Derivative of the function. The geometric meaning of the derivative. Derivative How to solve the graphs of the derivative of the ege function

Master - class in mathematics

in grade 11

on this topic

"DERIVATIVE FUNCTION

IN THE TASKS OF THE USE "

mathematic teacher

Martynenko E.N.

2017-2018 academic year

The purpose of the master class: develop students' skillsapplication of theoretical knowledge on the topic "Derivative of a function" for solving problems of a single state examination.

Tasks

Educational:generalize and systematize the knowledge of students on the topic

"Function derivative", consider prototypes tasks of the exam on this topic, provide students with the opportunity to test their knowledge when solving problems on their own.

Developing: promote the development of memory, attention, self-esteem and self-control skills; the formation of basic key competencies (comparison, juxtaposition, classification of objects, determination of adequate ways to solve an educational problem based on specified algorithms, the ability to act independently in a situation of uncertainty, control and evaluate their activities, find and eliminate the causes of difficulties).

Educational: contribute:

Formation of a responsible attitude towards learning among students;

developing a sustained interest in mathematics;

creating a positive intrinsic motivation to study mathematics.

Technologies : individually differentiated learning, ICT.

Teaching methods : verbal, visual, practical, problematic.

Forms of work: individual, frontal, in pairs.

Equipment and materials for the lesson: projector, screen, PC, simulator(Appendix # 1), presentation for the lesson(Appendix # 2), individually - differentiated cards for independent work in pairs(Appendix No. 3), list of Internet sites, individually differentiated homework (Appendix # 4).

Explanation for the master class.

This master class is held in grade 11 to prepare for the Unified State Exam. Aims at the application of theoretical material on the topic "Derivative of a function" in solving exam problems.

Duration of the master class - 20 minutes.

Master class structure

I. Organizational moment -1 min.

II. Communication of the topic, the goals of the master - class, motivation of educational activities - 1 min.

III. Frontal work. Training "Tasks No. 14 BASE, No. 7 PROFILE of the Unified State Exam". Analysis of work with the simulator - 7 min.

IV.Individually - differentiated work in pairs. Independent solution tasks No. 12. (PROFILE) Mutual check - 9 min. On - line testing. (BASE) Analysis of test results - 8 min

V. Checking individual homework. -1 min.

Vi. Individually - differentiated homework -1 min.

Vii. CONTROL TEST 20 MINUTES (4 OPTIONS)

Master class progress

I .Organizing time.

II . Communication of the topic, the goals of the master - class, motivation of educational activities.

(Slides 1-2, Appendix # 2)

The topic of our lesson is "The derivative of a function in uSE assignments". Everyone knows the saying "Small spool but expensive." Derivative is one of such “spools” in mathematics. The derivative is used to solve many practical tasks mathematics, physics, chemistry, economics and other disciplines. It allows you to solve problems simply, beautifully, and interestingly.

The topic "Derivative" is presented in the task No. 14 of the basic level and in the tasks of the profile level No. 7,12, 18 and the unified state examination.

You worked with documents regulating the structure and content of control measuring materials of the unified state exam in mathematics 2018. Make a conclusion about what knowledge and skills you need to successfully solve the USE problems on the topic "Derivative".

(Slides 3-4, Appendix # 2)

Have you learned "Codifier of content elements in MATHEMATICS for the preparation of control measuring materials for the unified state examination",

"Codifier of requirements for the level of training of graduates", "Specification of control measuring materials", "Demonstration version of control measuring materials of the unified state exam 2018" andfound out what knowledge and skills about the function and its derivative are needed to successfully solve problems on the topic "Derivative".

It is necessary

• KNOW

derivative calculation rules;

derivatives of basic elementary functions;

the geometric and physical meaning of the derivative;
equation of the tangent to the graph of the function;
investigation of a function using a derivative.

• Be able to

perform actions with functions (describe the behavior and properties of a function according to the graph, find its highest and lowest values).

• USE

acquired knowledge and skills in practice and everyday life.

You have theoretical knowledge of the Derivative topic. Today we willLEARN TO APPLY THE KNOWLEDGE ABOUT THE DERIVATIVE FUNCTION TO SOLVE THE USE PROBLEMS.(Slide 4, Appendix # 2)

It's not for nothing Aristotle said that“THE MIND IS NOT ONLY IN KNOWLEDGE, BUT ALSO IN THE ABILITY TO APPLY KNOWLEDGE IN PRACTICE”(Slide 5, Appendix No. 2)

At the end of the lesson we will return to the goal of our lesson and find out if we have achieved it?

