The largest value of the derivative 2.1.1 2. Derivative of a function

Problem B9 gives a graph of a function or derivative from which you need to determine one of the following quantities:

The value of the derivative at some point x 0,
Maximum or minimum points (extremum points),
Intervals of increasing and decreasing functions (intervals of monotonicity).

The functions and derivatives presented in this problem are always continuous, making the solution much easier. Despite the fact that the task belongs to the section of mathematical analysis, even the weakest students can do it, since no deep theoretical knowledge is required here.

To find the value of the derivative, extremum points and monotonicity intervals, there are simple and universal algorithms - all of them will be discussed below.

Read the conditions of problem B9 carefully to avoid making stupid mistakes: sometimes you come across quite lengthy texts, but there are few important conditions that affect the course of the solution.

Calculation of the derivative value. Two point method

If the problem is given a graph of a function f(x), tangent to this graph at some point x 0, and it is required to find the value of the derivative at this point, the following algorithm is applied:

Find two “adequate” points on the tangent graph: their coordinates must be integer. Let's denote these points as A (x 1 ; y 1) and B (x 2 ; y 2). Write down the coordinates correctly - this is a key point in the solution, and any mistake here will lead to an incorrect answer.
Knowing the coordinates, it is easy to calculate the increment of the argument Δx = x 2 − x 1 and the increment of the function Δy = y 2 − y 1 .
Finally, we find the value of the derivative D = Δy/Δx. In other words, you need to divide the increment of the function by the increment of the argument - and this will be the answer.

Let us note once again: points A and B must be looked for precisely on the tangent, and not on the graph of the function f(x), as often happens. The tangent line will necessarily contain at least two such points - otherwise the problem will not be formulated correctly.

Consider points A (−3; 2) and B (−1; 6) and find the increments:
Δx = x 2 − x 1 = −1 − (−3) = 2; Δy = y 2 − y 1 = 6 − 2 = 4.

Let's find the value of the derivative: D = Δy/Δx = 4/2 = 2.

Task. The figure shows a graph of the function y = f(x) and a tangent to it at the point with the abscissa x 0. Find the value of the derivative of the function f(x) at the point x 0 .

Consider points A (0; 3) and B (3; 0), find the increments:
Δx = x 2 − x 1 = 3 − 0 = 3; Δy = y 2 − y 1 = 0 − 3 = −3.

Now we find the value of the derivative: D = Δy/Δx = −3/3 = −1.

Task. The figure shows a graph of the function y = f(x) and a tangent to it at the point with the abscissa x 0. Find the value of the derivative of the function f(x) at the point x 0 .

Consider points A (0; 2) and B (5; 2) and find the increments:
Δx = x 2 − x 1 = 5 − 0 = 5; Δy = y 2 − y 1 = 2 − 2 = 0.

It remains to find the value of the derivative: D = Δy/Δx = 0/5 = 0.

From the last example, we can formulate a rule: if the tangent is parallel to the OX axis, the derivative of the function at the point of tangency is zero. In this case, you don’t even need to count anything - just look at the graph.

Calculation of maximum and minimum points

Sometimes, instead of a graph of a function, Problem B9 gives a graph of the derivative and requires finding the maximum or minimum point of the function. In this situation, the two-point method is useless, but there is another, even simpler algorithm. First, let's define the terminology:

The point x 0 is called the maximum point of the function f(x) if in some neighborhood of this point the following inequality holds: f(x 0) ≥ f(x).
The point x 0 is called the minimum point of the function f(x) if in some neighborhood of this point the following inequality holds: f(x 0) ≤ f(x).

