How to determine the largest derivative value from a graph. At what point is the value of the derivative greatest? Calculating the value of a derivative

In the interim ( but,b), but NS- is a randomly selected point of the given interval. Let's give an argument NS incrementΔx (positive or negative).

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

Δy = f (x + Δx) -f (x).

At infinitesimal Δх incrementΔy is also infinitely small.

For example:

Let us consider the solution of the derivative of a function using the example of a free fall of a body.

Since t 2 = t l + Δt, then

.

Having calculated the limit, we find:

The notation t 1 is introduced in order 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 dropped; 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 "produces" a function v... Hence the name " derived function».

Consider another 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 an argument NS increment Δ NS... The argument becomes equal 7 + Δ NS, and the function will receive the value (7 + Δ x) 2.

The derivative of a function is one of the difficult topics in the school curriculum. Not every graduate will answer the question what a derivative is.

This article explains simply and clearly what a derivative is and what it is for.... We will not now strive for mathematical rigor of presentation. The most important thing is to understand the meaning.

Let's remember the definition:

The derivative is the rate of change of the function.

The figure shows graphs of three functions. Which one do you think is growing faster?

The answer is obvious - the third. It has the highest rate of change, that is, the largest derivative.

Here's another example.

Kostya, Grisha and Matvey got a job at the same time. Let's see how their income has changed over the year:

You can see everything on the chart right away, isn't it? Kostya's income has more than doubled in six months. And Grisha's income also increased, but only slightly. And Matvey's income dropped to zero. The starting conditions are the same, but the rate of change of the function, that is derivative, - different. As for Matvey, the derivative of his income is generally negative.

Intuitively, we can easily estimate the rate of change of a function. But how do we do it?

We are actually looking at how steeply the function graph goes up (or down). In other words, how fast does y change with changing x. Obviously, the same function at different points can have different values ​​of the derivative - that is, it can change faster or slower.

The derivative of the function is denoted.

Let's show you how to find it using the graph.

A graph of some function is drawn. Let's take a point with an abscissa on it. Let us draw at this point the tangent to the graph of the function. We want to estimate how steeply up the function graph is. A convenient value for this is tangent of the angle of inclination of the tangent.

The derivative of the function at a point is equal to the tangent of the angle of inclination of the tangent drawn to the graph of the function at this point.

Pay attention - as the angle of inclination of the tangent, we take the angle between the tangent and the positive direction of the axis.

Sometimes students ask what a tangent function is. This is a straight line that has a single common point with the graph in this section, and as shown in our figure. It looks like a tangent to a circle.

We'll find it. We remember that the tangent of an acute angle in a right-angled triangle is equal to the ratio of the opposite leg to the adjacent one. From the triangle:

We found the derivative using the graph without even knowing the function formula. Such problems are often found in the exam in mathematics under the number.

There is another important relationship. Recall that the straight line is given by the equation

The quantity in this equation is called slope of the straight line... It is equal to the tangent of the angle of inclination of the straight line to the axis.

.

We get that

Let's remember this formula. It expresses the geometric meaning of the derivative.

The derivative of a function at a point is equal to the slope of the tangent drawn to the graph of the function at that point.

In other words, the derivative is equal to the tangent of the angle of inclination of the tangent.

We have already said that the same function can have different derivatives at different points. Let's see how the derivative is related to the behavior of the function.

Let's draw a graph of some function. Let this function increase in some areas, and decrease in others, and at different rates. And let this function have maximum and minimum points.

At a point, the function increases. A tangent to the graph drawn at a point forms an acute angle with the positive direction of the axis. This means that the derivative is positive at the point.

At the point, our function decreases. The tangent at this point forms an obtuse angle with the positive direction of the axis. Since the tangent of an obtuse angle is negative, the derivative at the point is negative.

Here's what happens:

If the function is increasing, its derivative is positive.

If it decreases, its derivative is negative.

And what will happen at the maximum and minimum points? We see that at the points (maximum point) and (minimum point) the tangent is horizontal. Therefore, the tangent of the angle of inclination of the tangent at these points is zero, and the derivative is also zero.

Point is the maximum point. At this point, the increase in the function is replaced by a decrease. Consequently, the sign of the derivative changes at the point from "plus" to "minus".