III ... Frontal work.Training "Tasks No. 14 BASE No. 7 PROFILE of the Unified State Exam" (Appendix No. 1). Analysis of work with the simulator.

Choose the correct answer from the four suggested.

What, in your opinion, is the difficulty of completing task # 7?

What do you think, what are the typical mistakes graduates make on the exam when solving this problem?

When answering the questions of task No. 14 BASE AND No. 7 PROFILE, you should be able to describe the behavior and properties of the function from the graph of the derivative, and from the graph of the function - the behavior and properties of the derivative of the function. And this requires good theoretical knowledge on the following topics: “Geometric and mechanical meaning of the derivative. Tangent to the graph of the function. Application of the derivative to the study of functions ”.

Analyze what tasks caused you difficulties?

What theoretical questions do you need to know?

IV. Оn - line testing on assignments No. 14 (BASE) Analysis of test results.

Site for testing in the lesson:http://www.mathb-ege.sdamgia.ru/

Who has not made mistakes?

Who experienced difficulty in testing? Why?

In what tasks were mistakes made?

Conclude, what theoretical questions do you need to know?

Individually - differentiated work in pairs. Independent solution of problems №12. (PROFILE) Mutual verification.(Appendix # 3)

Remember the algorithm for solving problems №12 of the USE for finding extremum points, extrema of a function, the largest and smallest values \u200b\u200bof a function on the interval using the derivative.

Solve problems with a derivative

The students are faced with a problem:

"Think, is it possible to solve some problems # 12 in a different way, without using a derivative?"

1 pair

2 pair

3 pair

4 pair

(Students defend their solution by writing the main steps for solving problems on the chalkboard. Students provide two ways to solve problem # 2).

Solution of a problem. Conclusion for students:

"Some problems No. 12 of the exam for finding the smallest and largest value of a function can be solved without using the derivative, relying on the properties of the functions."

Analyze what mistake you made in the task?

What theoretical questions do you need to repeat?

V. Checking individual homework. (Slides 7-8, Appendix No. 2)

V. Vegelman was given an individual homework assignment: from the training manuals for the exam number 18.

(The student gives the solution to the problem, relying on the functional-graphic method, as one of the methods for solving problems No. 18 of the exam and gives a short explanation of this method).

Vii. Individually - differentiated homework

(Slide 9, Appendix No. 2), (Appendix # 4).

I have prepared a list of Internet sites to prepare for the exam. You can also take online testing on these sites. For the next lesson you need to: 1) review the theoretical material on the topic "Derivative of a function";

2) on the site "Open bank of tasks in mathematics" (http://mathege.ru/ ) find prototypes of tasks No. 14 BASE AND No. 7 and 12 PROFILE and solve at least 10 PROFILE problems;

3) Vegelman V., to solve problems with parameters (APPENDIX 4). tasks 1-8 (option 1). A BASIC LEVEL OF

VIII. Lesson grades.

How would you rate yourself for a lesson?

Do you think you could have done better in class?

IX. Lesson summary. Reflection

Let's summarize our work. What was the purpose of the lesson? Do you think it has been achieved?

Look at the board and in one sentence, choosing the beginning of the phrase, continue with the sentence that suits you best.

I felt…

I learned…

I managed …

I was able to ...

I'll try …

I was surprised that …

I wanted…

Can you say that during the lesson there was an enrichment of your stock of knowledge?

So, you repeated the theoretical questions about the derivative of the function, applied your knowledge in solving the prototypes of the USE tasks (No. 14 BASIC LEVEL No. 7,12 PROFILE LEVEL), and V. Vegelman completed the task No. 18 with a parameter, which is a task of an increased degree difficulties.

It was a pleasure for me to work with you, and I hope that you will be able to successfully apply the knowledge gained in mathematics lessons not only for passing the exam, but also in his further studies.

I would like to end the lesson with the words of an Italian philosopher Thomas Aquinas "Knowledge is such a precious thing that it is not shameful to get it from any source"(Slide 10, Appendix # 2).

I wish you success in your preparation for the exam!

