Functionally graphical method for solving equations examples. Study of various methods for solving inequalities
Target: consider the tasks of ZNO using functional-graphic methods using the example exponential function y \u003d a x, a\u003e 0, a1
Lesson objectives:
repeat the property of monotonicity and boundedness of the exponential function;
repeat the algorithm for plotting function graphs using transformations;
find a set of values and a set of definitions of a function by the form of a formula and using a graph;
solve exponential equations, inequalities and systems using graphs and function properties.
work with graphs of functions containing a module;
look at charts complex function and their range;
1. introduction teachers. Motivation for studying this topic
slide 1 Exponential function. “Functional-graphical methods for solving equations and inequalities”
The functional-graphical method is based on the use of graphic illustrations, the application of function properties and allows solving many mathematical problems.
slide 2 Tasks for the lesson
Today we will consider the problems of ZNO of different levels of complexity using functional-graphic methods using the example of the exponential function y = a x, a > o, a1. With the help of a graphic program, we will perform illustrations for the tasks.
slide 3 Why is it important to know the properties of an exponential function?
According to the law of the exponential function, all life on Earth would multiply if there were favorable conditions, i.e. there were no natural enemies and there was plenty of food. Proof of this is the spread of rabbits in Australia, which were not there before. It was enough to release a couple of individuals, as after a while their offspring became a national disaster.
In nature, technology and economics, there are numerous processes in the course of which the value of a quantity changes by the same number of times, i.e. according to the law of exponential function. These processes are called processes organic growth or organic decay.
For instance, bacteria growth under ideal conditions corresponds to the process of organic growth; radioactive decay– the process of organic attenuation.
obeys the laws of organic growth contribution growth at the savings bank hemoglobin recovery in the blood of a donor or an injured person who has lost a lot of blood.
Give your examples
Application in real life(dose of medication).
Everyone knows that the pills recommended by the doctor for treatment must be taken several times a day, otherwise they will be ineffective. The need for repeated administration of the drug to maintain a constant concentration in the blood is caused by the destruction of the drug in the body. The figure shows how, in most cases, the concentration of drugs in the blood of a person or animal changes after a single injection. Slide 4.
The decrease in drug concentration can be approximated by an exponent whose exponent contains time. Obviously, the rate of destruction of the drug in the body should be proportional to the intensity of metabolic processes.
One tragic case is known, which occurred due to ignorance of this dependence. From a scientific point of view, the drug LSD is very interesting for psychiatrists and neurophysiologists, causing normal people kind of hallucinations. Some researchers decided to study the reaction of the elephant to this drug. To do this, they took the amount of LSD that infuriates cats and multiplied it by the factor that the mass of the elephant is greater than the mass of the cat, believing that the dose of the drug administered should be directly proportional to the mass of the animal. The introduction of such a dose of LSD to an elephant led to his death after 5 minutes, from which the authors concluded that elephants have an increased sensitivity to this drug. A review of this work that appeared later in the press called it an "elephant-like mistake" by the authors of the experiment.
2. Actualization of students' knowledge.
What does it mean to learn a function? (formulate a definition, describe properties, build a graph)
What is the exponential function? Give an example.
What are the main properties of an exponential function?
Scope (Limitation)
domain
monotonicity (ascending-decreasing condition)
slide 5 . Specify the set of function values (according to the finished drawing)
slide 6. Name the condition for the increase and decrease of the function and correlate the formula of the function with its graph
Slide 7. According to the finished drawing, describe the algorithm for plotting function graphs
b) y \u003d 3 x-2 - 2
3.Diagnostic independent work(using PC).
The class is divided into two groups. The main part of the class is doing test tasks. Strong students perform more difficult tasks.
Independent work in the programpower point(for the main part of the class by type test items from ZNO with a closed response form)
Which exponential function is increasing?
Find the scope of the function.
Find the range of the function.
The graph of the function is obtained from the graph of the exponential function by parallel translation along the axis ... by .. units ...
According to the finished drawing, determine the scope and scope of the function
Determine the value of a for which the exponential function passes through the point.
Which figure shows the graph of an exponential function with a base greater than one.
Match the graph of the function with the formula.
The graphical solution of which inequality is shown in the figure.
solve the inequality graphically (according to the finished drawing)
Independent work (for the strong part of the class)
slide 8. Write down the algorithm for plotting a graph of a function, name its domain of definition, range of value, intervals of increase, decrease.
slide 9. Match the formula of a function with its graph
Students check their answers without correcting mistakes, hand over independent work to the teacher
Slide 10. Answers to test tasks
5) D 6) C 7) B 8) 1-D 2-A 3-C 4-B
9) A 10)(2;+ 
)
Slide 11 (check task 8)
4. Study new topic. Application of the functional-graphical method for solving equations, inequalities, systems, determining the range of complex functions
Slide 12. Functionally graphical way to solve equations
To solve an equation of the form f (x) \u003d g (x) by the functional-graphic method, you need:
Construct graphs of functions y=f(x) and y=g(x) in one coordinate system.
Determine the coordinates of the intersection point of the graphs of these functions.
Write down the answer.
TASK №1 SOLUTION OF EQUATIONS
Slide 13.
Does the equation have a root, and if so, is it positive or negative?
6 x \u003d 1/6
(4/3) x = 4
SLIDE 14
5. Implementation of practical work.
slide 15.
This equation can be solved graphically. Students are invited to complete the task, and then answer the question: “Is it necessary to build graphs of functions to solve this equation?”. Answer: “The function is increasing on the entire domain of definition, and the function is decreasing. Therefore, the graphs of such functions have at most one intersection point, which means that the equation has at most one root. By selection, we find that .
Solve the equation:
slide 16. .Solution: we apply the functional method of solving equations:
because this system has a unique solution, then by selection we find x = 1
TASK № 2 SOLUTION OF INEQUALITIES
Graphical methods make it possible to solve inequalities containing different functions. To do this, after plotting the graphs of the functions on the left and right sides of the inequality and determining the abscissa of the intersection point of the graphs, it is necessary to determine the interval on which all points of one of the graphs lie above (below 0 points of the second.
Solve the inequality:
a) cos x 1 + 3 x
slide 1 8. Solution:
Answer: ( ; )
Solve graphically inequality.
Slide 19.
(The graph of the exponential function lies above the function written on the right side of the equation).
Answer: x>2. O
.
Answer: x>0.