At the point - the minimum point - the derivative is also zero, but its sign changes from "minus" to "plus".

Conclusion: using a derivative, you can learn everything that interests us about the behavior of a function.

If the derivative is positive, then the function is increasing.

If the derivative is negative, then the function decreases.

At the maximum point, the derivative is zero and changes sign from "plus" to "minus".

At the minimum point, the derivative is also zero and changes sign from "minus" to "plus".

Let's write these conclusions in the form of a table:

	is increasing	maximum point	decreases	minimum point	is increasing
+	0	-	0	+

Let's make two small clarifications. You will need one of them when solving the problems of the exam. Another - in the first year, with a more serious study of functions and derivatives.

The case is possible when the derivative of a function at some point is equal to zero, but the function has no maximum or minimum at this point. This is the so-called :

At a point, the tangent to the graph is horizontal and the derivative is zero. However, up to the point the function increased - and after the point it continues to increase. The sign of the derivative does not change - as it was positive, it remains.

It also happens that the derivative does not exist at the maximum or minimum point. On the graph, this corresponds to a sharp bend, when a tangent at a given point cannot be drawn.

And how to find the derivative if the function is given not by a graph, but by a formula? In this case, the

New tasks have appeared. Let's take a look at their solution.

Prototype Mission B8 (# 317543)

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

As we know, it is called

the limit of the ratio of the function increment to the argument increment when the argument increment tends to zero:

The derivative at the point shows function change rate at this point. The faster the function changes, that is, the larger the increment of the function, the greater the angle of inclination of the tangent. Since in the problem it is required to determine the point at which the value of the derivative is greatest, we exclude from consideration the points with abscissas -1 and 1 - at these points the function decreases, and the derivative at them is negative.

The function increases at points -2 and 2. However, it increases in them in different ways - at point -2 the graph of the function rises steeper than at point 2, and therefore, the increment of the function at this point, and hence the derivative, is greater.

Answer: -2

And a similar task:

B8 Mission Prototype (# 317544)

The figure shows the graph of the function and the points -2, -1, 1, 4 are marked. At which of these points is the value of the derivative the smallest? Indicate this point in your answer.

The solution to this problem is similar to the solution to the previous one "exactly the opposite"

We are interested in the point at which the derivative takes the smallest value, that is, we are looking for the point at which the function decreases most rapidly - on the graph this is the point at which the steepest "descent" is. This is a point with abscissa 4.

This section contains the problems of the exam in mathematics on topics related to the study of functions and their derivatives.

Demo versions Unified State Exam 2020 year they can meet under the number 14 for basic level and under number 7 for the profile level.

Take a close look at these three function graphs.
Have you noticed that these functions are in some sense "related"?
For example, in those areas where the graph of the green function is located above zero, the red function increases. In those areas where the graph of the green function is below zero, the red function decreases.
Similar remarks can be made for the red and blue graphs.
You can also notice that the zeros of the green function (dots x = −1 and x = 3) coincide with the extremum points of the red graph: at x = −1 on the red chart we see a local maximum, at NS = 3 on the red chart, the local minimum.
It is easy to see that the local highs and lows of the blue chart are reached at the same points where the red chart passes through the value y = 0.
Several more conclusions can be drawn about the peculiarities of the behavior of these graphs, because they are really related to each other. Look at the formulas of the functions located under each of the graphs, and by calculations make sure that each previous one is a derivative for the next one and, accordingly, each next one is one of the pre-forms of the previous function.

φ 1 (x ) = φ" 2 (x ) φ 2 (x ) = Φ 1 (x )
φ 2 (x ) = φ" 3 (x ) φ 3 (x ) = Φ 2 (x )

Let's remember what we know about the derivative:

Derivative of a function y = f(x) at the point NS expresses the rate of change of the function at the point x.

The physical meaning of the derivative lies in the fact that the derivative expresses the rate of the process described by the dependence y = f (x).

The geometric meaning of the derivative lies in the fact that its value at the point under consideration is equal to the slope of the tangent drawn to the graph of the differentiable function at this point.

Now let there be no red graph in the picture. Let us assume that we do not know the function formulas either.

