The monotonicity of function in the first grade of mathematics in Senior High School

The monotonicity of function in the first grade of mathematics in Senior High School

By substituting the endpoint of the domain into the calculation, the endpoint of the range can be obtained

1. If the maximum of F (x) = x & sup3; - 3A & sup2; X + a (a > 0) is positive and the minimum is negative, then the value range of a is?
The answer (root two, positive infinity)
2. It is known that the function f (x) = x & sup3; + MX & sup2; - M & sup2; X + 1 (M is a constant and M > 0) has a maximum. 9. Find the value of M
The answer is m = 2

(1) First, find the extreme point. F '(x) = 3x ^ 2-3a ^ 2 = 0, then x = a or x = - A
Then we find the second derivative, f '' (x) = 6x, from which we can see that x = a is the minimum point of F (x)
F (x) = f (a) = a-2a ^ 3, the root of two;
Similarly, we can see that x = - A is the maximum point of F (x), so f (- a) = a + 2A ^ 3 > 0
To sum up, a belongs to (2 / 2 root sign, positive infinity)
(2) If f '(x) = 3x ^ 2 + 2mx-m ^ 2 = 0, then x = m / 3 or x = - M
The second derivative, f '' (x) = 6x + 2m, because m > 0, so when x = - m, f '' (x) = - 4m

Who can help me talk about the plane vector triangle addition and subtraction problem
For example, AB vector + BC vector =?
CA direction + CB vector =?
AB vector - AC vector =?
I remember that our math teacher told us about the plane vector triangle rule, how to add the same starting point and how to add the same ending point. I can't remember clearly. Who can help me

You just have to remember two things
(1) Vector AB = vector ax + vector XB
(2) Vector AB = - vector Ba
For example: vector ab - vector AC = vector Ca + vector AB = vector CB

Given x > 0, Y > 0, x + y = 1, it is proved that x2 + Y2 is greater than or equal to 1 / 2. It is proved that because XY ≤ 1 / 4, X2 + Y2 ≥ 2XY, X2 + Y2 ≥ 1 / 2
I don't think so,

Yes, XY ≤ 1 / 4, the maximum value of XY is 1 / 4, X2 + Y2 ≥ 2XY, and the minimum value of x2 + Y2 is 2XY
x2+y2≥1/2

First year oral arithmetic practice

12-8= 23-12= 20-2= 10-4= 2+3=
3+5= 4+6= 7+6= 9-6= 11-2=
13-4= 20-5= 3+9= 4+7= 5+9=
7+3= 10-8= 15-6= 7+9= 8+8=
0+12= 10-0= 0+0= 0-0= 0+0-0=
2+9= 14-7= 19-9= 16-8= 13-7=

The solution of the system {x + 2Y = 3 2x-y = 6} is

Simulink x = 3, y = 0

Using five numbers of 1, 2, 4, 6 and 8, multiplying three digits by two digits, the formula to maximize the product is given

Use five numbers 1, 2, 4, 6 and 8, multiply three digits by two digits, and the formula to make the product maximum is 821 * 64 = 52544
Hope to adopt!

Find a hyperbolic equation with focus on X axis, distance between vertices 6 and asymptote equation y = ± 32x

Let the hyperbolic equation of y = ± 32x be x24 − Y29 = λ, λ≠ 0. ∵ the hyperbolic focus is on the x-axis, the distance between the vertices is 6, ∵ 2A = 6, that is, a = 3, ∵ 4, λ = 9, and the solution is λ = 94, ∵ the hyperbolic equation is x24 − Y29 = 94, that is, X29 − y2814 = 1

Is the common multiple of 8 and 9 only 72?

Firstly, we need to define the common multiple
Common multiple refers to two or more natural numbers. If they have the same multiple, these multiples are their common multiples. The smallest of these common multiples is called the lowest common multiple of these integers
Then:
The least common multiple of 8 and 9 is 72, but there are many common multiples, such as 720720072000
I hope I can help you

Let the image of quadratic function y = the square of AX + BX-1 (a greater than 0, b greater than 0) pass through two points (1,1) and (- 1, - 1)
Find the expression of (1) this quadratic function
(2) Let a (x1,0) and B (x2,0) be the two intersections of the image of this quadratic function and X axis, where X1 is less than x2 and C is the vertex, and then calculate the area of △ ABC
Speed

(1) Substituting (1,1) (- 1,1) into the square of y = ax + BX-1
We get a + B-1 = 1
a-b-1=1
The solution is a = 2, B = 0
So y = 2x ^ 2-1
(2) Let 2x ^ 2-1 = 0
x=±√2/2
So x1x2 = root 2
So s △ = 1 / 2 * radical 2 * 1 = √ 2 / 2