TASK №3 The exponential function contains the sign of the modulus in the exponent.
Let's repeat the definition of the module.
(writing on the board)
slide 20.
Make notes in your notebook:
1). 
2). 
A graphic illustration is presented on the slide. Explain how the graphs are built.
Slide 21.
To solve this equation, you need to remember the boundedness property of the exponential function. The function takes values >
1, a - 1 >
1, so equality is possible only if both sides of the equation are simultaneously equal to 1. Hence, Solving this system, we find that X = 0.
TASK 4. Finding the range of complex functions.
slide 22.
Using the ability to build a graph of a quadratic function, determine sequentially the coordinates of the top of the parabola, find the range.
slide 23.
, is the vertex of the parabola.
Question: determine the nature of the monotonicity of the function.
The exponential function y \u003d 16 t increases, since 16>1.
Functional-graphical method for solving the inequality f(x)< g(x). 1. Подбором найдем корень уравнения f(x)=g(x), используя свойства монотонных функций; 2. Построим схематически графики обеих функций, проходящие через точку с найденной абсциссой; 3. Выберем решение неравенства, соответствующее знаку неравенства; 4. Запишем ответ. :
Slide 9 from the presentation « exponential equations and inequalities". The size of the archive with the presentation is 174 KB.Algebra Grade 11
summary other presentations"Equations of the third degree" - (1). Tartaglia refuses. February 12 Cardano repeats his request. "Great Art". X3 + px + q \u003d 0. Example: x3 - 5 x2 + 8 x - 4 \u003d 0 x3 - 2 x2 -3 x2 + 8x - 4 \u003d 0 x2 (x - 2) - (3 x2 - 8x + 4) \u003d 0 3 x2 - 8x + 4 \u003d 0 x \u003d 2 x \u003d 2/3 x2 (x - 2) - (3 (x -2) (x - 2/3)) \u003d 0 x2 (x - 2) - (( x - 2) (3x - 2)) = 0 (x - 2) (x2 - 3x + 2) \u003d 0 x - 2 \u003d 0 x2 - 3x + 2 \u003d 0 x \u003d 2 x \u003d 2 x \u003d 1 Answer: x = 2; x = 1. Our formula gives: Municipality educational institution"Secondary school No. 24". X3 + ax = b (1). Here p = 6 and q = -2. First example:
"Application of a definite integral" - Chap. 4. Development of an elective on the topic "Definite Integral". Definite integral. §4. Properties definite integral. Ch. 2. Different approaches to the theory of the integral in teaching aids for schoolchildren. §one. The volume of the body of revolution. §6. Introduction. Darboux sums. §3. Mechanical work. Purpose: Approaches to the construction of integral theory: Introductory remarks. §2. Integration methods. §3. Conclusion. Chapter 3. Application of a definite integral. §one.
"Exponatory Equations and Inequalities" - 2) Equivalent to the inequality f(x)< g(x), 0<а<1. "Что значит решить задачу? Обоснование: 12). Сравните основание а с единицей: Если 0 The accuracy of such a solution is not high, but with the help of the graph, you can reasonably choose the first approximation, from which the further solution of the equation will begin. There are two ways to graphically solve equations. First way
. All members of the equation are transferred to the left side, i.e. the equation is presented as f(x) = 0. After that, the graph of the function y = f(x) is plotted, where f(x) is the left side of the equation. Abscissas of the points of intersection of the graph of the function y = f(x) with the axis Ox and are the roots of the equation, because at these points y = 0 . Second way
. All members of the equation are divided into two groups, one of them is written on the left side of the equation, and the other on the right, i.e. represent it in the form j(x) = g(x). After that, graphs of two functions y = j(x) and y = g(x) are built. The abscissas of the intersection points of the graphs of these two functions serve as the roots of this equation. Let the point of intersection of the graphs has the abscissa x o , the ordinates of both graphs at this point are equal to each other, i.e. j(x o) = g(x o). From this equality it follows that x 0 is the root of the equation. The process of finding approximate values of the roots of the equation is divided into two stages: 1) separation of roots; 2) refinement of the roots to a given accuracy. The root x of the equation f(x) = 0 is considered separated
on the segment if the equation f(x) = 0 has no other roots on this segment. To separate the roots means to divide the entire range of admissible values into segments, each of which contains one root. Graphical root separation method
- in this case, they do the same as with the graphical method of solving equations. If the curve touches the x-axis, then at this point the equation has a double root (for example, the equation x 3 - 3x + 2 \u003d 0 has three roots: x 1 \u003d -2; x 2 \u003d x 3 \u003d 1). If the equation has a threefold real root, then at the point of contact with the axis X
the curve y \u003d f (x) has an inflection point (for example, the equation x 3 - 3x 2 + 3x - 1 \u003d 0 has a root x 1 \u003d x 2 \u003d x 3 \u003d 1). Analytical root separation method
. To do this, use some properties of functions. Theorem 1
. If the function f(x) is continuous on a segment and takes values of different signs at the ends of this segment, then inside the segment there is at least one root of the equation f(x) = 0. Theorem 2.
