Ezhova Elena Sergeevna
Job title: math teacher
Educational institution: Municipal educational institution "Secondary school No. 77"
Locality: Saratov
Name of material: methodological development
Subject: Rationalization method for solving inequalities in preparation for the Unified State Exam"
Publication date: 16.05.2018
Chapter: complete education
Obviously, the same inequality can be solved in several ways. Successfully
in the chosen way or, as we used to say, in a rational way, any
inequality will be resolved quickly and easily, its solution will be beautiful and interesting.
I would like to consider in more detail the so-called rationalization method when
solving logarithmic and exponential inequalities, as well as inequalities containing
variable under the modulus sign.
The main idea of the method.
The method of replacing factors solves inequalities that can be reduced to the form
Where is the symbol "
" denotes one of four possible inequality signs:
When solving inequality (1), we are only interested in the sign of any factor in the numerator
or denominator, and not its absolute value. Therefore, if for some reason we
it is inconvenient to work with this multiplier, we can replace it with another
coinciding in sign with it in the domain of definition of inequality and having in this domain
the same roots.
This determines the main idea of the multiplier replacement method. It is important to record that
the fact that the replacement of factors is carried out only under the condition of bringing the inequality
to form (1), that is, when it is necessary to compare the product with zero.
The main part of the replacement is due to the following two equivalent statements.
Statement 1. The function f(x) is strictly increasing if and only if for
any values of t
) coincides with
sign with the difference (f(t
)), that is, f<=>(t
(↔ means sign coincidence)
Statement 2. The function f(x) is strictly decreasing if and only if for
any values of t
from the domain of definition of the function difference (t
) coincides with
sign with the difference (f(t
)), that is, f ↓<=>(t
The justification for these statements follows directly from the definition of strictly
monotonic function. According to these statements, it can be established that
The difference in degrees for the same base always coincides in sign with
the product of the difference between the indices of these powers and the deviation of the base from unity,
The difference of logarithms to the same base always coincides in sign with
the product of the difference between the numbers of these logarithms and the deviation of the base from unity, then
The fact that the difference of non-negative quantities coincides in sign with the difference
squares of these quantities allows the following substitutions:
Solve the inequality
Solution.
Let's move on to an equivalent system:
From the first inequality we get
The second inequality holds for all
From the third inequality we obtain
Thus, the set of solutions to the original inequality is:
Solve the inequality
Solution.
Let's solve the inequality:
ANSWER: (−4; −3)
Solve inequality
Let us reduce the inequality to a form in which the difference in the values of the logarithmic
Let's replace the difference between the values of the logarithmic function and the difference between the values of the argument. IN
the numerator is an increasing function, and the denominator is decreasing, so the inequality sign
will change to the opposite. It is important not to forget to take into account the domain of definition
logarithmic function, therefore this inequality is equivalent to a system of inequalities.
Numerator roots: 8; 8;
Root denominator: 1
Solve inequality
Let us replace in the numerator the difference between the moduli of two functions by the difference of their squares, and in
the denominator is the difference between the values of the logarithmic function and the difference in the arguments.
The denominator has a decreasing function, which means the inequality sign will change to
opposite.
In this case, it is necessary to take into account the domain of definition of the logarithmic
Let's solve the first inequality using the interval method.
Numerator roots:
Denominator roots:
Solve inequality
Let us replace the difference in the values of monotonic functions in the numerator and denominator with the difference
values of the arguments, taking into account the domain of definition of the functions and the nature of monotonicity.
Numerator roots:
Denominator roots:
The most frequently used replacements (excluding O D Z).
a) Replacement of constant sign factors.
b) Replacement of non-constant multipliers with modulus.
c) Replacing factors of unknown sign with exponential and logarithmic ones
expressions.
Solution. ODZ:
Replacing multipliers:
We have a system:
In this inequality it is no longer possible to factor
be considered as differences of non-negative quantities, since expressions 1
ODZ can take on both positive and negative values.
We have a system:
Replacing multipliers:
We have a system:
Replacing multipliers:
We have a system:
Replacing multipliers:
We have a system:
As a result we have: x
Rationalization method(decomposition method, multiplier replacement method, replacement method
functions, sign rule) consists in replacing the complex expression F(x) with a more
simple expression G(x), under which the inequality G(x)
0 is equivalent to the inequality F (x
0 in the domain of definition of the expression F(x).
