If there is a proposition that x belongs to R such that x2 + (A-1) x + 1

If there is a proposition that x belongs to R such that x2 + (A-1) x + 1

∵x∈R
∴△＜0
△=（a-1）^2-4*1*1＜0
a^2-2a+1-4＜0
a^2-2a-3＜0
（a+1）（a-3）＜0
∴-1＜a＜3
If the proposition "existence x belongs to R, x ^ 2 + (A-1) x + 1
The above proposition is false
It is shown that for any formula over x, it must be greater than 0
If the discriminant is less than 0, the range of a can be obtained
-1
If there is a real number such that | x-a | + | X-1 | ≤ 3, then the value range of real number a is
|X-a | denotes the distance from X to point a, and | X-1 | denotes the distance from X to 1
－2≤a≤4
－2≤a≤4
Given the function f (x) = - x2 + ax, X "1, AX-1, x > 1, if there are x1, X2, and X1 is not equal to X2, so that f (x1) = f (x2) holds, then the value range of real number a is
In the domain of definition, f (x) is not monotonic
According to the situation, the discussion is as follows:
1）x
Let f (x) = AX2 + BX + C + (a > 0) and f (1) = - A / 2, prove that f (x) has two zeros
Please
solution
f(1)=a+b+c=-a/2
therefore
b+c=-3a/2
Because a is greater than 0, - A / 2 is less than zero, the domain of this function is r, the opening of the image is upward, x = 1 is less than zero, so there are two zeros
The monotone increasing interval of function y = XX2 − 3x + 2 is ()
A. (- 2,1) ∪ (1,2) B. (- 2,1) and (1,2) C. (- 2,2) d. (- 2,1) ∪ (1,2)
Y ′ = x2 − 3x + 2 − x (2x − 3) (x2 − 3x + 2) 2 = 2 − X2 (x2 − 3x + 2) 2; the solution 2 − x2 ＞ 0x2 − 3x + 2 ≠ 0 is: − 2 ＜ x ＜ 1, or 1 ＜ x ＜ 2
It is known that X1 and X2 are the two zeros of function f (x) = ax ^ 2 + BX + 1 (a > 0). The minimum value of function f (x) is - A, P = {x | f (x)
The minimum value of F (x) = ax ^ 2 + BX + 1 is (4a-b ^ 2) / 4A = - A
Given f (x) = (X-2) square, X ∈ [- 1,3], find the monotone decreasing interval of function f (x + 1)
The domain of definition is [- 2,2]
Why is the monotone decreasing interval [- 2,1)
∵ f (x) = (X-2) & # 178; ∵ f (x + 1) = (x + 1-2) & # 178; = (x-1) & # 178; f (x + 1) the symmetry axis X = 1 of the image has the opening upward and the lowest point, i.e. (1,0). According to these, we can draw the image and get: (- ∞, 1) monotone decreasing, (1, + ∞) monotone increasing; we also know that the definition domain is [- 2
Move the original image one unit to the left, and the abscissa of the lowest point becomes 1, so the monotonic decreasing interval is [- 2,1]. When discussing increasing or decreasing the interval, the influence of the opening and closing interval is small, and generally the former is taken before the latter is not taken
If f (x) = (X-2) ^ 2, X ∈ [- 1,3], then f (x + 1) = (x-1) ^ 2 = x ^ 2-2x + 1, X ∈ [- 2,2]
The axis of symmetry is x = - 1, and the opening is upward
Let me explain why the domain is [- 2,2]
For X, f (x) = (X-2) ^ 2, X ∈ [- 1,3], and f (x + 1) = (x-1) ^ 2, the range of X + 1 is equivalent to the [- 1,3] solution of the original X
If f (x) = (X-2) ^ 2, X ∈ [- 1,3], then f (x + 1) = (x-1) ^ 2 = x ^ 2-2x + 1, X ∈ [- 2,2]
The axis of symmetry is x = - 1, and the opening is upward
Let me explain why the domain is [- 2,2]
For X, f (x) = (X-2) ^ 2, X ∈ [- 1,3], and f (x + 1) = (x-1) ^ 2, the range of X + 1 is equivalent to the [- 1,3] solution of the original x, and X ∈ [- 2,2] is retracted
It is known that X1 and X2 are the two zeros of function f (x) = ax ^ 2 + BX + 1 (a, B ∈ R, a > 0). The minimum value of function f (x) is - A, P = {X / F (x)
The monotone decreasing interval of function f (x) = x square - IXI is?
thank
There are two monotone decreasing intervals, which are (- ∞), - 1 / 2] and [0,1 / 2]