The first three extrema problems, the difference between extrema and extremum points

I don't know what a derivative is in the third year of junior high school.........

But the answer is indeed 4, and an inequality is fine.

a^3-3a+2

a^3-1-3a+3

a^3-1)-(3a-3)

a-1)(a^2+a+1)-3(a-1)(a-1)(a^2+a+1-3)

a-1)(a^2+a-2)

a-1）(a-1)(a+2)

a-1) 2(a+2), it can be seen that when a -2, the value is an integer.

Because a<=1 4

When a=-1, the maximum value is =4

Let f(a)=a 3-3a+2

f'(a)=3x^2-3

Order f'(a) = 0 and since a<=1 4 gives a=-1, there is a maximum value of 4 when a=-1

For a 3-3a+2, we get 3*a 2-3

Let 3*a2-3 0 get a=1ora=-1

When a -1, 3*A 2-3>0, f(a) increases; When -1 so there is a maximum value at a=-1.

This problem is not easy to do in the third year of junior high school (I didn't learn the derivative, and I didn't seem to learn the mean theorem) My approach is to make up a 3-3a+2= a 3-1-3(a-1)=(a-1)(a 2+a-1)-3(a-1)= (a-1) 2(a+2) where (a-1) 2>0 explains that a positive number multiplied by a small positive number will only get smaller and smaller To take the maximum value a+2>1 or =1a> or =-1 because (a-1) 2 in -1 or =a or = 1 4 than (a+2) The trend of change is even greater.

Therefore, when a=-1, the maximum value is 4 (this understanding should be able to be written on the answer sheet and find a better way, and the knowledge is forgotten, sorry).

Are you a secondary school student? Well, if you're in college, you can't ask such a question anymore!

Let's do it the same way as a middle school.

1, it is observed that for equation a 3-3a+2=0, there is a solution that is a=1, so the left equation can be given a factor of (a-1).

2, get: (a-1) (a 2 + a-2).

3, because a 1 4, so (a-1) <0, put aside.

4. Find the minimum value of the equation (a 2 + a-2) in the negative interval.

Isn't that easy? The negative interval is a (-2,1 4), but in fact it can only be solved here. The derivative has to be used below. But it seems that I have studied in the third year of high school.

Then the above content will be scrapped (but I hope that the majority of middle school students must master this method).

For the equation f(x)=x 3-3x+2, the derivative has:

f'(x)=3x 2-3, so at x=1, the original equation has an extremum. Obviously here a = 1

So is the original equation extremely extreme at this time? Just make a picture to illustrate!

Let y=a 3-3a+2 and derive: y'=3a 2-3 another y'=0, find the extreme point as x=-1, and x=1 to analyze y'Positive and negative can be seen that the original function increases on (negative infinity, -1), decreases on (-1, 1), and increases at (1, positive infinity).

And because a<=1 4

So the maximum value is 4 when a=-1

1. Different definitions.

1. Extreme point: If f(a) is the maximum or minimum value of the function f(x), then a is the extreme point of the function f(x), and the maximum point and the minimum point are collectively referred to as the extreme point.

2. Extremum: Extremum is the maximum or minimum value of a function. If a function has a definite value everywhere in a neighborhood of a point, and the value at that point is the maximum (small), the value of the function at that point is a maximum (small) value.

2. The meaning expressed is different.

The maximum and minimum points are the values of the abscissa; Whereas, the extreme value refers to the numerical value of the ordinate.

Third, the attributes are different.

The maximum point and the minimum point each refer to a point; An extreme value is a set of data that includes a maximum and a minimum.

...The extreme point is not a coordinate. It is the horizontal and vertical coordinates of the upper maximum or minimum point in a sub-interval of the function image, for example, y=x 2-2x, and after derivation, it is 2x-2

The minimum point of the function is x=1....As for the extreme value, it is the maximum (small) value, and you can bring the extreme point in to find it.

The extreme value is x=0, which is the maximum or minimum value corresponding to the derivative function; The extreme point is to bring this maximum or minimum (a) into the original function to find the f(x) value (b), and (a,b) is the extreme point.

To put it simply, when the derivative of f(x) is 0, you get the abscissa of the extreme point, and bring this point, i.e., x, to the original function, and you get the extreme value y, and (x,y) is the coordinate representation of the extreme point.

In layman's terms, the extreme value is y, which is a number. The extreme points are (x,y) are the coordinates and are two numbers.

An extremum is a value.

An extreme point is a point.

First find the first derivative of the function, and then find the value of the independent variable when the first derivative of the function is zero, that is, solve the equation f(x)=0, and get the solution of the equation as x=x1 (there may be other solutions), f(x1) is the extreme value of the function, and then determine whether f(x1) is the maximum or minimum.

How to judge: Use the increase or decrease of the function.

Isn't it possible to exit the continuum?

In the neighborhood, the second derivative, so the first derivative f'(x) is continuous.

Because f'(x) monotonically decreasing, so in the neighborhood, there are no non-derivatives. The points in the neighborhood are all derivable and derivable, so they are continuous.

f(x)=x 2(1-x)=x 2-x 3 derivative yields 2x-3x 2

Let the reciprocal be 0 and find the root as 0 and 2 3

This is the corresponding extreme point.

Have you learned derivatives?

There are no maximum and minimum values.