The rationalization method allows you to move from inequalities containing complex exponential, logarithmic, etc. expression, to its equivalent simpler rational inequality.
Therefore, before we start talking about rationalization in inequalities, let's talk about equivalence.
Equivalence
Equivalent or equivalent are called equations (inequalities) whose sets of roots coincide. Equations (inequalities) that do not have roots are also considered equivalent.
Example 1. The equations and are equivalent because they have the same roots.
Example 2. The equations and are also equivalent, since the solution to each of them is the empty set.
Example 3. The inequalities and are equivalent, since the solution to both is the set .
Example 4. and – are unequal. The solution to the second equation is only 4, and the solution to the first is both 4 and 2.
Example 5. Inequality is equivalent to inequality, since in both inequalities the solution is 6.
That is, in appearance, equivalent inequalities (equations) can be very far from similar.
In fact, when we solve complex, long equations (inequalities), like this one, and get the answer, what we have in our hands is nothing more than an equation (inequality) equivalent to the original one. The look is different, but the essence is the same!
Example 6. Let's remember how we solved inequality before getting acquainted with the interval method. We replaced the original inequality with a set of two systems:
That is, inequality and the latter aggregate are equivalent to each other.
Also, we could, having in our hands the totality
replace it with inequality, which can be solved in no time by the interval method.
We have come close to the rationalization method in logarithmic inequalities Oh.
Rationalization method in logarithmic inequalities
Let's consider inequality.
We represent 4 as a logarithm:
We are dealing with a variable base of the logarithm, therefore, depending on whether the base of the logarithm is greater than 1 or less than 1 (that is, we are dealing with an increasing or decreasing function), the inequality sign will remain the same or change to “”. Therefore, a combination (union) of two systems arises:
But, ATTENTION, this system must be decided taking into account DL! I deliberately did not load the ODZ system so that the main idea would not get lost.
Look, now we will rewrite our system like this (we will move everything in each line of the inequality to the left):
Does this remind you of anything? By analogy with example 6 We will replace this set of systems with the following inequality:
Having solved this inequality on the ODZ, we obtain a solution to the inequality.
Let us first find the ODZ of the original inequality:
Now let's decide
Solution of the last inequality taking into account the ODZ:
So, here it is, this “magic” table:
Note that the table works under the condition
where are functions of ,
– function or number,
- one of the signs
Note also that the second and third lines of the table are consequences of the first. In the second line, 1 is represented first as , and in the third line, 0 is represented as .
And a couple more useful consequences (I hope it’s easy for you to understand where they come from):
where are functions of ,
– function or number,
- one of the signs
Method of rationalization in exponential inequalities
Let's solve the inequality.
Solving the original inequality is equivalent to solving the inequality
Answer: .
Table for rationalization in exponential inequalities:
– functions of , – function or number, – one of the signs The table works under the condition . Also in the third, fourth lines – additionally –
Again, essentially, you need to remember the first and third lines of the table. Second line - special case the first, and the fourth line is a special case of the third.
Rationalization method in inequalities containing a modulus
When working with inequalities of type , where are functions of some variable, we can be guided by the following equivalent transitions:
Let's solve the inequality."
A Here I also suggest consider several examples on the topic “Rationalization of inequalities.”
Sections: Mathematics
Often, when solving logarithmic inequalities, there are problems with a variable logarithm base. Thus, an inequality of the form
is a standard school inequality. As a rule, to solve it, a transition to an equivalent set of systems is used:
The disadvantage of this method is the need to solve seven inequalities, not counting two systems and one population. Already with these quadratic functions, solving the population can take a lot of time.
It is possible to propose an alternative, less time-consuming way to solve this standard inequality. To do this, we take into account the following theorem.
Theorem 1. Let there be a continuous increasing function on a set X. Then on this set the sign of the increment of the function will coincide with the sign of the increment of the argument, i.e. , Where 
.
Note: if a continuous decreasing function on a set X, then .