If the function f(x) is continuous and monotonic on a segment and takes values of different signs at the ends of the segment, then the segment contains the root of the equation f(x) = 0, and this root is unique. Theorem 3
. If the function f (x) is continuous on a segment and takes values of different signs at the ends of this segment, and the derivative f "(x) retains a constant sign inside the segment, then inside the segment there is a root of the equation f (x) \u003d 0 and, moreover, unique. If the function f(x) is given analytically, then domain of existence (domain of definition) of the function
is called the set of all those real values of the argument for which the analytical expression that defines the function does not lose its numerical meaning and takes only real values. The function y = f(x) is called increasing
if the value of the function increases as the argument increases, and waning
if the value of the function decreases as the argument increases. The function is called monotonous
, if it either only increases or only decreases in a given interval. Let the function f (x) be continuous on the segment and take values of different signs at the ends of the segment, and the derivative f "(x) retains a constant sign on the interval. Then if at all points of the interval the first derivative is positive, i.e. f "(x) >0, then the function f(x) in this interval increases
. If at all points of the interval the first derivative is negative, i.e. f"(x)<0,
то функция в этом интервале decreases
. Let the function f(x) on an interval have a second-order derivative that retains a constant sign on the entire interval. Then if f ""(x)>0, then the graph of the function is convex down
; if f ""(x)<0, то
график функции является convex up
. The points at which the first derivative of a function is equal to zero, as well as those at which it does not exist (for example, goes to infinity), but the function remains continuous, are called critical
. The procedure for separating the roots by the analytical method:
1) Find f "(x) - the first derivative. 2) Make a table of signs of the function f(x), assuming X
equal to: a) critical values (roots) of the derivative or those closest to them; b) boundary values (based on the range of acceptable values of the unknown). Example. Separate the roots of the equation 2x - 5x - 3 = 0. We have f(x) = 2 x - 5x - 3 . The domain of the function f(x) is the entire numerical axis. Calculate the first derivative f "(x) = 2 x ln(2) - 5 . Equate this derivative to zero: 2 x ln(2) - 5 = 0 ; 2 x log(2) = 5 ; 2 x = 5/ln(2) ; xlg(2) = lg(5) - lg(ln(2)) . We compile a table of signs of the function f(x), assuming X
equal to: a) critical values (roots of the derivative) or closest to them; b) boundary values (based on the range of acceptable values of the unknown): The roots of the equation are in the intervals (-1.0) and (4.5). Ivanova Anastasia Task 15 of the Mathematics Profile Exam is an advanced task representing inequality. When solving these inequalities, students must show knowledge of theorems on the equivalence of inequalities of a certain type, the ability to use standard and non-standard methods of solving. An analysis of the content of school textbooks shows that in most of them, methods for solving inequalities using the properties of functions are not given due attention, and almost every year USE assignments offer inequalities, the solution of which is simplified if the properties of functions are applied. According to the statistics presented on the website of the Federal Institute for Pedagogical Measurements, in 2017, about 15% of the exam participants received non-zero points for this task; the maximum score is about 11%. All of the above indicates that students are experiencing great difficulties in solving task No. 15 of the USE. Target: explore different ways of solving inequalities.
: 1. Study the theoretical material on this topic. 2. Consider the examples offered in the bank of USE tasks on the website of the Federal Institute of Pedagogical Measurements. 3. To study functional-graphical methods for solving inequalities. 4. Compare different methods for solving inequalities. 5. Check experimentally which way of solving inequalities is the most rational. Research methods: survey, questioning, analysis, comparison and generalization of results. In our work, we studied functional-graphical methods for solving inequalities. We compared various methods for solving inequalities. We checked experimentally which way of solving inequalities is the most rational. And they came to the conclusion that the student must master several ways to solve inequalities in order to save time and reduce the risk of logical and computational errors. Study of various methods for solving inequalities Ivanova Anastasia Evgenievna 11b class Scientific article (job description) 1. Introduction Relevance. Task No. 15 of the profile exam in mathematics is a task of an increased level of complexity, representing inequality (rational, irrational, exponential, logarithmic). When solving these inequalities, students must show knowledge of theorems on the equivalence of inequalities of a certain type, the ability to use standard and non-standard methods of solving. Complete correct solution of this task is worth 2 points. When solving a problem, any mathematical methods are allowed - algebraic, functional, graphic, geometric, etc. According to the statistics presented on the website of the Federal Institute for Pedagogical Measurements, in 2017, about 15% of the exam participants received non-zero points for this task; the maximum score is about 11%. Typical errors are associated with inattentive reading of the mathematical notation of inequality, misunderstanding of the algorithm for solving populations and systems of logarithmic inequalities. A lot of mistakes were made by the exam participants when solving a fractional rational inequality (the denominator was forgotten). The results of assignment No. 15 by students of our school at the Unified State Examination in mathematics are presented in Table 1 and in the diagram (Fig. 1). Table 1 The results of the task number 15 students of our school Fig.1. The results of the task number 15 students of our school The results of completing task No. 15 at the trial city exam of grades 11a, b in 2017-2018. year are presented in table 2 and in the diagram (Fig. 2). table 2 The results of task number 15 on a trial city exam in 2017-2018 academic year year by students of our school Fig.2. The results of assignment No. 15 at the trial exam in 2017-2018 academic year. year by students of our school We conducted a survey of mathematics teachers of our school and identified the main problems that students have when solving inequalities: incorrect finding of the range of acceptable values of inequalities; consideration of not all cases of transition from logarithmic to rational inequality; transformation of logarithmic expressions; errors in using the interval method, etc. A number of typical errors are associated with the use of the interval method and the introduction of an auxiliary variable. For example, an error in determining the signs on the intervals or an incorrect arrangement of numbers on the coordinate line, according to the criteria, can be treated as computational errors. Others related to skipping the steps of the algorithm or their incorrect execution are evaluated with 0 points. All of the above indicates that students are experiencing great difficulties in solving task No. 15 of the USE in mathematics. In this regard, we have put forward hypothesis : if the student has several ways to solve inequalities, then he will be able to choose the most rational one. Object of study: inequalities. Subject of study: different ways of solving inequalities. Target : explore different ways of solving inequalities. To achieve this goal, the following tasks were solved:
2. Main body 2.1. Theoretical part 1. Linear inequalities Linear inequalitiesare inequalities of the form: ax + b0; ax+b≥0; ax+b≤0, where a and b - any numbers, and a≠0, x is an unknown variable. Rules for transforming inequalities: 1.
Any term of the inequality can be transferred from one part of the inequality to another, while changing the sign to the opposite. 2.
Both parts of the inequality can be multiplied/divided by the same positive number, which results in an inequality equivalent to the given one. 3.
Both parts of the inequality can be multiplied/divided by the same negative number by reversing the sign of the inequality. 2. Quadratic inequalities Inequality of the form where x is a variable, a, b, c are numbers, , is called square. When solving a quadratic inequality, it is necessary to find the roots of the correspondingquadratic equation
3. Rational inequalities Rational inequalitywith one variable x is called an inequality of the form f(x) expressions, i.e. algebraic expressions made up of numbers, variable x and using mathematical operations, i.e. operations of addition, subtraction, multiplication, division and raising to a natural power.Algorithm for solving rational inequalities by the interval method(Annex 1). 4. Exponential inequalities exponential inequality is an inequality , in which the unknown is in the exponent. Protozoaexponential inequality has the form: a x ‹ b or a x › b, where a > 0, a ≠ 1, x is unknown. 5. Logarithmic inequalities logarithmic inequalitycalled an inequality in which the unknown value is under the signlogarithm
.