May I ask you something related to the behavior of a function φ 2 (x ) if it is known that it is the derivative of the function φ 3 (x ) and the antiderivative function φ 1 (x )?
Can. And many questions can be answered exactly, because we know that the derivative is a characteristic of the rate of change of a function, so we can judge some of the behavior of one of these functions by looking at the graph of the other.

Before answering the following questions, scroll up the page so that the top figure containing the red graph is hidden. When the answers are given, put it back in to check the result. And only after that see my decision.

Attention: To enhance the teaching effect answers and solutions are loaded separately for each task by sequential pressing of buttons on a yellow background. (When there are a lot of tasks, the buttons may appear with a delay. If the buttons are not visible at all, check if your browser is allowed JavaScript.)

1) Using the graph of the derivative φ" 2 (x ) (in our case, this is a green graph), determine which of the 2 values ​​of the function is greater φ 2 (−3) or φ 2 (−2)?

The graph of the derivative shows that in the segment [−3; −2] its values ​​are strictly positive, which means that the function in this segment only increases, therefore the value of the function at the left end x = −3 is less than its value at the right end x = −2.

Answer: φ 2 (−3) φ 2 (−2)

2) Using the antiderivative graph Φ 2 (x ) (in our case, this is a blue graph), determine which of the 2 values ​​of the function is greater φ 2 (−1) or φ 2 (4)?

The antiderivative graph shows that the point x = −1 is in the increasing region, hence the value of the corresponding derivative is positive. Dot x = 4 is in the region of decreasing and the value of the corresponding derivative is negative. Since the positive value is greater than the negative one, we conclude that the value of the unknown function, which is precisely the derivative, is less at point 4 than at point −1.

Answer: φ 2 (−1) > φ 2 (4)

There are a lot of similar questions you can ask about the missing schedule, which leads to a large variety of problems with a short answer, built according to the same scheme. Try to solve some of them.

Tasks for determining the characteristics of a graph derivative of a function.

Picture 1.

Figure 2.

Problem 1

y = f (x ) defined on the interval (−10.5; 19). Determine the number of integer points at which the derivative of the function is positive.

The derivative of the function is positive in those areas where the function increases. The figure shows that these are the intervals (−10.5; −7.6), (−1; 8.2) and (15.7; 19). Let's list the whole points inside these intervals: "−10", "- 9", "−8", "0", "1", "2", "3", "4", "5", "6", "7", "8", "16", "17", "18". There are 15 points in total.

Answer: 15

Remarks.
1. When in problems about the graphs of functions it is required to name "points", as a rule, they mean only the values ​​of the argument x , which are the abscissas of the corresponding points located on the graph. The ordinates of these points are the values ​​of the function, they are dependent and can be easily calculated if necessary.
2. When listing the points, we did not take into account the edges of the intervals, since the function at these points does not increase or decrease, but "unfolds". The derivative at such points is neither positive nor negative, it is equal to zero, therefore they are called stationary points. In addition, we do not consider the boundaries of the domain of definition here, because the condition says that this is an interval.

Task 2

Figure 1 shows the graph of the function y = f (x ) defined on the interval (−10.5; 19). Determine the number of integer points at which the derivative of the function f " (x ) is negative.

The derivative of the function is negative in those areas where the function decreases. The figure shows that these are the intervals (−7.6; −1) and (8.2; ​​15.7). Integer points within these intervals: "−7", "- 6", "−5", "- 4", "−3", "- 2", "9", "10", "11", "12 "," 13 "," 14 "," 15 ". There are 13 points in total.

Answer: 13

See notes for the previous task.

To solve the following problems, you need to remember one more definition.

The maximum and minimum points of the function are united by a common name - extremum points .

At these points, the derivative of the function is either zero or does not exist ( necessary extremum condition).
However, a necessary condition is a sign, but not a guarantee of the existence of an extremum of a function. A sufficient condition for an extremum is the change of sign of the derivative: if the derivative at a point changes sign from "+" to "-", then this is the maximum point of the function; if the derivative at a point changes sign from "-" to "+", then this is the minimum point of the function; if the derivative of the function is equal to zero at a point, or does not exist, but the sign of the derivative does not change to the opposite when passing through this point, then the specified point is not the extremum point of the function. This can be an inflection point, a break point, or a break point in the graph of a function.