Preview:

To use the preview of presentations, create yourself a Google account (account) and log into it: https://accounts.google.com

Slide captions:

Preparing for the exam SIMULATOR on the topic "Derivative" Task number 14 basic level, number 7, 12 profile level

f (x) f / (x) x The figure shows the graph of the derivative of the function y \u003d f (x), specified on the interval (- 8; 8). Let's explore the properties of the graph and we will be able to answer many questions about the properties of the function, although the graph of the function itself is not presented! y \u003d f / (x) 1 2 3 4 5 6 7 -7 -6 -5 -4 -3 -2 -1 4 3 2 1 -1 -2 -3 -4 -5 yx 6 3 0 -5 Find points where f / (x) \u003d 0 (these are the zeros of the function). + - - + +

ASSIGNMENT number 14 Mathematics basic level

The figure shows a graph of the function y \u003d f (x) and points A, B, C and D on the Ox axis are marked. Using the graph, assign to each point the characteristics of the function and its derivative. ABCD 1) the value of the function at the point is negative, and the value of the derivative of the function at the point is positive 2) the value of the function at the point is positive, and the value of the derivative of the function at the point is negative 3) the value of the function at the point is negative, and the value of the derivative of the function at the point is negative 4) the value of the function at the point is positive, and the value of the derivative of the function at the point is positive

№ 1 The figure shows the graph of the function y \u003d f (x) and marked points A, B, C and D on the Ox axis. Using the graph, assign to each point the characteristics of the function and its derivative. 1) the value of the function at the point is positive, and the value of the derivative of the function at the point is negative 2) the value of the function at the point is negative, and the value of the derivative of the function at the point is negative 3) the value of the function at the point is positive, and the value of the derivative of the function at the point is positive 4) the value of the function at the point is negative, and the value of the derivative of the function at the point is positive ABCD

The figure shows the graph of the function y \u003d f (x). Points a, b, c, d, and e define intervals on the Ox axis. Using the graph, assign to each interval the characteristic of the function or its derivative. A) (a; b) B) (b; c) C) (c; d) D) (d; e) 1) the values \u200b\u200bof the function are positive at each point of the interval 2) the values \u200b\u200bof the derivative of the function are negative at each point of the interval 3) the values derivative of the function are positive at each point of the interval 4) the values \u200b\u200bof the function are negative at each point of the interval

The figure shows the graph of the function y \u003d f (x). The numbers a, b, c, d, and e define intervals on the Ox axis. Using the graph, assign to each interval the characteristics of the function or its derivative. A) (a; b) B) (b; c) C) (c; d) D) (d; e) 1) the values \u200b\u200bof the function are positive at each point of the interval 2) the values \u200b\u200bof the function are negative at each point of the interval 3) the values \u200b\u200bof the derivative functions are negative at each point of the interval 4) the values \u200b\u200bof the derivative of the function are positive at each point of the interval

The figure shows a graph of a function and tangents drawn to it at points with abscissas A, B, C and D. A B C D 1) - 1.5 2) 0.5 3) 2 4) - 0.3

The figure shows a graph of a function and tangents drawn to it at points with abscissas A, B, C and D. A B C D 1) 23 2) - 12 3) - 113 4) 123

ASSIGNMENT number 7 Mathematics profile level

Problems for the geometric meaning of the derivative

1) 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 at x 0. -2 -0.5 2 0.5 Think! Think! Right! Think! x 0 Geometric meaning derivative: k \u003d tg α The angle of inclination of the tangent to the Ox axis is obtuse, so k

5 11 8 2) The continuous function y \u003d f (x) is set on the interval (-6; 7). The figure shows her graph. Find the number of points at which the tangent to the graph of the function is parallel to the straight line y \u003d 6. Checking y \u003d f (x) y x 3 Think! Think! Think! Right! - 6 7 y \u003d 6. Breakpoint. The derivative does NOT exist at this point! О -4 3 5 1, 5

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

3) The figure shows the graph of the derivative of the function y \u003d f / (x), given on the interval (- 6; 8). Examine the function y \u003d f (x) for extremum and indicate the number of its extremum points. 2 1 4 5 Wrong! Not true! Right! Not true! Check (2) f (x) f / (x) -2 + - y \u003d f / (x) 1 2 3 4 5 6 7 -7 -6 -5 -4 -3 -2 -1 4 3 2 1 - 1 -2 -3 -4 -5 yx -5 + min max О

4 -3 -2 -1 1 2 3 4 5 x 5) The figure shows the graph of the derivative of a function set in the interval [-5; 5]. Examine the function for monotony and indicate highest point maximum. 3 2 4 5 Think! Think! Right! Think! y \u003d f / (x) + + + - - О - f / (x) - + - + - + f (x) -4 -2 0 3 4 Of the two maximum points, the largest x max \u003d 3 max max y

7) The figure shows the graph of the derivative of the function. Find the length of the increasing interval of this function. Check О -7 -6 -5 -4 -3 -2 -1 7 6 5 4 3 2 1 -1 -2 -3 -4 -5 -6 -7 4 2 3 5 THINK! + THINK! RIGHT! THINK! y x 3 y \u003d f / (x)

4 -3 -2 -1 1 2 3 4 5 x 6) The figure shows the graph of the derivative of a function specified in the interval [-5; 5]. Examine the function y \u003d f (x) for monotonicity and indicate the number of intervals of decrease. 3 2 4 1 Think! Think! Right! Think! y \u003d f / (x) f (x) -4 -2 0 4 f / (x) - + - + - + + О - - - y

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

The figure shows the graph of the differentiable function y \u003d f (x). Nine points are marked on the abscissa: x 1, x 2, ..., x 9. Find all the marked points at which the derivative of the function f (x) is negative. In the answer, indicate the number of these points.