In order to find the maximum and minimum points from the derivative graph, just follow these steps:

Redraw the derivative graph, removing all unnecessary information. As practice shows, unnecessary data only interferes with the decision. Therefore, we mark the zeros of the derivative on the coordinate axis - and that’s it.
Find out the signs of the derivative on the intervals between zeros. If for some point x 0 it is known that f'(x 0) ≠ 0, then only two options are possible: f'(x 0) ≥ 0 or f'(x 0) ≤ 0. The sign of the derivative is easy to determine from the original drawing: if the derivative graph lies above the OX axis, then f'(x) ≥ 0. And vice versa, if the derivative graph lies below the OX axis, then f'(x) ≤ 0.
We check the zeros and signs of the derivative again. Where the sign changes from minus to plus is the minimum point. Conversely, if the sign of the derivative changes from plus to minus, this is the maximum point. Counting is always done from left to right.

This scheme only works for continuous functions - there are no others in Problem B9.

Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−5; 5]. Find the minimum point of the function f(x) on this segment.

Let's get rid of unnecessary information and leave only the boundaries [−5; 5] and zeros of the derivative x = −3 and x = 2.5. We also note the signs:

Obviously, at the point x = −3 the sign of the derivative changes from minus to plus. This is the minimum point.

Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−3; 7]. Find the maximum point of the function f(x) on this segment.

Let's redraw the graph, leaving only the boundaries [−3; 7] and zeros of the derivative x = −1.7 and x = 5. Let us note the signs of the derivative on the resulting graph. We have:

Obviously, at the point x = 5 the sign of the derivative changes from plus to minus - this is the maximum point.

Task. The figure shows a graph of the derivative of the function f(x), defined on the interval [−6; 4]. Find the number of maximum points of the function f(x) belonging to the segment [−4; 3].

From the conditions of the problem it follows that it is enough to consider only the part of the graph limited by the segment [−4; 3]. Therefore, we build a new graph on which we mark only the boundaries [−4; 3] and zeros of the derivative inside it. Namely, points x = −3.5 and x = 2. We get:

On this graph there is only one maximum point x = 2. It is at this point that the sign of the derivative changes from plus to minus.

A small note about points with non-integer coordinates. For example, in the last problem the point x = −3.5 was considered, but with the same success we can take x = −3.4. If the problem is compiled correctly, such changes should not affect the answer, since the points “without a fixed place of residence” do not directly participate in solving the problem. Of course, this trick won’t work with integer points.

Finding intervals of increasing and decreasing functions

In such a problem, like the maximum and minimum points, it is proposed to use the derivative graph to find areas in which the function itself increases or decreases. First, let's define what increasing and decreasing are:

A function f(x) is said to be increasing on a segment if for any two points x 1 and x 2 from this segment the following statement is true: x 1 ≤ x 2 ⇒ f(x 1) ≤ f(x 2). In other words, the larger the argument value, the larger the function value.
A function f(x) is called decreasing on a segment if for any two points x 1 and x 2 from this segment the following statement is true: x 1 ≤ x 2 ⇒ f(x 1) ≥ f(x 2). Those. A larger argument value corresponds to a smaller function value.

Let us formulate sufficient conditions for increasing and decreasing:

In order for a continuous function f(x) to increase on the segment , it is sufficient that its derivative inside the segment be positive, i.e. f’(x) ≥ 0.
In order for a continuous function f(x) to decrease on the segment , it is sufficient that its derivative inside the segment be negative, i.e. f’(x) ≤ 0.

Let us accept these statements without evidence. Thus, we obtain a scheme for finding intervals of increasing and decreasing, which is in many ways similar to the algorithm for calculating extremum points:

Remove all unnecessary information. In the original graph of the derivative, we are primarily interested in the zeros of the function, so we will leave only them.
Mark the signs of the derivative at the intervals between zeros. Where f’(x) ≥ 0, the function increases, and where f’(x) ≤ 0, it decreases. If the problem sets restrictions on the variable x, we additionally mark them on a new graph.
Now that we know the behavior of the function and the constraints, it remains to calculate the quantity required in the problem.

Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−3; 7.5]. Find the intervals of decrease of the function f(x). In your answer, indicate the sum of the integers included in these intervals.