Problem 3

Figure 1 shows the graph of the function y = f (x ) defined on the interval (−10.5; 19). Find the number of points at which the tangent to the graph of the function is parallel to the straight line y = 6 or matches it.

Recall that the equation of the line has the form y = kx + b , where k- the coefficient of inclination of this straight line to the axis Ox... In our case k= 0, i.e. straight y = 6 not tilted but parallel to the axis Ox... This means that the required tangents must also be parallel to the axis Ox and must also have a slope coefficient of 0. Tangents have this property at the extremum points of functions. Therefore, to answer the question, you just need to calculate all the extreme points on the chart. There are 4 of them - two maximum points and two minimum points.

Answer: 4

Problem 4

Functions y = f (x ) defined on the interval (−11; 23). Find the sum of the extremum points of the function on the segment.

On the indicated segment, we see 2 extremum points. The maximum of the function is reached at the point x 1 = 4, minimum at point x 2 = 8.
x 1 + x 2 = 4 + 8 = 12.

Answer: 12

Problem 5

Figure 1 shows the graph of the function y = f (x ) defined on the interval (−10.5; 19). Find the number of points at which the derivative of the function f " (x ) is equal to 0.

The derivative of the function is equal to zero at the extremum points, of which 4 are visible on the graph:
2 points of maximum and 2 points of minimum.

Answer: 4

Tasks for determining the characteristics of a function from the graph of its derivative.

Picture 1.

Figure 2.

Problem 6

Figure 2 shows the graph f " (x ) - the derivative of the function f (x ) defined on the interval (−11; 23). At what point of the segment [−6; 2] the function f (x ) takes the largest value.

On the indicated interval, the derivative was nowhere positive, therefore the function did not increase. It decreased or passed through stationary points. Thus, the function reached its greatest value on the left border of the segment: x = −6.

Answer: −6

Comment: The graph of the derivative shows that on the segment [−6; 2] it is equal to zero three times: at the points x = −6, x = −2, x = 2. But at the point x = −2, it did not change sign, which means that there could not be an extremum of the function at this point. Most likely there was an inflection point in the original function graph.

Problem 7

Figure 2 shows the graph f " (x ) - the derivative of the function f (x ) defined on the interval (−11; 23). At what point of the segment the function takes the smallest value.

On the segment, the derivative is strictly positive, therefore, the function on this segment only increased. Thus, the function reached the smallest value on the left border of the segment: x = 3.

Answer: 3

Problem 8

Figure 2 shows the graph f " (x ) - the derivative of the function f (x ) defined on the interval (−11; 23). Find the number of maximum points of the function f (x ) belonging to the segment [−5; 10].

According to the necessary condition for an extremum, the maximum of the function may be at the points where its derivative is zero. On a given segment, these are points: x = −2, x = 2, x = 6, x = 10. But according to the sufficient condition, it will definitely be only in those of them where the sign of the derivative changes from "+" to "-". On the graph of the derivative, we see that of the listed points, only the point is such x = 6.

Answer: 1

Problem 9

Figure 2 shows the graph f " (x ) - the derivative of the function f (x ) defined on the interval (−11; 23). Find the number of extremum points of the function f (x ) belonging to the segment.

The extrema of a function can be at those points where its derivative is 0. On a given segment of the derivative graph, we see 5 such points: x = 2, x = 6, x = 10, x = 14, x = 18. But at the point x = 14 the derivative has not changed its sign, therefore it must be excluded from consideration. This leaves 4 points.

Answer: 4

Problem 10

Figure 1 shows the graph f " (x ) - the derivative of the function f (x ) defined on the interval (−10.5; 19). Find the intervals of increasing function f (x ). In the answer, indicate the length of the longest of them.

The intervals of increase of the function coincide with the intervals of positiveness of the derivative. On the graph we see three of them - (−9; −7), (4; 12), (18; 19). The longest of them is the second. Its length l = 12 − 4 = 8.

