In problem B9, a graph of a function or derivative is given, from which it is required to determine one of the following quantities:

1. The value of the derivative at some point x 0,
2. High or low points (extremum points),
3. Intervals of increasing and decreasing functions (intervals of monotonicity).

The functions and derivatives presented in this problem are always continuous, which greatly simplifies the solution. Despite the fact that the task belongs to the section of mathematical analysis, it is quite within the power of even the most weak students, 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.

Carefully read the condition of problem B9 in order not to make stupid mistakes: sometimes quite voluminous texts come across, but there are few important conditions that affect the course of the solution.

Calculation of the value of the derivative. Two point method

If the problem is given a graph of the 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:

1. 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 the key point of the solution, and any mistake here leads to the wrong answer.
2. 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 .
3. Finally, we find the value of the derivative D = Δy/Δx. In other words, you need to divide the function increment by the argument increment - and this will be the answer.

Once again, we note: points A and B must be sought precisely on the tangent, and not on the graph of the function f(x), as is often the case. The tangent will necessarily contain at least two such points, otherwise the problem is formulated incorrectly.

Consider the points A (−3; 2) and B (−1; 6) and find the increments:
Δx \u003d x 2 - x 1 \u003d -1 - (-3) \u003d 2; Δy \u003d y 2 - y 1 \u003d 6 - 2 \u003d 4.

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

Task. The figure shows the graph of the function y \u003d f (x) and the 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 increments:
Δx \u003d x 2 - x 1 \u003d 3 - 0 \u003d 3; Δy \u003d y 2 - y 1 \u003d 0 - 3 \u003d -3.

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

Task. The figure shows the graph of the function y \u003d f (x) and the 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 increments:
Δx \u003d x 2 - x 1 \u003d 5 - 0 \u003d 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 the rule: if the tangent is parallel to the OX axis, the derivative of the function at the point of contact is equal to zero. In this case, you don’t even need to calculate anything - just look at the graph.

Calculating High and Low Points

Sometimes instead of a graph of a function in problem B9, a derivative graph is given and it is required to find the maximum or minimum point of the function. In this scenario, the two-point method is useless, but there is another, even simpler algorithm. First, let's define the terminology:

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

In order to find the maximum and minimum points on the graph of the derivative, it is enough to perform the following steps:

1. Redraw the graph of the derivative, removing all unnecessary information. As practice shows, extra data only interfere with the solution. Therefore, we mark the zeros of the derivative on the coordinate axis - and that's it.
2. 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. Conversely, if the derivative graph lies below the OX axis, then f'(x) ≤ 0.
3. We again check the zeros and signs of the derivative. Where the sign changes from minus to plus, there is a 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 works only 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 - we will leave only the borders [−5; 5] and the zeros of the derivative x = −3 and x = 2.5. 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 the zeros of the derivative x = −1.7 and x = 5. 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) that belong to the interval [−4; 3].

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

On this graph, there is only one maximum point x = 2. It is in it 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 formulated correctly, such changes should not affect the answer, since the points "without a fixed place of residence" are not directly involved in solving the problem. Of course, with integer points such a trick will not work.

Finding intervals of increase and decrease of a function

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

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

We formulate sufficient conditions for increasing and decreasing:

1. 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.
2. 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.

We accept these assertions without proof. Thus, we obtain a scheme for finding intervals of increase and decrease, which is in many ways similar to the algorithm for calculating extremum points:

1. Remove all redundant information. On the original graph of the derivative, we are primarily interested in the zeros of the function, so we leave only them.
2. 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 has restrictions on the variable x, we additionally mark them on the new chart.
3. Now that we know the behavior of the function and the constraint, it remains to calculate the required value in the problem.

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

As usual, we redraw the graph and mark the boundaries [−3; 7.5], as well as the zeros of the derivative x = −1.5 and x = 5.3. Then we mark 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 segment [−10; 4]. Find the intervals of increasing function f(x). In your answer, write the length of the largest of them.

Let's get rid of redundant information. We leave only the boundaries [−10; 4] and zeros of the derivative, which this time turned out to be four: x = −8, x = −6, x = −3 and x = 2. Note the signs of the derivative and get the following picture:

We are interested in the intervals of increasing function, i.e. 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 it is required to find the length of the largest of the intervals, we write the value l 2 = 5 in response.

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

This article simply and clearly explains what a derivative is and why it is needed.. 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 grows the fastest?

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

Here is another example.

Kostya, Grisha and Matvey got jobs at the same time. Let's see how their income changed during the year:

You can see everything on the chart right away, right? Kostya's income has more than doubled in six months. And Grisha's income also increased, but just a little bit. And Matthew's income decreased to zero. The starting conditions are the same, but the rate of change of the function, i.e. 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?

What we are really looking at is how steeply the graph of the function goes up (or down). In other words, how fast y changes with x. Obviously, the same function at different points can have different meaning derivative - that is, it can change faster or slower.

The derivative of a function is denoted by .

Let's show how to find using the graph.

A graph of some function is drawn. Take a point on it with an abscissa. Draw a tangent to the graph of the function at this point. We want to evaluate how steeply the graph of the function goes up. A handy value for this is tangent of the slope of the tangent.

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

Please note - 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 is the tangent to the graph of a function. This is a straight line that has the only common point with the graph in this section, moreover, as shown in our figure. It looks like a tangent to a circle.