1. Inequality if reduces to the equivalent inequality. If - then to inequality.
Similarly, the inequalityis equivalent to the inequalities for: ; for : . The solutions of the obtained inequalities must intersect with the ODZ: 2. Solution logarithmic inequality kindis equivalent to solving the following systems: a) Inequality in each of the two cases is reduced to one of the systems: a) 6. Irrational inequalities If the inequality includes functions under the root sign, then such inequalities are called irrational.
2.2. Practical part Study #1 Target : to study the method of limited functions. Progress: 1. Study the method of limited functions. 2. Solve inequalities by this method. To use the boundedness of a function, it is necessary to be able to find the set of function values and to know the bounds on the range of standard functions (for example,) .
Example #1 . Solve the inequality: Solution: Domain: For all x from the resulting set we have: Therefore, the solution to the inequality Answer: Example #2. Solve the inequality: Solution: Because This inequality is equivalent to The first equation of the system has one root x = - 0.4, which also satisfies the second equation. Answer: - 0.4 Conclusion: this method is most efficient if the inequality contains functions such asand others whose ranges are limited from above or below. Study #2 Target : to study the rationalization method for solving inequalities. Progress: 1. Study the rationalization method. 2. Solve inequalities by this method. The rationalization method consists in replacing the complex expression F(x) with a simpler expression G(x), in which the inequality G(x) v 0 is equivalent to the inequality F(x) v 0 on the domain of the expression F(x) (symbol "v" replaces one of the inequality signs: ≤, ≥, >, We single out some typical expressions F and rationalizing expressions G corresponding to them (table 1), where f, g, h, p, q are expressions with the variable x (h>0, h≠1, f>0, g>0), a -fixed number (а>0, a≠1). (Appendix 2). Example #1. Solve the inequality: O.D.Z: Answer: Example #2. Solve the inequality: O.D.Z: Taking into account the domain of definition, we get Answer: Conclusion : inequalities with logarithms in a variable base cause the most difficulty. The rationalization method allows you to move from an inequality containing complex exponential, logarithmic, etc. expressions, to an equivalent simpler rational inequality. The rationalization method not only saves time, but also reduces the risk of logical and computational errors. Study #3 Target : in the process of solving inequalities, compare different methods. Progress: 1. Solve the inequalitydifferent methods. 2. Compare the results and draw a conclusion. Example #1. Solve the inequality Solution: 1 way. Algebraic method Solution of the first system: We solve the second inequality of the second system: 2 way . Using Function Scope Domain: For these x values, we get: The right side of the inequality is negative on its domain. Therefore, the inequality is valid for Answer: 3 way. Graphic method Conclusion : solving the inequality by the algebraic method, I came to the inequality of the sixth degree, spent a lot of time solving it, but could not solve it. A rational method, in my opinion, is the use of function scope or graphical. Example #2. Solve the inequality:.
Answer: Conclusion: I managed to solve this inequality only thanks to the rationalization method. Conclusion Content analysis school textbooks shows that in most of them, methods for solving inequalities using the properties of functions are not given due attention, and almost every year in the USE tasks, inequalities are proposed, the solution of which is simplified if the properties of functions are applied. Most students solve inequalities using standard, algorithmic methods, sometimes leading to cumbersome calculations. In this regard, the percentage of completion of assignment No. 15 on the USE is low. The scope of the properties of functions in solving inequalities is very wide. Using the properties (boundedness, monotonicity, etc.) of the functions included in the inequalities makes it possible to apply non-standard solution methods. In our opinion, the ability to use the necessary properties of functions when solving inequalities can allow students to choose a more rational way of solving. In our work, we studied functional-graphical methods for solving inequalities. We compared various methods for solving inequalities. We checked experimentally which way of solving inequalities is the most rational. And they came to the conclusion that the student must master several ways to solve inequalities in order to save time and reduce the risk of logical and computational errors. The tasks of our work have been fulfilled, the goal has been achieved, the hypothesis has been confirmed. Literature: To use the preview of presentations, create an account for yourself ( account) Google and sign in: https://accounts.google.com The study of various methods for solving inequalities Ivanova Anastasia Evgenievna MBOU "Secondary School No. 30 with in-depth study individual items" The results of the task number 15 students of our school The results of assignment No. 15 at the trial exam in 2017-2018 academic year. year by students of our school Hypothesis: if the student has several ways to solve inequalities, then he will be able to choose the most rational Object of study: inequalities Subject of study: various ways to solve inequalities Purpose: to study different ways of solving inequalities. To achieve this goal, the following tasks were solved: To study the theoretical material on this topic. Consider the examples offered in the bank of USE tasks on the website Federal Institute pedagogical dimensions. To study functional-graphical methods for solving inequalities. Compare different methods for solving inequalities. Check experimentally which way of solving inequalities is the most rational. Research No. 1 Purpose: to study the method of limited functions. Progress of work: 1. Study the method of limited functions. 2. Solve inequalities by this method. Example #1. Solve the inequality: Solution: Domain of definition: For all x from the resulting set we have: Therefore, the solution of the inequality Answer: Example #2. Solve the inequality: Solution: This inequality is equivalent to The first equation of the system has one root x = - 0.4, which also satisfies the second equation. Answer: - 0.4 Conclusion: this method is most effective if the inequality contains functions, like others, whose ranges are limited from above or below. Research No. 2 Purpose: to study the rationalization method for solving inequalities. Progress of work: 1. Study the method of rationalization. 2. Solve inequalities by this method. Example No. 1. Solve the inequality: O.D.Z: Given the domain of definition, we get the Answer: Example No. 2. Solve the inequality: O.D.Z: Considering the domain of definition, we get the Answer: Conclusion: inequalities with logarithms in a variable base cause the greatest complexity. The rationalization method allows you to move from an inequality containing complex exponential, logarithmic, etc. expressions, to an equivalent simpler rational inequality. The rationalization method not only saves time, but also reduces the risk of logical and computational errors. Research No. 3 Purpose: in the process of solving inequalities, compare different methods. Progress of work: 1. Solve the inequality by different methods. 2. Compare the results and draw a conclusion. Example No. 1. Solve inequality 1 way. Algebraic method Solution of the first system: We solve the second inequality of the second system: 2 way. Using the scope of a function Domain of definition: For these values of x, we get: The right side of the inequality is negative on its domain of definition. Therefore, the inequality is valid for 3 way. Graphical method Conclusion: solving the inequality using the algebraic method, I came to the inequality of the sixth degree, spent a lot of time solving it, but could not solve it. A rational method, in my opinion, is the use of function scope or graphical. Example No. 2. Solve the inequality: Answer: Conclusion: I managed to solve this inequality only thanks to the rationalization method. The scope of the properties of functions in solving inequalities is very wide. Using the properties (boundedness, monotonicity, etc.) of the functions included in the inequalities makes it possible to apply non-standard solution methods. In our opinion, the ability to use the necessary properties of functions when solving inequalities can allow students to choose a more rational way of solving. In our work, we studied functional-graphical methods for solving inequalities. We compared various methods for solving inequalities. We checked experimentally which way of solving inequalities is the most rational. And they came to the conclusion that the student must master several ways to solve inequalities in order to save time and reduce the risk of logical and computational errors. The tasks of our work have been fulfilled, the goal has been achieved, the hypothesis has been confirmed. Thank you for your attention!