Answer: 8

Assignment 11

Figure 2 shows the graph f " (x ) - the derivative of the function f (x ) defined on the interval (−11; 23). Find the number of points at which the tangent to the graph of the function f (x ) is parallel to the straight line y = −2x − 11 or matches it.

The slope (aka the tangent of the slope) of a given straight line k = −2. We are interested in parallel or coinciding tangents, i.e. straight lines with the same slope. Based on the geometric meaning of the derivative - the slope of the tangent at the considered point of the graph of the function, we recalculate the points at which the derivative is equal to −2. There are 9 such points in Figure 2. It is convenient to count them by the intersections of the graph and the grid line passing through the value −2 on the axis Oy.

Answer: 9

As you can see, using the same graph, you can ask a wide variety of questions about the behavior of a function and its derivative. Also, the same question can be attributed to the graphs of different functions. Be careful when solving this problem on the exam, and it will seem very easy to you. Other types of problems in this assignment - on the geometric meaning of the antiderivative - will be discussed in another section.

Hello! Let's hit the upcoming USE with high-quality systematic preparation, and persistence in crushing the granite of science !!! INat the end of the post there is a competition problem, be the first! In one of the articles of this heading, we are with you, 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 tasks included in the exam in mathematics, in which the graph of the derivative of a function is given, and the following questions are posed:

1. At what point of a given segment the function takes the largest (or smallest) value.

2. Find the number of maximum (or minimum) points of the function that belong to a given segment.

3. Find the number of extremum points of the function that belong 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) of the 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 the 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 the straight line of the form y = kx + b or coincides with it.

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 indicated to articles that provide the information necessary for a solution, I recommend repeating).

Basic information (briefly):

1. The derivative on the 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 increases on this interval.

2. On intervals of decrease, 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 the 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 ox axis.

This must be clearly understood and remembered !!!

Many are confused by the derivative graph. Some 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: a graph of a function or a graph of a derivative of a function?

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

Consider the task:

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

We will answer the following questions:

1. At what point of the segment is the function f(NS) takes the greatest value.

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

Answer: 7

2. At what point of the segment is the function f(NS)

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

Answer: 3

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

The maximum points correspond to the points of the change in the sign of the derivative from positive to negative. Let us 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(NS) belonging to the segment.

The minimum points correspond to the points of the change of the sign of the derivative from negative to positive. Our derivative on the interval (0; 3) is negative, on the interval (3; 4) it is positive.

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

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

Answer: 1

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

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

The extremum points correspond to the points of change of the sign of the derivative (from positive to negative or vice versa). On the given graph 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(NS)

Ascending intervals of this function f(NS) correspond to the intervals on which its derivative is positive, that is, to the intervals (3; 6) and (16; 18). Note that the borders of the interval are not included in it (parentheses - borders are not included in the interval, square - are 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(NS) at a given interval. In your answer, indicate the sum of whole points included in these intervals.

Decrease intervals of a function f(NS) correspond to intervals at 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 the 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(NS)

Ascending function intervals f(NS) 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(NS)... In the answer, indicate the length of the longest of them.

Decrease intervals of a function f(NS) correspond to intervals at 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 equal to 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(NS) parallel to the straight line y = 2x + 3 or coincides with it.

The value of the derivative at the tangent point 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 slopes are 2. Hence, 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 graph of the derivative with the straight line y = 2. On this interval there are 4 such points.

Answer: 4

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

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

Answer: 3

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

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 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 equal to zero at points 3, 6, 16, 18. The largest is 18.

You can build the reasoning this way:

The value of the derivative at the tangent point is equal to the slope of the tangent. Since the tangent is parallel or coincident with the abscissa, its slope is 0 (indeed the tangent of zero degrees is zero). Therefore, we are looking for a point at which the slope is equal to zero, and hence the derivative is equal to zero. The derivative is equal to zero at the point at which its graph crosses the abscissa axis, and these are points 3, 6, 16, 18.

Answer: 18

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

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

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

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

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

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

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

F The figure shows a graph

The condition of the problem is 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 person to give the correct answer will receive an incentive prize - 150 rubles. Write your answers in the comments. If this is your first comment on the blog, then it will not appear immediately, a little later (do not worry, the time of writing the comment is recorded).

Success to you!

Best regards, Alexander Krutitsikh.

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