The figure shows the graph of the function y \u003d f (x), defined on the interval (a; b). Determine the number of integer points at which the derivative of the function is positive. a) b) Decide for yourself! Decision. if it increases. Whole solutions for: x \u003d -2; x \u003d -1; x \u003d 5; x \u003d 6. Their number is 4. Whole solutions for: x \u003d 2; x \u003d 3; x \u003d 4; x \u003d 10; x \u003d 11. Their number is 5. Answer: 4. Answer: 5.

Problems for the physical meaning of the derivative

Answer: 3 Answer: 14

ASSIGNMENT number 12 Mathematics profile level

Independent work in pairs Task number 12 Profile level

Preview:

Appendix 3 individual cards No. 12

1. Find the maximum point of the function 1 Find the minimum point of the function

2.Find the maximum point of the function 2Find the minimum point of the function

Linnik D. Vovnenko I

1.Find the smallest function value 1. Find the largest value of the function on the segment

on the segment

Vegelman V.

A.

1. Find the maximum point of the function 1. Find the minimum point of the function

2. Find the smallest value of the function 2. Find the largest value of the function on the segment

On the segment

Leontyeva A. Isaenko K.

The line y \u003d 3x + 2 is tangent to the graph of the function y \u003d -12x ^ 2 + bx-10. Find b, given that the abscissa of the touch point is less than zero.

Show solution

Decision

Let x_0 be the abscissa of the point on the graph of the function y \u003d -12x ^ 2 + bx-10, through which the tangent to this graph passes.

The value of the derivative at the point x_0 is equal to the slope of the tangent, that is, y "(x_0) \u003d - 24x_0 + b \u003d 3. On the other hand, the tangent point belongs to both the graph of the function and the tangent, that is, -12x_0 ^ 2 + bx_0-10 \u003d 3x_0 + 2. We get the system of equations \\ begin (cases) -24x_0 + b \u003d 3, \\\\ - 12x_0 ^ 2 + bx_0-10 \u003d 3x_0 + 2. \\ end (cases)

Solving this system, we get x_0 ^ 2 \u003d 1, which means either x_0 \u003d -1, or x_0 \u003d 1. According to the condition, the abscissa of the touch point is less than zero, therefore x_0 \u003d -1, then b \u003d 3 + 24x_0 \u003d -21.

Answer

Condition

The figure shows the graph of the function y \u003d f (x) (which is a broken line made up of three straight line segments). Using the figure, calculate F (9) -F (5), where F (x) is one of the antiderivatives of f (x).

Show solution

Decision

According to the Newton-Leibniz formula, the difference F (9) -F (5), where F (x) is one of the antiderivatives of the function f (x), is equal to the area of \u200b\u200bthe curvilinear trapezoid bounded by the graph of the function y \u003d f (x), by the straight lines y \u003d 0 , x \u003d 9 and x \u003d 5. According to the graph, we determine that the indicated curved trapezoid is a trapezoid with bases equal to 4 and 3 and a height of 3.

Its area is \\ frac (4 + 3) (2) \\ cdot 3 \u003d 10.5.

Answer

Source: “Mathematics. Preparation for the exam-2017. Profile level ". Ed. FF Lysenko, S. Yu. Kulabukhova.

Condition

The figure shows the graph of y \u003d f "(x) - the derivative of the function f (x), defined on the interval (-4; 10). Find the intervals of decrease of the function f (x). In the answer, indicate the length of the largest of them.

Show solution

Decision

As you know, the function f (x) decreases on those intervals, at each point of which the derivative f "(x) is less than zero. Taking into account that it is necessary to find the length of the largest of them, three such intervals are naturally distinguished from the figure: (-4; -2) ; (0; 3); (5; 9).

The length of the largest of them - (5; 9) is equal to 4.

Answer

Source: “Mathematics. Preparation for the exam-2017. Profile level ". Ed. FF Lysenko, S. Yu. Kulabukhova.