Municipal educational institution Yuryevskaya basic comprehensive school Ostrovsky district Municipal stage of the regional methodological competition Nomination Toolkit Topic Functional-graphical method for solving equations and inequalities in the school course of high school algebra. Prepared by: mathematic teacher
Introduction Analysis of school textbooks USE analysis 1. General theoretical part 1.1. Graphic method 1.2. functional method 2. Solving Equations and Inequalities Using the Properties of the Incoming they have functions 2.1. Use of the DHS 2.2. Use of Limited Functions 2.3. Using function monotonicity 2.4. Using Function Graphs 2.5. Using the Even or Odd Properties and Periodicity of Functions .
3. Solving equations and inequalities 3.1. Solving Equations 3.2. Solving inequalities Workshop Bibliography The topic of my work is “A functional-graphical method for solving equations and inequalities in a school course in high school algebra”. One of the main topics of the high school algebra course. Solving equations and inequalities play an important role in the high school mathematics course. Schoolchildren begin to get acquainted with inequalities and equations in elementary school. The content of the topics "Equations" and "Inequalities" is gradually deepening and expanding. So, for example, the percentage of inequalities from all the studied material in grade 7 is 20%, in grade 8 - 25%, in grade 9 - 30%, in grades 10-11 - 35%. The final study of inequalities and equations takes place in the course of algebra and the beginning of analysis in grades 10-11. Some universities include equations and inequalities in exam papers, which are often quite complex and require different approaches to a decision. One of the most difficult sections in school school course Mathematics is considered only in a few elective classes. The focus of this work is to provide a more complete disclosure of the application of the functional-graphical method to the solution of equations and inequalities in high school algebra course. The relevance of this work is that this topic is included in the exam. Preparing this work, I set a goal to consider as many types of equations and inequalities as possible, solved by the functional-graphical method. Also, to study this topic more deeply, identifying the most rational solution that quickly leads to an answer. The object of research is algebra grades 10-11 under the editorship and variants of the exam. In this paper, frequently encountered types of equations and inequalities are considered, I hope that the knowledge I have gained in the process of work will help in passing school exams and entering a university. It can also serve methodological guide to prepare students for the exam. Analysis of school textbooks V methodological literature all methods on which the school line of equations and inequalities is based from grades 7 to 11 are divided into three groups: ü factorization method; the method of introducing new variables; ü functional-graphic method. Consider the third method, namely, the use of graphs of functions and various properties of functions. Schoolchildren must be taught to use the functional-graphic method from the very beginning of studying the topic "Equations". The solution of some problems can be based on the properties of monotonicity, periodicity, evenness or oddness, etc. of the functions included in them. After analyzing the textbooks, we can conclude that this topic is considered only in the textbooks of mathematics of the new generation,,, The construction of the course in these textbooks is carried out on the basis of the priority of the functional-graphic line. In other textbooks, the functional-graphic method for solving equations and inequalities is not singled out as a separate topic. The use of function properties to solve problems is mentioned in passing in other topics. The new textbooks also contain a sufficient number of tasks of this type. The textbook contains tasks advanced level. The most complete system of tasks, systematized for each property of the function, is given. Textbook "Algebra and the Beginnings of Analysis 10-11", a textbook for educational institutions, , "Algebra and the Beginnings of Analysis 11", textbook for educational institutions (profile level)
and others. "Algebra and the beginning of analysis 11", a textbook for educational institutions and others. "Algebra and the beginning of analysis 10-11", a textbook for educational institutions Place in the know Chapter 8 “Equations and inequalities. Systems of Equations and Inequalities" ( last topic course) Chapter 6 “Equations and inequalities. Systems of Equations and Inequalities” (last topic of the course) Chapter II "Equations, inequalities, systems" There is no separate topic. But in the topic "Decision trigonometric equations and inequalities” a root theorem is formulated, which is used in the further study No dedicated topic § §56 General methods for solving equations and inequalities (, functional-graphic method: root theorem, function boundedness) § §27 General methods for solving equations and inequalities (, functional-graphic method: root theorem, function boundedness) § Equations (inequalities) of the form ; § §12*Non-standard methods for solving equations and inequalities (use of domains of existence of functions, non-negativity of functions, boundedness, use of sin and cos properties, use of derivative) The property of the monotonicity of a function, even-odd (when deriving formulas for the roots of trigonometric equations) The property of monotonicity is mentioned when parsing an example in the topic "Exponential Function" Examples of considered equations and inequalities ( Solve the equation. How many roots belonging to this interval does the equation have? solve the equation Analysis of the USE (texts and results) The unified state exam as a form of attestation that has been put into practice Russian education in 2002, since 2009 it has switched from experimental to regular mode. An analysis of the USE texts showed that tasks in the solution of which the properties of functions are used occur every year. In 2003, in tasks A9 and C2, when solving, you can apply the properties of functions: A9. Specify the interval to which the roots of the equation belong C2. Find all values p, for which the equation · In 2004 – task В2. How many roots does the equation have In 2005, task C2 (solve the equation In 2007, when completing the task "Solve the equation" in part B, graduates considered two cases when solving the equation, habitually revealing the sign of the modulus..gif" width="81" height="24"> takes only positive values. Even well-prepared students often perform tasks using "template" solution methods that lead to cumbersome transformations and calculations. Obviously, when performing the above tasks, a well-trained graduate had to show not only knowledge of known methods for solving equations or transforming expressions, but also the ability to analyze the condition, correlate the data and the requirements of the task, derive various consequences from the condition, etc., that is show a certain level of development of mathematical thinking. Thus, when teaching well-performing students, it is necessary not only to take care of mastering the basic component of the algebra course and the beginnings of analysis (mastering the learned rules, formulas, methods), but also about the implementation of one of the main goals of teaching mathematics - the development of students' thinking, in particular, mathematical thinking. To achieve this goal, elective courses