If the function y = - x + M & sup2; intersects with the image of y = 4x-1 at the same point on the x-axis, the value of M is obtained 2, the first-order function y = 2 (1-k) x + 1 / 2k-1, the image does not pass through the first quadrant, find the value range of K

If the function y = - x + M & sup2; intersects with the image of y = 4x-1 at the same point on the x-axis, the value of M is obtained 2, the first-order function y = 2 (1-k) x + 1 / 2k-1, the image does not pass through the first quadrant, find the value range of K

1. Let Y1 = - X1 + m2, y2 = 4x2-1, when Y1 = 0, X1 = m2, when y2 = 0, X2 = 1 / 4, and because Y1 and Y2 intersect at the same point on the x-axis, X1 = X2, that is, M2 = 1 / 4m = ± 1 / 22. When y = 0, x = K / 4 (1-k) 0 = 2 (1-k) x + 1 / 2k-1x = (1-1 / 2K) / 2 (1-k) = K / 4 (1-k) when x = 0, y = 1 /
1:m=1/2
2: When k = 1, y = - 1 / 2 satisfies,
Take two points Q (0,1 / 2k-1) P (1 / (k-1) (k-1), 0)
If the function does not exceed the first quadrant, then
1/2k-1
If the function f (x) = 2x & # 178; - MX + 3 is an increasing function on [- 2, + ∞) and a decreasing function on (- ∞, - 2], then the value of M is
Because f (x) = 2x ^ 2-mx + 3 is an increasing function on [- 2, + 00], and a decreasing function on (- 00, - 2)
Then we can see that the axis of symmetry of function f (x) is x = - 2
That is - 2x2 / - M = - 2
So m = - 2
So the value of M is - 2
Let's discuss it according to the situation?.
4. Follow up: there is no such option,
If the image of the function y = - x + m and y = 4x-1 intersects at a point on the x-axis, then the value of M is a ± 1 / 2 B ± 1 / 4 C1 / 2 D1 / 4
Because the image intersects at a point on the x-axis
So y = 0
Substituting 0 = 4x-1
The solution is x = 1 / 4
So 0 = - 1 / 4 + M
The solution is m = 1 / 4
D correct
y = 4x - 1
When y = 0, x = 1 / 4
Substituting x = 1 / 4, y = 0 into y = - x + m, we get the following result:
0 = -1/4 + m
m = 1/4
So choose D
The function f (x) = 2x ^ 2-mx + 3, if x ∈ [- 2, + ∞) is an increasing function, if x ∈ (- ∞, - 2) is a decreasing function, find the value of F (1)
∵ f (x) is an increasing function in X ∈ [- 2, + ∞) and a decreasing function in X ∈ (- ∞, - 2)
The abscissa of the vertex of parabola f (x) is - 2
∴f(x)=2x^2-mx+3=2(x+2)^2+n=2x^2+8x+8+n
∴m=-8,n=-5
∴f(x)=2x^2+8x+3
∴f(1)=2*1^2+8*1+3=13
It is known that the function y = (M + 3) x to the power of M2 + 1 + 4x-5 is a first-order function. Find the value of M
m²+1=1
M=0
Or M + 3 = 0
m=-3
The range of the function y = x & # 178; - 2x, X ∈ [0,3] is?
Y = x & # 178; - 2x = (x-1) 2-1, so this function is a decreasing function on X ∈ [0,1], and an increasing function on X ∈ [1,3], so the minimum value of the function is f (1) = - 1, and the maximum value is f (3) = 3, and the range of values is y ∈ [- 1,3]
What are the image features of the function y = x ^ 2 + 4x + 3?
y=x^2+4x+3=(x+2)²-1=(x+1)(x+3)
The vertex is (- 2, - 1),
The axis of symmetry is a straight line x = - 2,
When x = - 2, y has a minimum value of - 1,
The intersection point with X axis is (- 1,0) (- 3,0)
The intersection point with y axis is (0,3)
That is, the center line is x = - 2, and the minimum value is - 1
If f (2x + 1) = x ^ 2-2x, what is f (3) equal to
Let 2x + 1 = t, then x = (t-1) / 2
f(t)=(t-1)²/4-(t-1)
f(3)=1-2=-1
3^2/4-3x/2+5/4
-1
When 2x + 1 = 3, x = 1
Then f (3) = 1 ^ 2-2x1 = - 1
f（2x+1)=x^2-2x
f(3)=x^2-2x
Left 2x + 1 = 3, x = 1
Substituting x = 1 into the right, then the right x ^ 2-2x = 1
So f (3) = 1
Because we know that the function is f (2x + 1), then 2x + 1 = 3 in F (3) can solve the equation x = 1. From this we know that
F (2x + 1 = 3) = x ^ 2-2x can be written as f (3) = 1 ^ 2 - (2 * 1) = - 1
Let 2x + 1 = t, then x = (t-1) / 2, f (T) = ((t-1) / 2) ^ 2 - 2 * (t-1) / 2 = (t-1) ^ 2 / 4 - (t-1) = (t-1) ^ 2 / 4-T + 1, f (3) = 1-3 + 1 = - 1, 2. F (2x + 1) = x ^ 2-2x, f (2x + 1) = 1 / 4 (2x + 1) ^ 2-3 / 2 (2x + 1) - 5 / 2, f (3) = 1
There are two algorithms
1. Conventional algorithm
Let 2x + 1 = t, then x = (t-1) / 2
f(t)=(t/2-1/2)^2-(t-1)
The answer is - 1
2. Simple method
2x＋1＝3，x＝1
Then f (3) = 1 ^ 2-2 * 1 = - 1
Let t = 2x + 1, then x = (t-1) / 2, f (x) = 0.25x ∧ 2-1.5x + 1.25, and then carry 3 into - 1
Let 2x + 1 = 3, then x = 1
So f (3) = 3 ^ 2-2 * 2 = - 1
Because 2x + 1 = 3, the solution is x = 1, so f (3) = 1 ^ 2-2 * 2 = - 1
Given the function f (x) = x ^ 2 + ax + 3, when x ∈ [1,2], f (x) ≥ A is constant, and the minimum value of a is obtained
When a ≥ - 2, - A / 2 ≤ 1, f (x) increases on [1,2], a ≤ f (x), so a ≤ f (x) min = f (1), that is, a ≤ 1 + A + 3, it is constant, so a ≥ - 2 meets the requirements; when - 4 ≤ a ≤ - 2, 1 ≤ - A / 2 ≤ 2, a ≤ f (x)
Let f (x) = e ^ 2x, then what is the N derivative of F (0)
f(x) = e^(2x)
f^(n)(x)= 2^n.e^(2x)
f^(n)(0) = 2^n
F (0) = e ^ 0 = 1, so any derivative of F (0) is 0