-3) > 0 can be transformed into f (2a + 1) > - f (4a-3) = f (3-4A),
And f (x) is the subtraction function on the definition field [- 2,2],
So there are
2a+1<3−4a
−2≤2a+1≤2
−2≤4a−3≤2 , Solution 1
4≤a<1
3,
So the value range of real number a is 1
4≤a<1
3.

It is known that the odd function f (x) is a subtractive function on the definition field [- 2,2]. If f (2a + 1) + F (- 4a-3) > 0, find the value of the real number a It is known that the odd function f (x) is a subtractive function on the definition field [- 2,2]. If f (2a + 1) + F (- 4a-3) > 0, find the value range of the real number a

Because it's an odd function
So f (- 4a-3) = - f (4a + 3)
So f (2a + 1) > F (4a + 3) is a subtractive function
So - 2

The function f (x) = x2 + ax + 3. When x ∈ [- 2,2], f (x) ≥ A is constant. Find the value range of real number a

∵ function f (x) = x2 + ax + 3. When x ∈ [- 2,2], f (x) ≥ A is constant,
‡ (x-1) a ≥ - x2-3 is constant when x ∈ [- 2,2],
① When x ∈ (1,2],
∴a≥−x2−   three
X − 1 is constant in X ∈ (1,2]
Let g (x) = − x2 − 3
X − 1, X ∈ (1, 2], i.e. a ≥ g (x) max
∵g′(x)=−(x−3)(x+1)
(x − 1) 2, ‡ (1, 2] is the increasing range, G (2) is the largest and - 7
∴a≥-7;
② When x ∈ [- 2, 1),
∴a≤−x2−   three
X − 1 is constant in X ∈ [- 2, 1)
Let g (x) = − x2 − 3
x−1,x∈[-2,1),
I.e. a ≤ g (x) min
And G (x) = − x2 − 3
The minimum value of X − 1 on ∈ [- 2, 1) is g (- 1) = 2,
∴a≤2;
To sum up, the value range of real number A: [- 7, 2]

Given the function f (x) = (x ^ 2-ax + 3) / (2 ^ x + 1), when x ∈ [2,3], f (x) > = 0 is always true, find the value range of real number a

Because the denominator is always greater than 0, only molecules can be considered
y=x^2-ax+3=(x-a/2)^2+3-a^2/4
When 4 = A6 or a = 0, --- > A = 0, --- > A

Known inequality 1 x+1 y+m If x + y ≥ 0 is constant for any positive real number x and y, the minimum value of real number m is __

∵ inequality 1
x+1
y+m
X + y ≥ 0 is constant for any positive real number x and y,
‡ inequality (x + y) (1)
x+1
y) ≥ - M is constant for any positive real number x and y
And (x + y) (1)
x+1
y)=2+y
x+x
y≥4
‡ - M ≤ 4, i.e. m ≥ - 4
So the answer is: - 4

A function and equation problem Find the number of zeros of function y = LNX + 2x-6

From the question, we can get LNX + 2x-6 = 0, that is, LNX = 6-2x
So
Draw the images of y = LNX and y = 6-2x in the same coordinate system
It can be seen from the figure that the two images have only one intersection
So there's only one zero