Stationary points of a function. A necessary condition for a local extremum of a function

The first sufficient condition for a local extremum

Second and third sufficient conditions for a local extremum

The smallest and largest values ​​of a function on a segment

Convex functions and inflection points

1. Stationary points of the function. A necessary condition for a local extremum of a function

Definition 1 . Let the function be defined on
. Dot called the stationary point of the function
, If
differentiated at a point And
.

Theorem 1 (necessary condition for a local extremum of a function) . Let the function
determined on
and has at the point
local extremum. Then one of the conditions is satisfied:

Thus, in order to find points that are suspicious for an extremum, it is necessary to find stationary points of the function and points at which the derivative of the function does not exist, but which belong to the domain of definition of the function.

Example . Let
. Find points for it that are suspicious for extremum. To solve the problem, first of all, we find the domain of definition of the function:
. Let us now find the derivative of the function:

Points at which the derivative does not exist:
. Stationary function points:

Since and
, And
belong to the domain of definition of the function, then both of them will be suspicious for an extremum. But in order to conclude whether there will really be an extremum there, it is necessary to apply sufficient conditions for the extremum.

2. The first sufficient condition for a local extremum

Theorem 1 (first sufficient condition for local extremum) . Let the function
determined on
and differentiated on this interval everywhere except, perhaps, the point
, but at this point function
is continuous. If there are such right and left semi-neighborhoods of a point , in each of which
retains a certain sign, then

1) function
has a local extremum at the point , If
takes values ​​of different signs in the corresponding semi-neighborhoods;

2) function
does not have a local extremum at the point , if to the right and left of the point
has the same sign.

Proof . 1) Suppose that in a semi-neighborhood
derivative
, and in

.

Thus at the point function
has a local extremum, namely a local maximum, which was what needed to be proved.

2) Suppose that to the left and right of the point the derivative retains its sign, for example,
. Then on
And
function
increases strictly monotonically, that is:

Thus the extremum at the point function
does not have, which was what needed to be proven.

Note 1 . If the derivative
when passing through a point changes sign from “+” to “-”, then at the point function
has a local maximum, and if the sign changes from “-” to “+”, then there is a local minimum.

Note 2 . An important condition is the continuity of the function
at the point . If this condition is not met, then Theorem 1 may not hold.

Example . The function is considered (Fig. 1):

This function is defined on and is continuous everywhere except a point
, where it has a removable gap. When passing through a point

changes sign from “-” to “+”, but the function does not have a local minimum at this point, but has a local maximum by definition. Indeed, near the point
it is possible to construct a neighborhood such that for all arguments from this neighborhood the function values ​​will be less than the value
. Theorem 1 did not work because at the point
the function had a gap.

Note 3 . The first sufficient condition for a local extremum cannot be used when the derivative of the function
changes its sign in each left and each right semi-neighborhood of a point .

Example . The function being considered is:

Because the
, That
, and therefore
, But
. Thus:

,

those. at the point
function
has a local minimum by definition. Let's see if the first sufficient condition for a local extremum works here.

For
:

For the first term on the right side of the resulting formula we have:

,

and therefore in a small neighborhood of the point
the sign of the derivative is determined by the sign of the second term, that is:

,

which means that in any neighborhood of the point

will take both positive and negative values. Indeed, consider an arbitrary neighborhood of the point
:
. When

,

That

(Fig. 2), and changes its sign here infinitely many times. Thus, the first sufficient condition for a local extremum cannot be used in the example given.

Domain of a function, calculate its derivative, find the domain of a derivative of a function, find points turning the derivative to zero, prove that the found points belong to the domain of definition of the original function.

Example 1 Identify critical points functions y = (x - 3)²·(x-2).

Solution Find the domain of definition of the function, in this case there are no restrictions: x ∈ (-∞; +∞); Calculate the derivative y’. According to the rules for differentiating the product of two, we have: y' = ((x - 3)²)'·(x - 2) + (x - 3)²·(x - 2)' = 2·(x - 3)·(x - 2) + (x - 3)²·1. Afterwards it turns out quadratic equation: y’ = 3 x² – 16 x + 21.

Find the domain of definition of the derivative of the function: x ∈ (-∞; +∞). Solve the equation 3 x² – 16 x + 21 = 0 in order to find at which it becomes zero: 3 x² – 16 x + 21 = 0.