can serve. Indeed, students of educational institutions traditionally get acquainted with the graphical method of solving equations, inequalities and their systems when studying mathematics. However, in last years in the content of teaching mathematics, new classes of equations (inequalities) and new functional methods for solving them appear. However, contained in the control and measuring materials of the unified state exam(USE) tasks (the so-called combined equations), the solutions of which require the use of only the functional-graphic method, cause difficulties for students. Let X and Y be two arbitrary numerical sets. The elements of these sets will be denoted x and y, respectively, and will be called variables. Definition. A numerical function defined on the set X and taking values in the set Y is a correspondence (rule, law) that matches each x from the set X with one and only one value y from the set Y. The variable x is called the independent variable or argument, and the variable y is the dependent variable. We also say that the variable y is function from the variable x. The values of the dependent variable are called the values of the function. Introduced concept numeric function is a special case general concept functions as correspondences between elements of two or more arbitrary sets. Let X and Y be two arbitrary sets. Definition. A function defined on the set X and taking values in the set Y is a correspondence that relates to each element of the set X one and only one element from the set Y. Definition. Specifying a function means specifying the scope of its definition and the correspondence (rule) by which the values of the function corresponding to it are found from a given value of an independent variable. Two methods of solving equations are associated with the concept of a function: graphic and functional. A special case of the functional method is the method functional, or universal substitutions. Definition. To solve this equation means to find the set of all its roots (solutions). The set of roots (solutions) can be empty, finite or infinite. In the following chapters of the theoretical section, we will analyze the above methods for solving equations, and in the Practicum section we will show their application in various situations. 1.1. Graphic method. In practice, to plot some functions, a table of function values is compiled for some argument values, then plotted corresponding points on the coordinate plane and connect them in series with a line. It is assumed that the points accurately show the progress of the change in the function. Definition. The graph of a function y = f(x) is the set of all points (x, f(x) | x https://pandia.ru/text/78/500/images/image024_0.jpg" width="616" height="403"> The intersection point of the graphs has coordinates (0.5; 0). Hence, x=0.5 Answer: x=0.5 Example 2 10|
sinx|=10|cosx|-1 This equation is rationally solved by the graphical-analytical method. Since 10>1, then this equation is equivalent to the following: The intersection points of the graphs have coordinates ();. Therefore, x=. Answer: x= 1.2. functional method Not every equation of the form f(x)=g(x) as a result of transformations can be reduced to an equation of one or another standard form, for which the usual solution methods are suitable. In such cases, it makes sense to use such properties of the functions f(x) and g(x) as monotonicity, boundedness, evenness, periodicity, etc. So, if one of the functions increases and the other decreases over a certain interval, then the equation f(x) = g(x) cannot have more than one root, which, in principle, can be found by selection. Further, if the function f(x) is bounded from above and the function g(x) is bounded from below so that f(x) max=A g(x) min=A, then the equation f(x)=g(x) is equivalent to the system of equations Also, when using the functional method, it is rational to use some theorems given below. To prove and use them, the following equations are needed general view: Theorem 1. The roots of equation (1) are the roots of equation (2). Theorem 2. If f(x) is an increasing function on the interval a The last theorem implies a corollary, which is also used in the solutions: Corollary 1. If f(x) increases over its entire domain of definition, then equations (1) and (2) are equivalent on this interval. If f(x) decreases over its entire domain of definition, n is odd, then equations (1) and (2) are equivalent on the given interval. Theorem 3. If the equation f(x)=g(x) for any admissible x satisfies the conditions f(x)≥a, g(x)≤a, where a is some real number, then the given equation is equivalent to the system Consequence 2. If in the equation f(x)+g(x)=a+b for any admissible x f(x)≤a, g(x)≤b, then this equation is equivalent to the system The functional method of solving equations is often used in combination with the graphical one, since both of these methods are based on the same properties of functions. Sometimes a combination of these methods is called graph-analytical method. Example 1 coshttps://pandia.ru/text/78/500/images/image033_3.gif" width="64" height="41 src=">≤1 x2+1≥1 => coshttps://pandia.ru/text/78/500/images/image035_3.gif" width="121" height="48"> Answer: x=π A special case of the functional method is the method of functional substitution - perhaps the most common method of solving challenging tasks mathematics. The essence of the method is the introduction of a new variable y=ƒ(x), the application of which leads to a simpler expression. A separate case of functional substitution is trigonometric substitution. R(sin kx, cos nx, tg mx.ctg lx) = 0 (3) where R is rational function, k,n,m,lОZ, using trigonometric formulas double and triple argument, as well as addition formulas, can be reduced to a rational equation with respect to the arguments sin x, cos x, tg x.ctg x, after which equation (3) can be reduced to a rational equation for t=tg( x/2) using the universal trigonometric substitution formulas 2tg(x/2) 1-tg²(x/2) Sin x= cos x= 1+tg²(x/2) 1+tg²(x/2) 2tg(x/2) 1-tg²(x/2) Tg x=ctg x= 1-tg²(x/2) 2tg(x/2) It should be noted that the application of formulas (4) can lead to a narrowing of the ODZ of the original equation, since tg(x/2) is not defined at the points x=π+2πk, kОZ, so in such cases it is necessary to check whether the angles x=π+ 2πk, kОZ as roots of the original equation. Example 1 sin x+√2-sin² x+ sin x√2-sin² x = 3
Let now r = u+v and s=uv, then it follows from the system of equations Since, u = sin x and u = 1, then sin x= 1 and x = π/2+2πk, kн Z Answer: x = π/2+2πk, kОZ Example 2 5
sin x-5
tg x
+4(1-
cos x)=0
sin x+
tg x
It is rational to solve this equation by the method of functional substitution. Since tg x is not defined for x = π/2+πk, kн Z, and sin x+tg x=0 for x = πk, kн Z, then the angles x = πk/2, kн Z are not included in the ODZ equation. We use the formulas for the tangent of a half angle and denote t=tg( x/2), while according to the condition of the problem t≠0;±1, then we get https://pandia.ru/text/78/500/images/image055_2.gif" width="165"> +4 1- =0 Since t≠0;±1, then this equation is equivalent to the equation 5t² + = 0 ó-5-5t² + 8 = 0 whence t = ± ..gif" width="27" height="47"> +2πk, kн Z Example 3 tg x+
ctg x+
tg²x+
ctg²x+
tg³x+
ctg³x=6
This equation is rationally solved by the method of functional substitution. Let y=tg x+ctg x, then tg² x+ctg² x=y²-2, tg³ x+ctg³ x=y³-3y Since tg x+ctg x=2, then tg x+1/tg x=2. Hence it follows that tg x=1 and x = π/4+πk, kн Z Answer: x = π/2+2πk, kн Z 2. Solving equations and inequalities using the properties of the functions included in them 2.
