Let the quadratic function f (x) = x2 + ax + 5 have f (T) = f (- 4-T) for any t, and have the maximum value 5 and the minimum value 1 on the closed interval [M, 0], then the value range of M is______ ．

Let the quadratic function f (x) = x2 + ax + 5 have f (T) = f (- 4-T) for any t, and have the maximum value 5 and the minimum value 1 on the closed interval [M, 0], then the value range of M is______ ．

Because the known condition: for any t, there is f (T) = f (- 4-T), so the symmetry axis of quadratic function is x = - 2, so − A2 = - 2, so a = 4, so f (x) = x2 + 4x + 5, because f (- 2) = 1, f (0) = 5, because there is a maximum value 5 and a minimum value 1 on the closed interval [M, 0], so - 4 ≤ m ≤ - 2, so the answer is - 4 ≤ m ≤ - 2

A question about the range of value of mathematical function
Y = 1-x / x under the root sign to find the value range of X
1.1-x / X is greater than or equal to 0
2. X doesn't equal 0. I'll only be here

In this way, from (1-x / x) > = 0, then X-1 / x = means greater than or equal to)

Value range of mathematical function
When x > 0, f (x) = ln x-ax, the equation f (x) = 0 has exactly five different real solutions on R,
1. Find f (x)

The second problem is to use the separation constant, that is, by dividing the numerator of 1 / (2 ^ X - 1) with the denominator

If f (a + b) = f (a) + F (b) for any real number a and B, and if x > 0, f (x)

Set X10
f(x2)
=f[x1+(x2-x1)]
=f(x1)+f(x2-x1)
So f (x2) - f (x1) = f (x2-x1)
For any x > 0, there is always f (x) 0, f (x2-x1)

It is known that the definition field of function FX is all real numbers with X ≠ 0. For any x1, X2 in the definition field, f (x1x2) = f (x1) + F (x2), and when x > 1, FX > 0, f (2) = 1
1. Prove that FX is an even function
2. Prove that FX increases monotonically at (0, positive infinity)
3. Solve the inequality f (2x & # 178; - 1)

1. The assignment method, f (x1x2) = f (x1) + F (x2) X1 = - 1, X2 = 0 substitution, get f (- 1) = 0x1 = - 1, X2 = x2 substitution, get f (- x2) = f (x2) + 0 proof 2. Use the title form to deformation, let 0 < X1 < X2F (x1) - f (x2) = f (x1) - f (x1 × x2 / x1) = f (x1) - [f (x1) + F (x2 / x1)] = - f (x2 / x1) because 0 < X1 < X2

As shown in the figure, the axial section △ SAB of cone so is an equilateral triangle with side length of 4, M is the midpoint of bus sb, and plane β⊥ SAB is made through straight line am. Suppose the intersection line between β and the side of cone is ellipse C, then the short half axis of ellipse C is ()
A. 2B. 102C. 3D. 2

Make a section (circle) parallel to the bottom of the cone through ellipse C, intersect as and BS at R and T, intersect ellipse C at two points P and Q, then p and Q are the vertices of the short half axis of the ellipse. In the circle, RT is the diameter, as shown in the figure, ∵ axis section △ SAB is an equilateral triangle with side length of 4, C is the midpoint of am, ∵ TC = 12ab = 2, RC = 14ab = 1, ∵ P

Given the function f (x) = X3 + 1, find the tangent equation of the curve y = f (x) through P (1,2)

f'(x)=3x^2
f'(1)=3
The tangent equation is y = 3 (x-1) + 2 = 3x-1

The image of quadratic function y = the square of AX + BX + C, as shown in the figure, the following conclusion is wrong: 1, ab

If the opening is downward, then A0, then b > 0, so ab

Quadratic function y = ax'2 + BX + C (AF (3) C.F (- 1) > F (0) D.F (2) > F (3)

Because in A1, the larger x is, the smaller f (x) is. So f (2) > F (3)

Define (ABC) as the characteristic number of the function y = ax ^ 2 + BX + C. some conclusions of the functions with characteristic numbers (2m, 1-m, - 1-m) are given below
1 when m = - 3, the vertex coordinates of the function image are (1 / 3,8 / 3)
2 when m > 0, the line segment of the function image which cuts the x-axis is greater than 3 / 2
When m ＜ 0, y decreases with the increase of X when x ＞ 1 / 4
When m ≠ 0, the graph of the function passes through the same point
A1234
B124
C134
D24
Ask for every step to prove

1 when m = - 3, the axis of symmetry is x = 1 / 3, then substitute x = 1 / 3 to get y = 8 / 3, that is, the vertex coordinates of function image are (1 / 3,8 / 3) 2. When m > 0, let the intersection of function and X axis be x1, X2, (x1-x2) ^ 2 = (- B / a) ^ 2-4c / a = (3 / 2 + 1 / 2m) ^ 2