D = 256 – 252 = 4x1 = (16 + 2)/6 = 3; x2 = (16 - 2)/6 = 7/3. So, the derivative goes to zero at values ​​of x equal to 3 and 7/3.

Determine whether the found ones belong points domain of definition of the original function. Since x (-∞; +∞), then both of these points are critical.

Example 2: Identify critical points functions y = x² – 2/x.

SolutionDomain of the function: x ∈ (-∞; 0) ∪ (0; +∞), since x is in the denominator. Calculate the derivative y’ = 2 x + 2/x².

The domain of definition of the derivative of the function is the same as that of the original: x ∈ (-∞; 0) ∪ (0; +∞). Solve the equation 2 x + 2/x² = 0: 2 x = -2/x² → x = -1.

So, the derivative goes to zero at x = -1. The necessary but not sufficient condition for criticality is met. Since x=-1 falls into the interval (-∞; 0) ∪ (0; +∞), then this point is critical.

Sources:

• Critical sales volume, pcsThreshold

Many women suffer from premenstrual syndrome, which manifests itself not only painful sensations, but also increased appetite. As a result, critical days can significantly slow down the weight loss process.

Reasons for increased appetite during menstrual periods

The reason for the increase in appetite during menstrual periods is a change in the general hormonal levels in the female body. A few days before the onset of menstruation, the level of the hormone progesterone rises, the body adjusts to the possibility and tries to make additional energy reserves in the form of fat deposits, even if the woman is sitting. Thus, weight changes on critical days are normal.

How to eat during your period

Try not to eat sweets, confectionery and other high-calorie foods containing “fast” foods these days. Their excess will immediately be deposited in fat. During this period, many women really want to eat chocolate; in this case, you can buy dark chocolate and treat yourself to a few slices, but no more. Do not use during menstruation alcoholic drinks, marinades, pickles, smoked meats, seeds and nuts. In general, pickles and smoked foods should be limited in the diet 6-8 days before the start of menstruation, since such products increase water reserves in the body, and this period is characterized by increased fluid accumulation. To reduce the amount of salt in your diet, add it to minimum quantity in ready meals.

It is recommended to consume low-fat dairy products, plant foods, and cereals. Beans, boiled potatoes, rice - products that contain “slow” carbohydrates will be useful. Seafood, liver, fish, beef, poultry, eggs, legumes, and dried fruits will help replenish iron losses. Wheat bran will be useful. A natural reaction during menstruation is swelling. Light diuretic herbs will help correct the condition: basil, dill, parsley, celery. They can be used as a seasoning. In the second half of the cycle it is recommended to use protein products(lean meats and fish, dairy products), and the amount of carbohydrates in the diet should be reduced as much as possible.

Economic concept critical volume sales corresponds to the position of the enterprise in the market, in which revenue from the sale of goods is minimal. This situation is called the break-even point, when demand for products falls and profits barely cover costs. To determine the critical volume sales, use several methods.

Instructions

The work cycle is not limited to its activities - production or services. This is a complex work of a certain structure, including the work of the main personnel, management staff, management staff, etc., as well as economists, whose task is the financial analysis of the enterprise.

The purpose of this analysis is to calculate certain quantities that, to one degree or another, affect the size of the final profit. This different kinds volumes of production and sales, full and average, demand indicators, etc. The main task is to identify the volume of production at which a stable relationship between costs and profits is established.

Minimum volume sales, at which income completely covers costs, but does not increase the company’s equity capital, is called critical volume sales. There are three methods for calculating the method of this indicator: the method of equations, marginal income and graphical.

To determine the critical volume sales according to the first method, create an equation of the form: Вп – Zper – Зpos = Пп = 0, where: Вп – revenue from sales and ;Zper and Zpos – variable and constant costs; Pp – profit from sales And.

According to another method, the first term, revenue from sales, present it as the product of marginal income per unit of goods and volume sales, the same applies to variable costs. Fixed costs apply to the entire batch of goods, so leave this component common: MD N – Zper1 N – Zpos = 0.

Express the value of N from this equation and you get the critical volume sales:N = Zpos/(MD – Zper1), where Zper1 is the variable costs per unit of goods.

The graphical method involves constructing. Draw two lines on the coordinate plane: the revenue function from sales minus both the cost and profit function. On the abscissa axis, plot the volume of production, and on the ordinate axis, plot the income from the corresponding quantity of goods, expressed in monetary units. The intersection point of these lines corresponds to the critical volume sales, break-even position.