As usual, let's redraw the graph and mark the boundaries [−3; 7.5], as well as zeros of the derivative x = −1.5 and x = 5.3. Then we note the signs of the derivative. We have:

Since the derivative is negative on the interval (− 1.5), this is the interval of decreasing function. It remains to sum all the integers that are inside this interval:
−1 + 0 + 1 + 2 + 3 + 4 + 5 = 14.

Task. The figure shows a graph of the derivative of the function f(x), defined on the interval [−10; 4]. Find the intervals of increase of the function f(x). In your answer, indicate the length of the largest of them.

Let's get rid of unnecessary information. Let us leave only the boundaries [−10; 4] and zeros of the derivative, of which there were four this time: x = −8, x = −6, x = −3 and x = 2. Let’s mark the signs of the derivative and get the following picture:

We are interested in the intervals of increasing function, i.e. such where f’(x) ≥ 0. There are two such intervals on the graph: (−8; −6) and (−3; 2). Let's calculate their lengths:
l 1 = − 6 − (−8) = 2;
l 2 = 2 − (−3) = 5.

Since we need to find the length of the largest of the intervals, we write down the value l 2 = 5 as an answer.

In the interim ( A,b), A X- is a randomly selected point in a given interval. Let's give the argument X incrementΔx (positive or negative).

The function y =f(x) will receive an increment Δу equal to:

Δy = f(x + Δx)-f(x).

At infinitesimal Δx incrementΔy is also infinitely small.

For example:

Let's consider solving the derivative of a function using the example of a freely falling body.

Since t 2 = t l + Δt, then

Having calculated the limit, we find:

The notation t 1 is introduced to emphasize the constancy of t when calculating the limit of the function. Since t 1 is an arbitrary time value, index 1 can be discarded; then we get:

It can be seen that the speed v, like the way s, There is function time. Function type v depends entirely on the type of function s, so the function s as if “producing” a function v. Hence the name " derivative function».

Consider another one example.

Find the value of the derivative of the function:

y = x 2 at x = 7.

Solution. At x = 7 we have y=7 2 = 49. Let's give the argument X increment Δ X. The argument will become equal 7 + Δ X, and the function will receive the value (7 + Δ x) 2.

Dear friends! The group of tasks related to the derivative includes tasks - the condition gives a graph of a function, several points on this graph and the question is:

At what point is the derivative greatest (smallest)?

Let's briefly repeat:

The derivative at a point is equal to the slope of the tangent passing throughthis point on the graph.
UThe global coefficient of the tangent, in turn, is equal to the tangent of the angle of inclination of this tangent.

*This refers to the angle between the tangent and the x-axis.

1. At intervals of increasing function, the derivative has a positive value.

2. At intervals of its decrease, the derivative has a negative value.

Consider the following sketch:

At points 1,2,4, the derivative of the function has a negative value, since these points belong to decreasing intervals.

At points 3,5,6, the derivative of the function has a positive value, since these points belong to increasing intervals.

As you can see, everything is clear with the meaning of the derivative, that is, it is not at all difficult to determine what sign it has (positive or negative) at a certain point in the graph.

Moreover, if we mentally construct tangents at these points, we will see that straight lines passing through points 3, 5 and 6 form angles with the oX axis ranging from 0 to 90 o, and straight lines passing through points 1, 2 and 4 form with the oX axis the angles range from 90 o to 180 o.

*The relationship is clear: tangents passing through points belonging to intervals of increasing functions form acute angles with the oX axis, tangents passing through points belonging to intervals of decreasing functions form obtuse angles with the oX axis.

Now the important question!

How does the value of the derivative change? After all, the tangent at different points on the graph of a continuous function forms different angles, depending on which point on the graph it passes through.

*Or, in simple terms, the tangent is located more “horizontally” or “vertically”. Look:

Straight lines form angles with the oX axis ranging from 0 to 90 o

Straight lines form angles with the oX axis ranging from 90° to 180°

Therefore, if you have any questions:

— at which of the given points on the graph does the derivative have the smallest value?