Condition

The figure shows the graph y \u003d f "(x) - the derivative of the function f (x), defined on the interval (-8; 7). Find the number of maximum points of the function f (x) belonging to the interval [-6; -2].

Show solution

Decision

The graph shows that the derivative f "(x) of the function f (x) changes sign from plus to minus (it is at such points that there will be a maximum) at exactly one point (between -5 and -4) from the interval [-6; -2 ]. Therefore, there is exactly one maximum point on the interval [-6; -2].

Answer

Source: “Mathematics. Preparation for the exam-2017. Profile level ". Ed. FF Lysenko, S. Yu. Kulabukhova.

Condition

The figure shows a graph of the function y \u003d f (x), defined on the interval (-2; 8). Determine the number of points at which the derivative of the function f (x) is 0.

Show solution

Decision

The equality to zero of the derivative at a point means that the tangent to the graph of the function drawn at this point is parallel to the Ox axis. Therefore, we find points at which the tangent to the graph of the function is parallel to the Ox axis. On this chart, such points are extreme points (points of maximum or minimum). As you can see, there are 5 extremum points.

Answer

Source: “Mathematics. Preparation for the exam-2017. Profile level ". Ed. FF Lysenko, S. Yu. Kulabukhova.

Condition

Line y \u003d -3x + 4 is parallel to the tangent to the graph of the function y \u003d -x ^ 2 + 5x-7. Find the abscissa of the touch point.

Show solution

Decision

The slope of the straight line to the graph of the function y \u003d -x ^ 2 + 5x-7 at an arbitrary point x_0 is equal to y "(x_0). But y" \u003d - 2x + 5, so y "(x_0) \u003d - 2x_0 + 5. Angular the coefficient of the straight line y \u003d -3x + 4, specified in the condition, is equal to -3. Parallel lines have the same slope. Therefore, we find such a value of x_0 that \u003d -2x_0 + 5 \u003d -3.

We get: x_0 \u003d 4.

Answer

Source: “Mathematics. Preparation for the exam-2017. Profile level ". Ed. FF Lysenko, S. Yu. Kulabukhova.

Condition

The figure shows the graph of the function y \u003d f (x) and the points -6, -1, 1, 4 are marked on the abscissa axis. At which of these points is the value of the derivative the smallest? Indicate this point in your answer.

Function derivative is one of the tricky topics in 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 strive now 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're 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 meaning 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. Draw at this point a 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 the only common point with a graph, 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 at right triangle is equal to the ratio 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 may 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 makes 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. Consequently, 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 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".

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, before 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

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

IN demonstration options 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 \u003d −1 and x \u003d 3) coincide with the extremum points of the red graph: at x \u003d −1 on the red chart we see a local maximum, at x \u003d 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 x 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 \u003d 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 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 answers are given, put it back in to check the result. And only after that see my solution.

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 \u200b\u200bof the function is greater φ 2 (−3) or φ 2 (−2)?

The graph of the derivative shows that on the segment [−3; −2] its values \u200b\u200bare strictly positive, which means that the function on this segment only increases, therefore the value of the function at the left end x \u003d −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 \u200b\u200bof the function is greater φ 2 (−1) or φ 2 (4)?

The antiderivative graph shows that the point x \u003d −1 is in the increasing region, hence the value of the corresponding derivative is positive. Dot x \u003d 4 is in the decreasing region and the value of the corresponding derivative is negative. Insofar as positive value is greater than negative, we conclude that the value of the unknown function, which is precisely the derivative, at point 4 is less 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 tasks 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". Only 15 points.

Answer: 15

Remarks.
1. When in problems about graphs of functions it is required to name "points", as a rule, they mean only the values \u200b\u200bof the argument x , which are the abscissas of the corresponding points located on the graph. The ordinates of these points are the values \u200b\u200bof 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; \u200b\u200b15.7). 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 a function extremum. A sufficient condition for an extremum is the change in 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 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 the extremum point of the function. This can be an inflection point, a break point, or a break point in the function graph.

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 \u003d 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 \u003d 0, i.e. straight y \u003d 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

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 indicated segment, we see 2 extremum points. The maximum of the function is reached at the point x 1 \u003d 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 maximum points and 2 minimum points.

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 ) - 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 \u003d 2. But at the point x \u003d −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 ) - 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 ) - 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 \u003d 10. But according to the sufficient condition, it will definitely beonly 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 ) - 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 \u003d 18. But at the point x \u003d 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 ) - 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 ) - 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 \u003d −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 task - on the geometric meaning of the antiderivative - will be discussed in another section.