Let's find . We remember that the tangent of an acute angle in right triangle equal to the ratio of the opposite leg to the adjacent one. From triangle:

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

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

The quantity in this equation is called slope of a 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. She expresses geometric meaning 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 slope 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 is increasing. The tangent to the graph drawn at the point forms sharp corner with positive axis direction. So the derivative is positive at the point.

At the point, our function is decreasing. 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 a 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 (maximum point) and (minimum point) the tangent is horizontal. Therefore, the tangent of the slope of the tangent at these points is zero, and the derivative is also zero.

The point is the maximum point. At this point, the increase of 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 equal to zero, but its sign changes from "minus" to "plus".

Conclusion: with the help of the derivative, you can find out everything that interests us about the behavior of the function.

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

If the derivative is negative, then the function is decreasing.

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.

We write these findings in the form of a table:

	increases	maximum point	decreasing	minimum point	increases
+	0	-	0	+

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

A case is possible when the derivative of a function at some point is equal to zero, but the function has neither a maximum nor a minimum at this point. This so-called :

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

It also happens that at the point of maximum or minimum, the derivative does not exist. On the graph, this corresponds to a sharp break, when it is impossible to draw a tangent at a given point.

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

Sergei 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 attached 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 it be argued 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 service

The value of the derivative at a point is not directly related to the increase of the function on the interval. Consider, for example, functions - they all increase 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 the points a and b, then it is increasing on the segment . Those. the point x=2 is included in the given interval.

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

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

Considering that the first part of the exam for " middle group kindergarten", then perhaps such nuances are too much.

Separately, thank you very much for "I will decide the exam" for all employees - an excellent benefit.

Sergei 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 the interval if the larger argument of the function corresponds to a larger/smaller value of the function. Such a definition does not use the concept of a 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 borders should be included. Let's say I agree with this. But look, please, at your solution to problem 7089. There, when specifying intervals of increase, the boundaries are not included. And that affects the response. Those. the solutions of tasks 6429 and 7089 contradict each other. Please clarify this situation.

Alexander Ivanov

Tasks 6429 and 7089 have completely different questions.

In one, there are intervals of increase, and in the other, there are intervals with a positive derivative.

There is no contradiction.

The extrema are included in the intervals of increase and decrease, but the points at which the derivative is equal to zero do not enter the intervals at 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 the point 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 given point x=2 is a special process.

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

Alexander Ivanov

The function y=f(x) is called increasing on some interval if the larger value of the argument from this interval corresponds to the 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 that its values ​​​​are strictly positive on the interval, which means that the function only increases in this section, so the value of the function at the left end x = −3 less than its value at the right end x = −2.

Answer: φ 2 (−3) φ 2 (−2)

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

From the graph of the antiderivative, it can be seen that the point x = −1 is in the increasing area, hence the value of the corresponding derivative is positive. Dot x = 4 is in the decreasing area 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)

You can ask a lot of similar questions on the missing graph, which leads to a wide 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 the derivative according to the graph of a function.

Picture 1.

Figure 2.

Task 1

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

The derivative of the function is positive in those areas where the function increases. It can be seen from the figure that these are the intervals (−10.5;−7.6), (−1;8.2) and (15.7;19). Let's list integer 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 tasks about function graphs it is required to name "points", as a rule, only the values ​​of the argument are meant 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 here, because the condition says that this is an interval.

Task 2

Figure 1 shows a 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 a function is negative in those areas where the function is decreasing. The figure shows that these are the intervals (−7.6;−1) and (8.2;15.7). Integer points inside 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 on the previous problem.

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

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

At these points, the derivative of the function either vanishes or does not exist ( necessary condition for an extremum).
However, a necessary condition is a sign, but not a guarantee of the existence of an extremum of the function. A sufficient condition for an extremum is the sign change 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 at a point the derivative of the function is equal to zero, 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 an extremum point of the function. It can be an inflection point, a break point, or a break point in a function graph.

Task 3

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

Recall that the equation of a straight line has the form y = kx + b , where k- 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. So the desired tangents must also be parallel to the axis Ox and must also have a slope coefficient of 0. Tangents have this property at the points of extrema of functions. Therefore, to answer the question, you just need to count all the extreme points on the chart. There are 4 of them - two maximum points and two minimum points.

Answer: 4

Task 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 specified 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

Task 5

Figure 1 shows a graph of the function y = f (x ) defined on the interval (−10.5;19). Find the number of points where 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 can be seen on the graph:
2 highs and 2 lows.

Answer: 4

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

Picture 1.

Figure 2.

Task 6

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

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

Answer: −6

Comment: From the graph of the derivative, it can be seen 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 at this point there could not be an extremum of the function. Most likely there was an inflection point in the graph of the original function.

Task 7

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

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

Answer: 3

Task 8

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

According to the necessary extremum condition, the maximum of the function may be at the points where its derivative is equal to zero. On a given segment, these are the points: x = −2, x = 2, x = 6, x = 10. But according to the sufficient condition, it 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

Task 9

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

The extrema of the function can be at those points where its derivative is equal to 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 sign, so it must be excluded from consideration. Thus, 4 points remain.

Answer: 4

Task 10

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

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

Answer: 8

Task 11

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

The slope coefficient (aka the tangent of the slope angle) 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 −2. There are 9 such points in Figure 2. It is convenient to count them at 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 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 you will find it very easy. Other types of tasks of this task - on the geometric meaning of the antiderivative - will be considered in another section.