Judge the number of zeros of function f (x) = - 2 / 3x3 + x2 + 4x on [- 1,4], and explain the reason

Judge the number of zeros of function f (x) = - 2 / 3x3 + x2 + 4x on [- 1,4], and explain the reason

DF / DX = - 2x ^ 2 + 2x + 4 = 0 = > x ^ 2-x - 2 = 0, so x = 2, x = - 1D ^ 2F / DX ^ 2 = - 4x + 2F '(2) = 0, f' (- 1) = 0 between (- 1,2), f '(x) > 0 is monotone increasing function, on (2,4), monotone decreasing, because f (- 1) = 2 / 3 + 1 + 4 > 0f (2) = 12-16 / 3 > 0, and the function is monotone, so on (- 1,2)

If α and β are the two real roots of the equation x ^ 2 + 3x-2006 = 0, then α ＾ 2 + β ＾ 2 + 3 α + 3 β

α. β is the two real roots of the equation x ^ 2 + 3x-2006 = 0
α+β=-3 α*β=-2006
α＾2+β＾2+3α+3β
=（α+β）＾2-2αβ+3（α+β）
=9+4012-9
=4012

When m takes what value, the equation 2x2 - (M + 2) x + 2m-2 = 0 about X has two equal real roots? Find out that this is the root of the equation

According to the meaning of the title:
b²-4ac=[-(m+2)]²-4*2*(2m-2)
=m²+4m+4-16m+16
=m²-12m+20=0
So (m-2) (M-10) = 0
m1=2,m2=10
When M1 = 2, the equation is 2x & # 178; - 4x + 2 = 0, X1 = x2 = 1
When M2 = 10, the equation is 2x & # 178; - 12x + 18 = 0, X1 = x2 = 3

It is known that the minimum value of the function f (x) = | x + 1 | + | X-2 | over R is a. (1) find the value of a; (2) if P, Q, R are positive real numbers and P + Q + r = a, prove that P2 + Q2 + R2 ≥ 3

(1) ∵|x + 1 | + |x-2 | ≥ | (x + 1) - (X-2) | = 3, if and only if - 1 ≤ x ≤ 2, the equal sign holds, and the minimum value of ∵ f (x) is 3, that is, a = 3; (2) it is proved that from (1) we know that P + Q + r = 3, and P, Q, R are positive real numbers, (P2 + Q2 + R2) (12 + 12) ≥ (P × 1 + Q × 1 + R × 1) 2 = (P + Q + R) 2 = 32 = 9, that is, P2 + Q2 + R2 ≥ 3

Let f (x) = (e ^ x-a) ^ 2 + (e ^ - x-a) ^ 2 (a ≥ 0). Problem (1) express f (x) as a function of u = (e ^ x + e ^ - x) / 2. (2) find the minimum value of f (x)
It is known that f (x) = (e ^ x-a) ^ 2 + (e ^ x-a) ^ 2 (a ≥ 0)
Problem (1) express f (x) as a function of u = (e ^ x + e ^ - x) / 2
(2) find the minimum value of F (x)
Don't copy. What's the u in the online answer?
It's not u on my paper, it's t, so I read it wrong, sorry~
But I want to know how to do it, especially the second step,

First question

There are seven points in a plane, four of which are straight lines. The remaining three points are not in the same line. How many triangles and quadrangles can you draw with seven points as the vertex?