Let's return to inequality. Let's move on to the decimal logarithm (you can move on to any with a constant base greater than one).
Now you can use the theorem, noticing the increment of functions in the numerator 
and in the denominator. So it's true
As a result, the number of calculations leading to the answer is reduced by approximately half, which saves not only time, but also allows you to potentially make fewer arithmetic and careless errors.
Example 1.
Comparing with (1) we find 
, 
, .
Moving on to (2) we will have:
Example 2.
Comparing with (1) we find , , .
Moving on to (2) we will have:
Example 3.
Since the left side of the inequality is an increasing function as and 
, then the answer will be many.
The many examples in which Theme 1 can be applied can easily be expanded by taking into account Theme 2.
Let on the set X the functions , , , are defined, and on this set the signs and coincide, i.e. , then it will be fair.
Example 4.
Example 5.
With the standard approach, the example is solved according to the following scheme: the product is less than zero when the factors are of different signs. Those. a set of two systems of inequalities is considered, in which, as indicated at the beginning, each inequality breaks down into seven more.
If we take into account theorem 2, then each of the factors, taking into account (2), can be replaced by another function that has the same sign in this example O.D.Z.
The method of replacing the increment of a function with an increment of argument, taking into account Theorem 2, turns out to be very convenient when solving typical tasks C3 Unified State Exam.
Example 6.
Example 7.
. Let's denote . We get
. Note that the replacement implies: . Returning to the equation, we get
.
Example 8.
In the theorems we use there are no restrictions on classes of functions. In this article, as an example, the theorems were applied to solving logarithmic inequalities. The following several examples will demonstrate the promise of the method for solving other types of inequalities.
Sections: Mathematics
The practice of checking examination papers shows that the greatest difficulty for schoolchildren is solving transcendental inequalities, especially logarithmic inequalities with a variable base. Therefore, the lesson summary offered to your attention is a presentation of the rationalization method (other names - decomposition method (Modenov V.P.), method of replacing factors (Golubev V.I.)), which allows you to reduce complex logarithmic, exponential, combined inequalities to a system of simpler ones rational inequalities. As a rule, the method of intervals as applied to rational inequalities is well understood and practiced by the time the topic “Solving logarithmic inequalities” is studied. Therefore, students perceive with great interest and enthusiasm those methods that allow them to simplify the solution, make it shorter and, ultimately, save time on the Unified State Exam for solving other tasks.
Lesson objectives:
- Educational: updating basic knowledge when solving logarithmic inequalities; introduction of a new way to solve inequalities; improving solution skills
- Developmental: development of mathematical outlook, mathematical speech, analytical thinking
- Educational: education of accuracy and self-control.
PROGRESS OF THE LESSON
1. Organizational moment. Greetings. Setting lesson goals.
2. Preparatory stage:
Solve inequalities:
3. Checking homework(No. 11.81*a)
When solving the inequality
You had to use the following scheme for solving logarithmic inequalities with a variable base:
Those. We need to consider 2 cases: the base is greater than 1 or the base is less than 1.
4. Explanation of new material
If you look at these formulas carefully, you will notice that the sign of the difference g(x) – h(x) coincides with the sign of the difference log f(x) g(x) – log f(x) h(x) in the case of an increasing function ( f(x) > 1, i.e. f(x) – 1 > 0) and is opposite to the sign of the difference log f(x) g(x) – log f(x) h(x) in the case of a decreasing function (0< f(x) < 1, т.е. f(x) – 1 < 0)
Consequently, this set can be reduced to a system of rational inequalities:
This is the essence of the rationalization method - replacing the more complex expression A with a simpler expression B, which is rational. In this case, inequality B V 0 will be equivalent to inequality A V 0 on the domain of definition of expression A.
Example 1. Let us rewrite the inequality in the form of an equivalent system of rational inequalities.
Note that conditions (1)–(4) are conditions for the domain of definition of the inequality, which I recommend finding at the beginning of the solution.
Example 2. Solve inequality using the rationalization method:
The domain of definition of inequality is specified by the conditions:
We get:
It remains to write inequality (5)
Taking into account the domain of definition
Answer: (3; 5)
5. Consolidation of the studied material
I. Write the inequality as a system of rational inequalities:
II. Present the right side of the inequality as a logarithm to the desired base and go to the equivalent system:
The teacher calls to the board the students who wrote down the systems from groups I and II, and invites one of the strongest students to solve home inequality (No. 11.81 * a) using the rationalization method.