- at which of the given points on the graph does the derivative have the greatest value?

then to answer it is necessary to understand how the value of the tangent of the tangent angle changes in the range from 0 to 180 o.

*As already mentioned, the value of the derivative of the function at a point is equal to the tangent of the angle of inclination of the tangent to the oX axis.

The tangent value changes as follows:

When the angle of inclination of the straight line changes from 0° to 90°, the value of the tangent, and therefore the derivative, changes accordingly from 0 to +∞;

When the angle of inclination of the straight line changes from 90° to 180°, the value of the tangent, and therefore the derivative, changes accordingly –∞ to 0.

This can be clearly seen from the graph of the tangent function:

In simple terms:

At a tangent inclination angle from 0° to 90°

The closer it is to 0 o, the greater the value of the derivative will be close to zero (on the positive side).

The closer the angle is to 90°, the more the derivative value will increase towards +∞.

With a tangent inclination angle from 90° to 180°

The closer it is to 90 o, the more the derivative value will decrease towards –∞.

The closer the angle is to 180°, the greater the value of the derivative will be close to zero (on the negative side).

317543. The figure shows a graph of the function y = f(x) and the points are marked–2, –1, 1, 2. At which of these points is the derivative greatest? Please indicate this point in your answer.

We have four points: two of them belong to the intervals on which the function decreases (these are points –1 and 1) and two to the intervals on which the function increases (these are points –2 and 2).

We can immediately conclude that at points –1 and 1 the derivative has a negative value, and at points –2 and 2 it has a positive value. Therefore, in this case, it is necessary to analyze points –2 and 2 and determine which of them will have the largest value. Let's construct tangents passing through the indicated points:

The value of the tangent of the angle between straight line a and the abscissa axis will be greater than the value of the tangent of the angle between straight line b and this axis. This means that the value of the derivative at point –2 will be greatest.

Let's answer the following question: at which point –2, –1, 1 or 2 is the value of the derivative most negative? Please indicate this point in your answer.

The derivative will have a negative value at points belonging to the decreasing intervals, so let’s consider points –2 and 1. Let’s construct tangents passing through them:

We see that the obtuse angle between straight line b and the oX axis is “closer” to 180 O , therefore its tangent will be greater than the tangent of the angle formed by the straight line a and the oX axis.

Thus, at the point x = 1, the value of the derivative will be greatest negative.

317544. The figure shows the graph of the function y = f(x) and the points are marked–2, –1, 1, 4. At which of these points is the derivative the smallest? Please indicate this point in your answer.

We have four points: two of them belong to the intervals at which the function decreases (these are points –1 and 4) and two to the intervals at which the function increases (these are points –2 and 1).

We can immediately conclude that at points –1 and 4 the derivative has a negative value, and at points –2 and 1 it has a positive value. Therefore, in this case, it is necessary to analyze points –1 and 4 and determine which of them will have the smallest value. Let's construct tangents passing through the indicated points:

Answer: 4

I hope I haven’t “overloaded” you with the amount of writing. In fact, everything is very simple, you just need to understand the properties of the derivative, its geometric meaning and how the value of the tangent of the angle changes from 0 to 180 o.

1. First, determine the signs of the derivative at these points (+ or -) and select the necessary points (depending on the question posed).

2. Construct tangents at these points.

3. Using the tangesoid graph, schematically mark the angles and displayAlexander.

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Hello! Let's hit the upcoming Unified State Exam with high-quality systematic preparation and persistence in grinding the granite of science!!! INThere is a competition task at the end of the post, be the first! In one of the articles in this section, you and me, in which a graph of the function was given and various questions were raised regarding extrema, intervals of increase (decrease) and others.