If it's a question of permutation and combination in senior high school, you can choose any three of the seven points, which means c73 {read as C 73} {sorry, I can't express this formula, I didn't learn computer well in junior high school}, where c73 = 35, that is, there are 35 connection methods, and because there are four points on the same line, it's impossible to form a triangle, so to subtract C43 = 1, that is, 35-1 = 34 triangles
In the same way, 2. Select 4 of the 7 points c74 = 35, and then subtract the 4 points on the straight line to form c44 = 1, which will form 35-1 = 34 quadrilateral

There are 10 points on the plane, and any three points are not in a straight line,
So that any two triangles have at most one common vertex, how many triangles can be drawn at most

N = 3, the total number of non edge triangles = 1,
N = 4, the total number of non common edge triangles = 1. [if there are three non common edge triangles, the triangle must contain the fourth point. The other two points of the triangle must come from the first three points. In this way, the edge connected by the other two points must be one edge of the first triangle. Therefore, the four points can only form one non common edge triangle]
When n = 5, the total number of triangles that are not in common with the first triangle is 1 + 1. [after selecting any three points to form a triangle, it is known from the discussion when n = 4 that the other triangles that are not in common with the first triangle can only contain at most one of the first three points. In this way, the two points in the other triangles that are not in common with the first triangle must be the fourth and fifth points, The last point of the triangle comes from one of the first three points. However, the other three triangles share the edge of the fourth and fifth points. Therefore, except for the first triangle, there is only one triangle that does not share the same edge
Let n = 6, the total number of non collinear triangles = 2 + 2. [from the discussion of n = 5, any five points can form two non collinear triangles. Let the vertices of the two triangles be [P (1-2-3)] and [P (1-4-5)]. If there are any non collinear triangles, the triangle must contain the remaining point P (6), and the other two vertices cannot come from the same triangle in the previous two triangles, Only one vertex can be selected from each of the two triangles [because there are edges between P (1) and P (2) ~ p (4), so p (1) cannot be selected. Therefore, other triangles are [P (2-4-6)], [P (3-5-6)]
Let n = 7, the total number of non equilateral triangles = 4 + 3. [from the discussion of n = 6, any six points can form four non equilateral triangles. Let the vertices of the four triangles be [P (1-2-3)], [P (1-4-5)], [P (2-4-6)] and [P (3-5-6)]. If there are any non equilateral triangles, then the triangle must contain the remaining points P (7), The other two vertices cannot come from the same triangle in the previous triangle. Therefore, other triangles have edges between [P (1-6-7)] [P (1) and P (2) ~ p (5), so only p (6)], [P (2-5-7)], [P (3-4-7)]
Let the vertices of these triangles be [P (1-2-3)], [P (1-4-5)], [P (1-6-7)], [P (2-4-6)], [P (2-5-7)], [P (3-4-7)], [P (3-5-6)], The other two vertices cannot come from the same triangle in the previous triangle. Therefore, there are no other triangles
Let the vertices of these triangles be [P (1-2-3)], [P (1-4-5)], [P (1-6-7)], [P (2-4-6)], [P (2-5-7)], [P (3-4-7)], [P (3-5-6)], The other two vertices cannot come from the same triangle in the previous triangle. Therefore, the other triangles are [P (1-8-9)]
Let the vertices of these triangles be [P (1-2-3)], [P (1-4-5)], [P (1-6-7)], [P (1-8-9], [P (2-4-6)], [P (2-5-7)], [P (3-4-7)], [P (3-5-6)], The other two vertices cannot come from the same triangle in the previous triangle. Therefore, the other triangles are [P (2-8-10)], [P (3-9-10)]

There are 5 points on the plane, and any 3 points are not in a straight line. Please answer: how many triangles are there with these points as vertices?
How many acute triangles are there at most?

(1) How many triangles are there with these points as their vertices? 5 * 4 * 3 / (1 * 2 * 3) = 10. A: there are 10 triangles with these points as their vertices. (2) how many acute angle triangles are there at most? Take four of the five points to form a quadrilateral, and then three of the internal angles of the quadrilateral are acute angles at most

Given six points on the plane, any three points are not on the same straight line. Please explain that at least one of all triangles with these six points as vertices

That is to say, choose three of the six to combine, a total of 6 * 5 * 4 = 120

There are five points on the plane, among which there are only three points on the same line. How many triangles can you draw with three points as the vertices?

Countless