Sources:

• how to define critical work

Critical thinking is a set of judgments on the basis of which certain conclusions are formed and an assessment of the objects of criticism is made. It is especially characteristic of researchers and scientists of all branches of science. Critical thinking occupies a higher level compared to ordinary thinking.

The value of experience in developing critical thinking

It’s difficult to analyze and draw conclusions about something you don’t understand well. Therefore, in order to learn to think critically, it is necessary to study objects in all sorts of connections and relationships with other phenomena. And great importance in this case, has knowledge of information about such objects, the ability to build logical chains of judgments and draw reasonable conclusions.

For example, judging the value work of art is possible only by knowing quite a lot of other fruits of literary activity. At the same time, it’s good to be an expert in the history of human development, the formation of literature and literary criticism. In isolation from the historical context, a work may lose its intended meaning. In order for the assessment of a work of art to be sufficiently complete and justified, it is also necessary to use your literary knowledge, which includes the rules for constructing a literary text within individual genres, a system of various literary techniques, classification and analysis of existing styles and trends in literature, etc. At the same time, it is also important to study the internal logic of the plot, the sequence of actions, the arrangement and interaction of characters in a work of art.

Features of critical thinking

Other features of critical thinking include the following:
- knowledge about the object under study is only a starting point for further brain activity associated with the construction of logical chains;
- consistently built and based on common sense reasoning leads to the identification of true and erroneous information about the object being studied;
- critical thinking is always associated with the assessment of available information about a given object and the corresponding conclusions, the assessment, in turn, is associated with existing skills.

Unlike ordinary thinking, critical thinking is not subject to blind faith. Critical thinking allows, with the help of a whole system of judgments about the object of criticism, to comprehend its essence, to identify true knowledge about it and refute false ones. It is based on logic, depth and completeness of study, truthfulness, adequacy and consistency of judgments. In this case, obvious and long-proven statements are accepted as postulates and do not require repeated proof and evaluation.

Consider the following figure.

It shows the graph of the function y = x^3 – 3*x^2. Let's consider some interval containing the point x = 0, for example from -1 to 1. Such an interval is also called the neighborhood of the point x = 0. As can be seen in the graph, in this neighborhood the function y = x^3 – 3*x^2 takes highest value exactly at the point x = 0.

Maximum and minimum function

In this case, the point x = 0 is called the maximum point of the function. By analogy with this, the point x = 2 is called the minimum point of the function y = x^3 – 3*x^2. Because there is a neighborhood of this point in which the value at this point will be minimal among all other values ​​from this neighborhood.

Dot maximum function f(x) is called the point x0, provided that there is a neighborhood of the point x0 such that for all x not equal to x0 from this neighborhood, the inequality f(x) holds< f(x0).

Dot minimum function f(x) is called the point x0, provided that there is a neighborhood of the point x0 such that for all x not equal to x0 from this neighborhood, the inequality f(x) > f(x0) holds.

At the points of maximum and minimum of functions, the value of the derivative of the function is zero. But this is not a sufficient condition for the existence of a function at a maximum or minimum point.

For example, the function y = x^3 at the point x = 0 has a derivative equal to zero. But the point x = 0 is not the minimum or maximum point of the function. As you know, the function y = x^3 increases along the entire numerical axis.

Thus, the minimum and maximum points will always be among the roots of the equation f’(x) = 0. But not all the roots of this equation will be maximum or minimum points.

Stationary and critical points

The points at which the value of the derivative of the function is zero are called stationary points. There may also be points of maximum or minimum at points at which the derivative of the function does not exist at all. For example, y = |x| at the point x = 0 has a minimum, but the derivative does not exist at this point. This point will be the critical point of the function.

The critical points of a function are the points at which the derivative is equal to zero, or the derivative does not exist at this point, that is, the function at this point is non-differentiable. In order to find the maximum or minimum of a function, a sufficient condition must be met.

Let f(x) be some differentiable function on the interval (a;b). Point x0 belongs to this interval and f’(x0) = 0. Then:

1. if, when passing through a stationary point x0, the function f(x) and its derivative changes sign, from “plus” to “minus”, then the point x0 is the maximum point of the function.

2. if, when passing through a stationary point x0, the function f(x) and its derivative changes sign, from “minus” to “plus”, then the point x0 is the minimum point of the function.