If the inequality ax square + BX + C > 0, the solution set is {X - 3}

If the inequality ax square + BX + C > 0, the solution set is {X - 3}

If the inequality ax square + BX + C > 0, the solution set is {X - 3}
{x| -3
The inequality ax ^ 2-2ax + 2A + 3 about X is known
If a = 0, then 30
And the minimum value is greater than or equal to 0
So there is at most one intersection with the X axis
The discriminant is less than or equal to 0
4a²-4a(2a+3)0
So a > 0
To sum up
a≥0
4a²-8a²-12a≤0
-4a²-12a≤0
a²+3a≥0
a≥0
A is less than minus 3
Function f (x) = Log1 / 2 (AX ^ 2-2x + 4), a ∈ R. Q: if function f (x) is an increasing function in (- ∞, 3), find the value range of real number a
F (x) Log1 / 2 u (x) is a decreasing function of U (x)
So only u (x) = ax & # 178; - 2x + 4 is a decreasing function in (- ∞, 3]
If u (x) > 0, it is always true [true number is greater than 0]
When a = 0, u (x) = - 2x + 4 is a decreasing function
U (x) > U (3) = - 2 does not satisfy U (x) > 0
When a ≠ 0, the opening of quadratic function should be upward, that is, a > 0
Axis of symmetry x = 1 / a ≥ 3
The minimum value u (3) = 9A - 2 > 0
The solution is 2 / 9 < a ≤ 1 / 3
The range of F (x) = 2x-1 / 3x + 4
f(x)=(2x-1)/(3x+4)
y=(2x-1)/(3x+4)
2x-1)=3yx+4y
(2-3y)x=(4y+1)
x=(4y+1)/(2-3y)
Because the function of Y makes sense,
2-3y≠0
y≠2/3
The range of the original function is:
(-∞,2/3)∪(2/3,+∞)
If the function f (x) = Log1 / 2 (AX2 ax + 4) is defined as R, then the value range of a is
F (x) = ax ^ 2 - ax + 4 > 0 for any R
When a = 0, 4 > 0 holds
If a is not zero, it is a quadratic function and holds for any R, that is to say, it has
A>0
If f (- - A / 2a) = f (1 / 2) = A / 4 - A / 2 + 4 > 0, then 0
Given f (x) = (- 3x-2) / (2x + 1), we can find the domain of definition and the range of value
•  the domain of definition of this kind of problems is mainly to consider the special case that the denominator is not 0. Under this condition, the original formula is transformed into the case that f (x) = - 2 / 3-0.5 / (2x + 1) to find the value that the denominator is 0
F (x) = - 2 / 3-0.5 / (2x 1)
Let f (x) = x power of E / 1 + ax * 2, where a is a positive real number one. When a = 4 / 3, find the extreme point of F (x),
1) If f '(x) = e ^ x {1 + (4 / 3) x ^ 2 - (8 / 3) x} / {1 + (4 / 3) x ^ 2} ^ 2
Because for the extreme point, then x = 0.5 or 1.5
The solution is x = 0.5 or 1.5
So the extreme point is x = 0.5 or 1.5
(2)f'(x)=e^x(ax^2-2ax+1)/(1+ax^2)^2
Because it is a monotone function, so as long as ax ^ 2-2ax + 1 is always greater than 0 or less than 0
When a = 0, the condition is satisfied
When a > 0, the minimum value 4ac-b ^ 2 / 4A > 0 is 0
Seeking derivative
The function y = x & # 178; - 4x + 6, when x ∈ [1,4], the value range of the function is?
A.【3,6】
B.【2,6】
C.【2,6）
D.【3,6）
y=(x-2)²+2
X = 2, y is the minimum of 2
X = 4, y max = 6, but can't be taken
SO 2 ≤ y
y=(x-2)²+2
When x = 2, there is a minimum of 2
When x = 4, y = 6.2
It is known that the domain of definition of function f (x) is r, and for all real numbers x, f (x + 2) = f (2-x), f (x + 7) = f (7-x). If f (5) = 9, then f (- 5) = find the great God
F (x + 2) = f (2-x), f (x + 7) = f (7-x), the image is symmetrical about x = 2, x = 7, the period is 10, so f (- 5) = f (5) = 9, do not understand can ask!
Points (2,1) and (1,2) are on the image of function f (x) = 2 ^ (AX = b),
2^(2a+b)=1
2^(a+b)=2
therefore
2a+b=0
a+b=1
therefore
A=1
B=0