1. Use of the ODZ. Sometimes knowledge of the ODZ allows one to prove that an equation (or inequality) has no solutions, and sometimes it allows one to find solutions to an equation (or an inequality) by directly substituting numbers from the ODZ. Example 1
solve the equation Solution. The ODZ of this equation consists of all x that simultaneously satisfy the conditions 3-x0 and x-3>0, that is, the ODZ is an empty set. This completes the solution of the equation, since it has been established that no number can be a solution, that is, that the equation has no roots. Answer: There are no solutions. Example 2
solve the equation Solution. The ODZ of this equation consists of all x that simultaneously satisfy the conditions, that is, the ODZ is Substituting these values of x into equation (1), we find that its left and right sides are 0, which means that all https://pandia.ru/ text/78/500/images/image065_2.gif" width="93 height=21" height="21"> Example 3
Solve the inequality Solution. The ODZ of inequality (2) consists of all х simultaneously satisfying the conditions Answer: x=1. Example 4
Solve the inequality Solution. The ODZ of inequality (3) is all x satisfying the condition 0<х1. Ясно, что х=1 не является решением неравенства (3). Для х из промежутка 0 Answer: 0 Example 5
Solve the inequality Solution..gif" width="73" height="19"> and . For x from the interval https://pandia.ru/text/78/500/images/image082_1.gif" width="72" height="24 src=">.gif" width="141 height=24" height= "24"> on this interval, and therefore inequality (4) has no solutions on this interval. Let x belong to the interval , then https://pandia.ru/text/78/500/images/image087_1.gif" width="141 height=24" height="24"> for such x, and, therefore, on Inequality (4) also has no solutions on this interval. So, inequality (4) has no solutions. Answer: There are no solutions. Remarks. When solving equations, it is not necessary to find the ODZ. Sometimes it is easier to go to the investigation and check the found roots. When solving inequalities, sometimes it is possible not to find the ODZ, but to solve the inequality by passing to an equivalent system of inequalities, in which either one of the inequalities has no solutions, or knowledge of its solution helps to solve the system of inequalities. Example 6
Solve the inequality Solution. Finding the ODZ inequality is not an easy task, so let's do it differently. Inequality (5) is equivalent to the system of inequalities The third inequality of this system is equivalent to the inequality that has no solutions. Consequently, the system of inequalities (6) has no solutions, which means that inequality (5) has no solutions either. Answer: There are no solutions. Example 7
Solve the inequality Solution. Finding the ODZ of inequality (7) is a difficult task. So let's do it differently. Inequality (7) is equivalent to the system of inequalities The third inequality of this system has solutions for all x from the interval -1 2.2. Use of limited functions. When solving equations and inequalities, the property of being bounded from below or above by a function on a certain set often plays a decisive role. For example, if for all x from some set M the inequalities f(x)>A and g(x) Note that the role of the number A is often played by zero; in this case, one speaks of the preservation of the sign of the functions f(x) and g(x) on the set M. Example 1
solve the equation Solution..gif" width="191" height="24 src="> Since for any value of x the left side of the equation does not exceed one, and the right side is always at least two, this equation has no solutions. Answer: no solutions. Example 2
solve the equation Solution. Obviously, x=0, x=1, x=-1 are solutions to equation (9)..gif" width="36" height="19">, because if is its solution, then (-) is also his decision. We divide the set x>0, , into two intervals (0;1) and (1;+∞). Let's rewrite equation (9) as https://pandia.ru/text/78/500/images/image103_1.gif" width="104" height="28">.gif" width="99" height="25 src=">positive only. Consequently, equation (9) has no solutions on this interval. Let x belong to the interval (1;+∞). For each of these x values, the function If х>2 , then , and this means that equation (9) also has no solutions on the interval (2;+∞). So, x=0, x=1 and x=-1 and only they are solutions to the original equation. Answer: Example 3
Solve the inequality Solution. The ODZ of inequality (10) is all real x, except for x=-1. Let's divide the ODZ into three sets: -∞<х<-1, -1 Let -∞<х<-1..gif" width="93" height="24 src=">. Therefore, all these х are solutions of inequality (10). Let -1 Let 0 Answer: -∞<х<-1; 0 Example 4
solve the equation Solution. Denote Consider x from the interval . On this interval, equation (11) can be rewritten in the form , that is, in the form It is clear that x=0 is a solution to equation (12), and hence the original equation..gif" width="39" height="19"> equation (12) is equivalent to the equation For any value Answer: x=0, ; https://pandia.ru/text/78/500/images/image139_0.gif" width="211" height="41">. (13) Solution. Let there be a solution to equation (13), then the equality and inequalities https://pandia.ru/text/78/500/images/image142_1.gif" width="68" height="27 src=">. the same sign as and , that is, the same sign as , and the right side has the same sign as . But since and satisfy equality (14), they have the same signs. Let us rewrite equality (14) as https://pandia.ru/text/78/500/images/image147_0.gif" width="284" height="24"> Let us rewrite equality (15) as Since and have the same signs, then ..gif" width="95" height="24">. (17) Obviously, any solution to equation (17) is a solution to equation (13). Therefore, equation (13) is equivalent to equation (17). Equation (17) has solutions Answer: Comment. In the same way as in Example 5, we can prove that the equation where n, m are any integers, is equivalent to the equation , and then solve this simpler equation. 2. 3. Using the monotonicity of a function. The solution of equations and inequalities using the monotonicity property is based on the following statements. Let f(x) be a continuous and strictly monotonic function on the interval L, then the equation f(x)=C, where C is a given constant, can have at most one solution on the interval L. Let f(x) and g(x) are functions continuous on the interval L, f(x) is strictly increasing, and g(x) is strictly decreasing on this interval, then the equation f(x)=g(x) can have no more than one solution on the interval L. Note that the interval L can be an infinite interval (-∞; +∞), semi-infinite intervals (a; +∞), (-∞; a), [a; +∞), (-∞; a], segments, intervals and half-intervals. Example 1
