Given the function f (x) = 2x square minus MX plus 2, if x belongs to (negative 2, positive infinity), it is an increasing function and if x belongs to (negative infinity, negative 2), it is a decreasing rare number, then f (1) =?

Given the function f (x) = 2x square minus MX plus 2, if x belongs to (negative 2, positive infinity), it is an increasing function and if x belongs to (negative infinity, negative 2), it is a decreasing rare number, then f (1) =?

When x = 2, it is the axis of symmetry of the parabola
So f (1) = f (3), that is 4-m = 20-3m, so m = 8
f(x)=2x^2 - 8x +2
f(1)=4-8+2=-2

F (x) = 2x square - MX + 3 in the range of finding m by (negative infinity, 3]

F (x) = 2x square - MX + 3 the axis of symmetry is x = m / 4
The function decreases on (negative infinity, 3]
So 3 = 12

If the function y = 2x square + MX + 5 is a decreasing function on (– infinity, 1), then the value range of real number M

Opening up
So the axis of symmetry x = - M / 4 decreases to the left
So - M / 4 ≥ 1
m≤-4

Given that a (2, a) is on the image of function y = 2x + m, then A-M=

Substituting x = 2 and y = a into the equation y = 2x + m, we get
a=4+m,
So, A-M = 4

It is known that a (2, a) is the intersection point of the image of the function y = 2x + m and y = mx-2, and m, a can be obtained

m=6,a=10

If the image of quadratic function y = 2x2 + MX + 8 is shown in the figure, then the value of M is ()
A. -8B. 8C. ±8D. 6

It can be seen from the figure that there is only one intersection point between the parabola and the x-axis. Therefore, △ = M2-4 × 2 × 8 = 0, the solution is m = ± 8, ∵ the axis of symmetry is a straight line, x = - M2 × 2 < 0, ∵ m > 0, and the value of∵ m is 8

The function y = ax (third power) - 3x (square) + 1, if there is a unique zero point of the function, and the zero point is greater than zero, then the range of A

When a = 0, y = - 3x ^ 2 + 1, there are two zeros;
When a is not zero, y '= 3ax ^ 2-6x = 3x (AX-2), and the extreme point is x = 0,2 / A
When a > 0, f (0) = 1 is the maximum point, because in X

If the function f (x) = x2 + ax + 2B has a zero point in the interval (0,1) and (1,2), find the value range of B − 2A − 1

It is known that: F (0) > 0f (1) < 0f (2) > 0 {(4 points) 2B > 0A + 2B + 1 < 02A + 2B + 4 > 0 {b > 0A + 2B + 1 < 0A + B + 2 > 0 (6 points), which represents the region m as shown in the figure: ((9 points) B − 2A − 1 represents the slope of the line between C (1,2) and points (a, b) in the region m. A (- 3,1), B (- 1,0) KCa = 14, KCB = 1. From the figure, we can see that B − 2A − 1 ∈ (14,1)

What kind of real number is t if the minimum value of quadratic function y = x + TX + T + 3 is nonnegative?

Because the coefficient of quadratic term is 1, the opening of quadratic function image is upward, and because the minimum value is nonnegative, y = 0 has a root or no root, that is, Δ≤ 0, t-4t-12 ≤ 0 (T-6) (T + 2) ≤ 0 - 2 ≤ t ≤ 6

The function f (x) = lgx SiNx has ()
A. 1B. 2C. 3D. 4

Function f (x) = the number of zeros of lgx SiNx in the interval (0, 10), that is, the number of intersections of the image of function y = lgx and the image of function y = SiNx in the interval (0, 10), as shown in the figure: according to the number of intersections of the image of function y = lgx and the image of function y = SiNx in the interval (0, 10), C