In this article, we will consider the problems included in the Unified State Examination in mathematics, in which a graph of the derivative of a function is given and the following questions are posed:

1. At what point of a given segment does the function take on the largest (or smallest) value.
2. Find the number of maximum (or minimum) points of the function belonging to a given segment.
3. Find the number of extremum points of the function belonging to a given segment.
4. Find the extremum point of the function belonging to the given segment.
5. Find the intervals of increasing (or decreasing) function and in the answer indicate the sum of integer points included in these intervals.
6. Find the intervals of increase (or decrease) of the function. In your answer, indicate the length of the largest of these intervals.
7. Find the number of points at which the tangent to the graph of the function is parallel to or coincides with a line of the form y = kx + b.
8. Find the abscissa of the point at which the tangent to the graph of the function is parallel to the abscissa axis or coincides with it.

There may be other questions, but they will not cause you any difficulties if you understand and (links are provided to articles that provide the information necessary for the solution, I recommend repeating them).

Basic information (briefly):

1. The derivative at increasing intervals has a positive sign.

If the derivative at a certain point from a certain interval has a positive value, then the graph of the function on this interval increases.

2. At decreasing intervals, the derivative has a negative sign.

If the derivative at a certain point from a certain interval has a negative value, then the graph of the function decreases on this interval.

3. The derivative at point x is equal to the slope of the tangent drawn to the graph of the function at the same point.

4. At the points of extremum (maximum-minimum) of the function, the derivative is equal to zero. The tangent to the graph of the function at this point is parallel to the x axis.

This must be clearly understood and remembered!!!

The derivative graph “confuses” many people. Some people inadvertently mistake it for the graph of the function itself. Therefore, in such buildings, where you see that a graph is given, immediately focus your attention in the condition on what is given: the graph of the function or the graph of the derivative of the function?

If it's a graph of the derivative of a function, then treat it as a "reflection" of the function itself, which simply gives you information about that function.

Consider the task:

The figure shows a graph y =f'(X)- derivative of a function f(X), defined on the interval (–2;21).

We will answer the following questions:

1. At what point on the segment is the function f(X) takes the greatest value.

On a given interval, the derivative of a function is negative, which means that the function on this interval decreases (it decreases from the left boundary of the interval to the right). Thus, the greatest value of the function is achieved on the left border of the segment, i.e. at point 7.

Answer: 7

2. At what point on the segment is the function f(X)

From this derivative graph we can say the following. On a given interval, the derivative of the function is positive, which means that the function on this interval increases (it increases from the left boundary of the interval to the right). Thus, the smallest value of the function is achieved on the left border of the segment, that is, at the point x = 3.

Answer: 3

3. Find the number of maximum points of the function f(X)

The maximum points correspond to the points where the derivative sign changes from positive to negative. Let's consider where the sign changes in this way.

On the segment (3;6) the derivative is positive, on the segment (6;16) it is negative.

On the segment (16;18) the derivative is positive, on the segment (18;20) it is negative.

Thus, on a given segment the function has two maximum points x = 6 and x = 18.

Answer: 2

4. Find the number of minimum points of the function f(X), belonging to the segment.

Minimum points correspond to points where the derivative sign changes from negative to positive. Our derivative is negative on the interval (0;3), and positive on the interval (3;4).

Thus, on the segment the function has only one minimum point x = 3.

*Be careful when writing down the answer - the number of points is recorded, not the x value; such a mistake can be made due to inattention.

Answer: 1

5. Find the number of extremum points of the function f(X), belonging to the segment.

Please note what you need to find quantity extremum points (these are both maximum and minimum points).

Extremum points correspond to points where the sign of the derivative changes (from positive to negative or vice versa). In the graph given in the condition, these are the zeros of the function. The derivative vanishes at points 3, 6, 16, 18.

Thus, the function has 4 extremum points on the segment.

Answer: 4

6. Find the intervals of increasing function f(X)

Intervals of increase of this function f(X) correspond to the intervals on which its derivative is positive, that is, the intervals (3;6) and (16;18). Please note that the boundaries of the interval are not included in it (round brackets - boundaries are not included in the interval, square brackets - included). These intervals contain integer points 4, 5, 17. Their sum is: 4 + 5 + 17 = 26

Answer: 26

7. Find the intervals of decreasing function f(X) at a given interval. In your answer, indicate the sum of integer points included in these intervals.