solve the equation Solution. Obviously, x0 cannot be a solution to equation (18), since then Answer: x=1. Example 2
Solve the inequality Solution. Each of the functions , , is continuous and strictly increasing on the entire axis. So the original function is the same Answer: -∞ Example 3
solve the equation Solution. The range of admissible values of equation (20) is the interval . On the range of admissible values of the function Answer: x=2. Example 4
Solve the inequality The solution..gif" width="253 height=27" height="27"> is continuous and strictly increasing. Since f(1)=4, then all x values from the set increase on the interval. Since on the interval .. gif" width="95" height="25 src="> are shown in Figure 7. It follows from the figure that inequality (26) is valid for all x from the ODZ. Let's prove it. For each we have , and for each such x we have Example 2
solve the equation Solution..gif" width="123" height="24"> and Answer: There are no solutions. Example 3
solve the equation The solution..gif" width="95" height="25 src="> are presented in Figure 9. It is easy to check that the point (-1; -2) is the intersection point of the graphs of the functions f(x) and g(x), that is, x=-1 is the solution of equation (28).Let's draw a straight line y=x-1.It follows from the figure that it is located between the graphs of the functions y=f(x) and y=g(x).This observation helps to prove that Eq. (28) has no other solutions. To do this, we prove that x from the interval (-1; +∞) satisfies the inequalities and , and for x from the interval (-∞; -1) the inequalities are true https://pandia.ru/text/78/500/images/image229_1 .gif" width="89" height="21 src=">. Obviously, the inequality is valid for x>-1, and the inequality https://pandia.ru/text/78/500/images/image228_1.gif" width="93" height="24">..gif" width="145" height="25">. Solutions to this inequality are all x<-1. Точно так же показывается, что решениями неравенства являются все х>-1. Therefore, the required statement is proved, and equation (28) has a unique root x=-1. Answer: x=-1. Example 4
Solve the inequality Solution..gif" width="39" height="19 src=">, that is, the ODZ consists of three gaps, , https://pandia.ru/text/78/500/images/image234_1.gif" width= "52" height="41">, equivalent to inequality and in the region x>0 it is equivalent to the inequality Sketches of function graphs Therefore, inequality (31) has no solutions, and inequality (30) will have solutions for all x from the interval . Let's prove it. A) Let. Inequality (29) is equivalent on this interval to inequality (30). It is easy to see that for each x from this interval, the inequalities Consequently, inequality (30) and, together with it, the original inequality (29) have no solutions on the interval . B) Let . Then inequality (29) is also equivalent to inequality (30). For every x in this interval Therefore, any such x is a solution to inequality (30), and therefore also to the original inequality (29). C) Let x>0. On this set, the original inequality is equivalent to inequality (31). Obviously, for any x from this set, the inequalities This implies: 1) inequality (31) has no solutions on the set where 2)
inequality (31) has no solutions on the set where https://pandia.ru/text/78/500/images/image253_1.gif" width="60" height="19">. It remains to find solutions to inequality (31 ) belonging to the interval 1 Root separation
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. To do this, you need to finddiscriminant
given quadratic equation. You can get 3 cases: 1) D=0 , quadratic equation has one root; 2) D>0 a quadratic equation has two roots; 3)D a quadratic equation has no roots. Depending on the obtained roots and the sign of the coefficient a one of six locations possiblefunction graph
(Annex 1).
b) 
b) 
.
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Appendix
Introduction
;
.
has no roots.
.
) were completed by 37% of students.1. General theoretical part
(2)
=> x=π, for k=01.3. Function substitution method
Trigonometric equation of the form
(1)
that is, ODZ consists of two numbers and . Substituting into inequality (2), we get that its left side is equal to 0, the right one is equal to https://pandia.ru/text/78/500/images/image070_1.gif" width="53" height="23">. gif" width="117 height=41" height="41">.
(3)
(6)
. (7)
(8)
(9)
takes positive values, the function https://pandia.ru/text/78/500/images/image105_1.gif" width="99" height="25 src="> is non-positive. Therefore, on this interval, equation (9) is not has solutions.
(11)
through f(x). It follows from the definition of the absolute value that f(x)= at , https://pandia.ru/text/78/500/images/image120_1.gif" width="84" height="41 src=">.gif" width="43" height="41 src=">. Therefore, if , then equation (11) can be rewritten in the form , that is, in the form ..gif" width="53" height="41"> satisfy only 
. If , then equation (11) can be rewritten as https://pandia.ru/text/78/500/images/image128_0.gif" width="73 height=41" height="41">. This equation has solutions 
. Of these x values, only 
.
, the function takes only positive values, so equation (12) has no solutions on the set 
.
(14)
, they and only they are solutions of Eq. (13).
(18)
. For x>0 the function 
is continuous and strictly increasing as the product of two continuous positive strictly increasing functions f=x for these x and https://pandia.ru/text/78/500/images/image157_0.gif > takes on each of its values at exactly one point It is easy to see that x=1 is a solution to Eq. (18), hence it is its only solution.
. (19)
. It is easy to see that for x=0 the function takes the value 3. Due to the continuity and strict monotonicity of this function for x>0, we have 
, at x<0 имеем . Consequently, the solutions of inequality (19) are all x<0.
(20)
and 
are continuous and strictly decreasing; therefore, the function is continuous and decreasing. Therefore, the function h(x) takes each value only at one point. Since h(2)=2, then x=2 is the only root of the original equation.
. Therefore, the solutions of inequality (26) will be all x from the interval [-1;1].
. (27)
are shown in Figure 8. Let's draw a straight line y=2. It follows from the figure that the graph of the function f(x) does not lie below this line, and the graph of the function g(x) does not lie above. Moreover, these graphs touch the straight line y=2 at different points. Therefore, the equation has no solutions. Let's prove it. For each we have , a . In this case, f(x)=2 only for x=-1, and g(x)=2 only for x=0. This means that equation (27) has no solutions.
. (28)
.
(29)
, (30)
. (31)
and are shown in Figure 10..gif" width="56" height="45"> and .
,
.
,
,
, that is, inequality (31) has no solutions on the set ;
