Given the complete set u = R, nonempty set a = {x | (X-2) / [x - (3a + 1)]

Given the complete set u = R, nonempty set a = {x | (X-2) / [x - (3a + 1)]

1) When = 1 / 2, a = {x | x is greater than 2 and less than 2.5}; b = {x | x is greater than 0.5 and less than 2.25} then cub = {x is less than or equal to 0.5 or greater than or equal to 2.25}, (cub) ∩ a = {x | x is greater than or equal to 2.25 and less than 2.5} 2)
Let the complete set be a set of real numbers and a = {x | x | X|
The complement of B is x ≥ a
A is - 3 ≤ x ≤ 3
So a < - 3
The tangent of the image of the function f (x) = ax-6 / x ^ 2 + B at the point (- 1, f (- 1)) is known in the process of finding the detailed solution
High school derivative, detailed solution process
Given that the tangent equation of the image of the function f (x) = ax-6 / x ^ 2 + B at the point (- 1, f (- 1)) is x + 2Y + 5 = 0, find the analytic expression of y = f (x)
The analytic formula F (x) = 23 / 2x-6 / x ^ 2 + 33 / 2, the abscissa of the tangent point is substituted into the tangent, and the ordinate of the tangent point is obtained; - 1 + 2Y + 5 = 0, y = - 1; the tangent point is (- 1, - 1); the tangent slope is (- 1 / 2); a relation is obtained by substituting it into f (x); - 1 = - a-6 + B; A-B = - 5; derivation; f '(x) = a + 12 / x ^ 3; f' (- 1) = A-12 = - 1 / 2; a = 23 / 2; b = 33
Detailed explanation: F '(x) = a + 12 / x ^ 3,
f′（-1）=a-12=-1/2,
The tangent equation at (- 1, f (- 1)) is y = f ′ (- 1) (x + 1) + F (- 1)) = - 1 / 2 (x + 1) - a-6 + B
A and B can be obtained by combining known tangent equations
There may be a problem with the title, but there must be no problem with the method
Derivation Bring in x = - 1 An equation with AB is obtained. Compared with the tangent equation, AB is obtained
If f (x) = x / [(x-1) × (2x + a)] is an odd function, then a is a real number=
The characteristic of odd function is f (x) = - f (- x)
f（-x）=（-x）/(-x+1)(-2x+a)=[-x/(2x^2+(2-a)x-a]
And f (x) = x / [(2x ^ 2 - (2-A) x-a]
To make f (x) = - f (- x)
There should be 2-A = - (2-A)
So a = 2
Hope to be helpful to Ning. If you have any questions, please ask
F (- x) = - f (x) is an odd function. If the coefficients of corresponding terms on both sides of the equation with - X / [(- x-1) * (- 2x + a)] = - X / [(x-1) * (2x + a)] are equal, a = 2 can be obtained
Given the function FX = alnx-ax-3 (a belongs to R), find the monotone interval of function FX
f'(x)=a/x-a
=(a-ax)/x
=a(1-x)/x
The domain is x > 0
When a > 0
Let f '(x) > = 0
Zero
F (x) = x & # 178; - 2aX, X belongs to [- 2,2], find the range of this function
f(x)=x²-2ax=(x-a)²-a²
When a ≤ - 2, the range is [f (- 2), f (2)]
When - 2 < a ≤ 0, the range is [f (a), f (2)]
When 0 ＜ a ≤ 2, the range is [f (a), f (- 2)]
When a > 2, the range is [f (2), f (- 2)]
4(1+a)≤f(x)≤4（1-a)
If the function f (x) = x & # 178; + (a + 2) x + 3, where x is on [a, b] (a
If the image of quadratic function is symmetric with respect to the line x = 1, then the axis of symmetry of F (x) = x & # 178; + (a + 2) x + 3 is x = 1
That is - (a + 2) / 2 = 1  a = - 4
The maximum value of function f (x) in the interval [a, b] is f (- 4) = 27
The minimum value is f (1) = 2
The line x = 1 is the symmetry axis of quadratic function
Symmetry axis formula x = - B / (2a)
∴1=-(a+2)/2
a=-4
f(x)=x^2-2x+3=(x-1)^2+2 x∈[-4,b]
∵ on x = 1 symmetry
∴b=6
∴max=27 min=2
Y = x & # 178; - 2aX + 3, find the range of Y in [- 1,2]
I know there are three situations, but I don't know how to do it
And the teacher said that when - 1 < a < 2, there are three situations I'm dizzy. Please explain the steps as clearly as possible
Y = x & # 178; - 2aX + 3 is a quadratic function, the axis of symmetry is x = a, Let f (x) = x & # 178; - 2aX + 3
In three cases
① A > = 2, then it is monotonically decreasing in [- 1,2], the maximum is f (- 1), and the minimum is f (2)
② A = f (- 1), the maximum is f (2)
② A ∈ (1 / 2,2), where a is closer to 2, then f (2)
If the image of the function f (x) is symmetric to the x power image of G (x) = 2 with respect to the line y = x, then the maximum value of F (4x-x2) is
If the image of function f (x) and the image of power X of G (x) = 2 are symmetric with respect to the line y = x, then the maximum value of F (4x-x & # 178;) is (2). Why?
F (x) is the inverse function of G (x). Find out f (x) = log2 (4x-x ^ 2)
-X ^ 2 + 4x = - (x + 2) ^ 2 + 4, when x = 2, take the maximum value as 4, bring in, f (x) max = 2
It is known that the range of the function f (x) = ax2-2ax + 2 + B (a > 0) in the interval [2,3] is [2,5] (I) find the value of a and B; (II) if the function g (x) = f (x) - (M + 1) x in the interval [2,4] is monotone, find the range of the real number M
(I) ∵ a ＞ 0, so the opening of parabola is upward and the axis of symmetry is x = 1. The function f (x) increases monotonically on [2,3]. From the condition, f (2) = 2F (3) = 5, that is, 2 + B = 23a + 2 + B = 5, the solution is a = 1, B = 0. So a = 1, B = 0
(1) When a > 0, a = 1, B = 0
When a