If the axis of symmetry equation of quadratic function y = - x ^ 2 + MX-3 is x = - 1, then its maximum value is () A-4 B-3 C-2 D-1

If the axis of symmetry equation of quadratic function y = - x ^ 2 + MX-3 is x = - 1, then its maximum value is () A-4 B-3 C-2 D-1

According to the title, B of - 2A = - 1
That is, M = - 1 of - 2 * (- 1)
So m = - 2
The quadratic function is y = - x ^ 2-2x-3
When x = - 1
y=-2
Choose C

When x ∈ (0, + ∞), the power function y = (m2-m-1) · x-5m-3 is a decreasing function, and the value of real number m is obtained

Because when x ∈ (0, + ∞), the power function y = (m2-m-1) · x-5m-3 is a decreasing function, so m2-m-1 = 1 and - 5m-3 < 0, the solution is m = 2 or - 1, and M > 35, that is & nbsp; & nbsp; m = 2

1, define the operation a & B = (when a is greater than or equal to B, it is equal to B; when a is less than B, it is equal to a), then the range of function f (x) = 3 ^ - X & 3 ^ x is ()
2, if the loga root 2 is less than 1, then the value range of a is () a is the base number, and the root 2 is the true number

1. When x > = 0, the function is 3 ^ - x, the function value is greater than 0, less than or equal to 1;
When x

If we define the operation a ⊗ B = B, a ≥ Ba, a < B, then the range of function f (x) = x ⊗ (2-x) is______ ．

From a ⊗ B = B, a ≥ Ba, a < B, f (x) = x ⊗ (2-x) = 2 − x, X ≥ 1 x, x < 1, ⊗ f (x) is an increasing function on (- ∞, 1), a decreasing function on [1, + ∞), and ⊗ f (x) ≤ 1, then the range of F (x) is: (- ∞, 1), so the answer is: (- ∞, 1]

Given that the domain of definition and value of function f (x) = 1 / 2 (x-1) square is [M, n] (M is less than n), find the value of real number m, n

Is the title: F (x) = 1 / 2 (x-1) ^ 2 + 1?
F (x) axis of symmetry = 1
① When m

Let f (x) = MCOs ^ 2x + root sign 3msinxcosx + n (M > 0) be defined as [0, Pai / 6] and [3,4]. Find the value of M.N

(x)=mcos²x+√3*m*sinxcosx+n
=m*[cos2x/2+√3/2*sin2x]+n+m/2
=m*sin(2x+π/6)+n+m/2
The value range is [N-M / 2, N + 3m / 2] = [3,4]
n-m/2=3
n+3m/2=4
m=1/2,n=13/4

It is known that the definition field of the function f (x) = 2x-a / X (A & # 160; is a real number) is (0,1]
1. When a = - 1, find the range of function y = f (x)
2. If the function y = f (x) is a decreasing function in the domain of definition, find the value range of A

1. When a = - 1, f (x) = 2x + 1 / x, because x is a positive number, f (x) ≥ 2 times the root sign 2. If and only if 2x = 1 / x, the equal sign holds, that is, x = 2 / 2 of the root sign 2. At this time, we should pay attention to whether 2 / 2 of the root sign 2 is in the definition field. Because the definition field is (0,1), we can get the minimum value of 2 times the root sign 2

It is known that the definition domain of function f (x) = 2msin2x-23msinx · cosx + n is [0, π 2], and the value domain is [- 5, 4]. Try to find the minimum positive period and the maximum value of function g (x) = msinx + 2ncosx (x ∈ R)

This is the case that when m ＞ 0, f (x) max = -2m (2m (- 12) + m + n = 4, and f (x) min = - M + n (x) min = - M + n = - M + N + n = - 5, solve the solution to get m = 3, n = -2, and then, G (x) = 3ssinx-4cosx = 5sin (x (x) (g (x) = 3, n = -2, so that, G (x) = 3ssinx-4cosx x x = 5sin (x) x (x) x (x) (5sin (x (x (x + x) x (x) (x (x (x (x (x) = 5sinx-4cosx) x = 5sin (5sin (x) (x (x (x (x (x (x) (x) (x + \\\\\\\\\\theng (x) = - 3sinx + 2cosx= 13 sin (x + φ), t = 2 π, the maximum is 13, the minimum is - 13

A car transports goods back and forth along the east-west highway on a certain day. The distance to the East is recorded as positive, and the distance to the west is recorded as negative. The records are as follows: (unit: kilometer) 30, - 28, - 13,15, - 20, - 30,45, - 27. At the end of the day, where is the starting point? How many kilometers has the car traveled? If the fuel consumption per kilometer is 0.2L, the fuel consumption per kilometer is 0.2L, How many liters of fuel does the car consume from the start to the end of the shipment and back to the starting point?

The first question is how to calculate the displacement
X=30-28-13+15-20-30+45-27=-28
According to the journey to the west, the number is negative, so the car is 28km west at the starting point
The second question is how to calculate the distance
S=30+28+13+15+20+30+45+27=208
So the car has traveled a total of 208 kilometers
Question 3: obviously, it is also calculated by distance
Fuel consumption = 0.2 × 208 = 41.6l

When using the discriminant method to calculate the range of value, after changing the original formula into the binary linear equation about X, let Δ ≥ 0, find out the range of Y, and then? Y will have an interval, as long as the two endpoints of this interval are brought into the original function, find out whether the value of X is in the domain of definition? Is it possible that there will be a value in this interval (not two endpoints) of y that makes x not in its domain of definition?
It's a little abstract,

It seems to understand that you want to say that for continuous functions in a closed interval (elementary functions all satisfy this condition), you can rest assured that the situation you said will not occur
However, some piecewise functions and non elementary functions may appear, but after thinking about them for a long time, I can't think of counter examples,
Please refer to Baidu Hi to send you a few pages