Decreasing intervals of a function f(X) correspond to intervals on which the derivative of the function is negative. In this problem these are the intervals (–2;3), (6;16), (18:21).

These intervals contain the following integer points: –1, 0, 1, 2, 7, 8, 9, 10, 11, 12, 13, 14, 15, 19, 20. Their sum is:

(–1) + 0 + 1 + 2 + 7 + 8 + 9 + 10 +

11 + 12 + 13 + 14 + 15 + 19 + 20 = 140

Answer: 140

*Pay attention to the condition: whether the boundaries are included in the interval or not. If boundaries are included, then in the intervals considered in the solution process these boundaries must also be taken into account.

8. Find the intervals of increasing function f(X)

Intervals of increasing function f(X) correspond to intervals on which the derivative of the function is positive. We have already indicated them: (3;6) and (16:18). The largest of them is the interval (3;6), its length is 3.

Answer: 3

9. Find the intervals of decreasing function f(X). In your answer, indicate the length of the largest of them.

Decreasing intervals of a function f(X) correspond to intervals on which the derivative of the function is negative. We have already indicated them; these are the intervals (–2;3), (6;16), (18;21), their lengths are respectively 5, 10, 3.

The length of the largest is 10.

Answer: 10

10. Find the number of points at which the tangent to the graph of the function f(X) parallel to or coincides with the straight line y = 2x + 3.

The value of the derivative at the point of tangency is equal to the slope of the tangent. Since the tangent is parallel to the straight line y = 2x + 3 or coincides with it, their angular coefficients are equal to 2. This means that it is necessary to find the number of points at which y′(x 0) = 2. Geometrically, this corresponds to the number of points of intersection of the derivative graph with the straight line y = 2. There are 4 such points on this interval.

Answer: 4

11. Find the extremum point of the function f(X), belonging to the segment.

The extremum point of a function is the point at which its derivative is equal to zero, and in the vicinity of this point the derivative changes sign (from positive to negative or vice versa). On the segment, the derivative graph intersects the x-axis, the derivative changes sign from negative to positive. Therefore, the point x = 3 is an extremum point.

Answer: 3

12. Find the abscissa of the points at which the tangents to the graph y = f (x) are parallel to the abscissa axis or coincide with it. In your answer, indicate the largest of them.

The tangent to the graph y = f (x) can be parallel to the abscissa axis or coincide with it, only at points where the derivative is equal to zero (these can be extremum points or stationary points in the vicinity of which the derivative does not change its sign). This graph shows that the derivative is zero at points 3, 6, 16,18. The largest is 18.

You can structure your reasoning this way:

The value of the derivative at the point of tangency is equal to the slope of the tangent. Since the tangent is parallel to or coincides with the x-axis, its slope is 0 (indeed, the tangent of an angle of zero degrees is zero). Therefore, we are looking for the point at which the slope is equal to zero, and therefore the derivative is equal to zero. The derivative is equal to zero at the point at which its graph intersects the x-axis, and these are points 3, 6, 16,18.

Answer: 18

The figure shows a graph y =f'(X)- derivative of a function f(X), defined on the interval (–8;4). At what point of the segment [–7;–3] is the function f(X) takes the smallest value.

The figure shows a graph y =f'(X)- derivative of a function f(X), defined on the interval (–7;14). Find the number of maximum points of the function f(X), belonging to the segment [–6;9].

The figure shows a graph y =f'(X)- derivative of a function f(X), defined on the interval (–18;6). Find the number of minimum points of the function f(X), belonging to the segment [–13;1].

The figure shows a graph y =f'(X)- derivative of a function f(X), defined on the interval (–11; –11). Find the number of extremum points of the function f(X), belonging to the segment [–10; -10].

The figure shows a graph y =f'(X)- derivative of a function f(X), defined on the interval (–7;4). Find the intervals of increasing function f(X). In your answer, indicate the sum of integer points included in these intervals.

The figure shows a graph y =f'(X)- derivative of a function f(X), defined on the interval (–5;7). Find the intervals of decreasing function f(X). In your answer, indicate the sum of integer points included in these intervals.

The figure shows a graph y =f'(X)- derivative of a function f(X), defined on the interval (–11;3). Find the intervals of increasing function f(X). In your answer, indicate the length of the largest of them.

F The figure shows a graph

The conditions of the problem are the same (which we considered). Find the sum of three numbers:

1. The sum of the squares of the extrema of the function f (x).

2. The difference between the squares of the sum of the maximum points and the sum of the minimum points of the function f (x).

3. The number of tangents to f (x) parallel to the straight line y = –3x + 5.

The first one to give the correct answer will receive an incentive prize of 150 rubles. Write your answers in the comments. If this is your first comment on the blog, it will not appear immediately, but a little later (don’t worry, the time the comment was written is recorded).

Good luck to you!

Best regards, Alexander Krutitsikh.

P.S: I would be grateful if you tell me about the site on social networks.

Sergey Nikiforov

If the derivative of a function is of constant sign on an interval, and the function itself is continuous on its boundaries, then the boundary points are added to both increasing and decreasing intervals, which fully corresponds to the definition of increasing and decreasing functions.

Farit Yamaev 26.10.2016 18:50

Hello. How (on what basis) can we say that at the point where the derivative is equal to zero, the function increases. Give reasons. Otherwise, it's just someone's whim. By what theorem? And also proof. Thank you.

Support

The value of the derivative at a point is not directly related to the increase in the function over the interval. Consider, for example, functions - they are all increasing on the interval

Vladlen Pisarev 02.11.2016 22:21

If a function is increasing on the interval (a;b) and is defined and continuous at points a and b, then it is increasing on the interval . Those. point x=2 is included in this interval.

Although, as a rule, increase and decrease are considered not on a segment, but on an interval.

But at the point x=2 itself, the function has a local minimum. And how to explain to children that when they are looking for points of increase (decrease), we do not count the points of local extremum, but enter into intervals of increase (decrease).

Considering that the first part of the Unified State Exam is for the “middle group of kindergarten”, then such nuances are probably too much.

Separately, many thanks to all the staff for “Solving the Unified State Exam” - an excellent guide.

Sergey Nikiforov

A simple explanation can be obtained if we start from the definition of an increasing/decreasing function. Let me remind you that it sounds like this: a function is called increasing/decreasing on an interval if a larger argument of the function corresponds to a larger/smaller value of the function. This definition does not use the concept of derivative in any way, so questions about the points where the derivative vanishes cannot arise.

Irina Ishmakova 20.11.2017 11:46

Good afternoon. Here in the comments I see beliefs that boundaries need to be included. Let's say I agree with this. But please look at your solution to problem 7089. There, when specifying increasing intervals, boundaries are not included. And this affects the answer. Those. the solutions to tasks 6429 and 7089 contradict each other. Please clarify this situation.

Alexander Ivanov

Tasks 6429 and 7089 have completely different questions.

One is about increasing intervals, and the other is about intervals with a positive derivative.

There is no contradiction.

The extrema are included in the intervals of increasing and decreasing, but the points in which the derivative is equal to zero are not included in the intervals in which the derivative is positive.

A Z 28.01.2019 19:09

Colleagues, there is a concept of increasing at a point

(see Fichtenholtz for example)

and your understanding of the increase at x=2 is contrary to the classical definition.

Increasing and decreasing is a process and I would like to adhere to this principle.

In any interval that contains the point x=2, the function is not increasing. Therefore, the inclusion of a given point x=2 is a special process.

Usually, to avoid confusion, inclusion of the ends of intervals is discussed separately.

Alexander Ivanov

A function y=f(x) is said to be increasing over a certain interval if a larger value of the argument from this interval corresponds to a larger value of the function.

At the point x=2 the function is differentiable, and on the interval (2; 6) the derivative is positive, which means on the interval )