6. Test work
Option 1
Option 2
1. Write down a system of rational inequalities to solve the inequalities:
2. Solve inequality using the rationalization method
Grading criteria:
3-4 points – “satisfactory”;
5-6 points – “good”;
7 points – “excellent”.
7. Reflection
Answer the question: which of the methods you know for solving logarithmic inequalities with a variable base will allow you to use your time more efficiently during the exam?
8. Homework: No. 11.80* (a,b), 11.81*(a,b), 11.84*(a,b) solve by rationalization method.
List of used literature:
- Algebra and the beginnings of analysis: Textbook. For 11th grade. general education Institutions /[S.M. Nikolsky, M.K. Potapov, N.N. Reshetnikov, A.V. Shevkin] – 5th ed. – M.: Education, OJSC “Moscow Textbooks”, 2006.
- A.G. Koryanov, A.A. Prokofiev. Materials of the course “Preparing good and excellent students for the Unified State Exam”: lectures 1-4. – M.: Pedagogical University"First of September", 2012.
Municipal Autonomous General Educational Institution"Yarkovskaya secondary school"
Educational project
Solving logarithmic inequalities using the rationalization method
MAOU "Yarkovskaya Secondary School"
Shanskikh Daria
Head: mathematics teacher
MAOU "Yarkovskaya Secondary School"
Yarkovo 2013
1) Introduction……………………………………………………….2
2) Main part……………………………………………………………………..3
3) Conclusion……………………………………………………..9
4) List of references…………….10
5) Applications……………………………………………………………11-12
1. Introduction
Often, when solving USE tasks from part “C”, and especially in tasks C3, you encounter inequalities containing logarithmic expressions with an unknown in the base of the logarithm. For example, here is a standard inequality:
As a rule, to solve such problems the classical method is used, that is, a transition to an equivalent set of systems is used 
With the standard approach, the example is solved according to the following scheme: the product is less than zero when the factors are of different signs. That is, a set of two systems of inequalities is considered, in which each inequality is divided into seven more. Therefore, a less time-consuming method for solving this standard inequality can be proposed. This is a rationalization method known in mathematical literature as decomposition.
When completing the project, I set the following goals :
1) Master this decision technique
2) Practice solving skills on tasks C3 from training and diagnostic work 2013
Project objectiveis to study the theoretical basis of the rationalization method.
Relevancework is that this method allows you to successfully solve logarithmic inequalities of part C3 of the Unified State Examination in mathematics.
2. Main part
Consider a logarithmic inequality of the form
font-size:14.0pt; line-height:150%">, (1)
where font-size:14.0pt;line-height:150%"> The standard method for solving such an inequality involves analyzing two cases into the range of acceptable values of the inequality.
In the first case, when the bases of logarithms satisfy the condition
font-size:14.0pt; line-height:150%">, the inequality sign is drawn: font-size:14.0pt;line-height:150%">In the second case , when the base satisfies the condition, the inequality sign is preserved: .
At first glance, everything is logical, let’s consider two cases and then combine the answers. True, when considering the second case, a certain discomfort arises - you have to repeat 90 percent of the calculations from the first case (transform, find the roots of auxiliary equations, determine the intervals of monotonicity of the sign). A natural question arises: is it possible to somehow combine all this?
The answer to this question is contained in the following theorem.
Theorem 1. Logarithmic inequality
font-size:14.0pt;line-height:150%">equivalent to the following system of inequalities :
font-size:14.0pt; line-height:150%"> (2)
Proof.
1. Let's start with the fact that the first four inequalities of system (2) define the set of admissible values of the original logarithmic inequality. Let us now turn our attention to the fifth inequality. If font-size:14.0pt; line-height:150%">, then the first factor of this inequality will be negative. When reducing by it, you will have to change the inequality sign to the opposite one, then you get the inequality .
If , That the first factor of the fifth inequality is positive, we cancel it without changing the sign of the inequality, we get the inequality font-size:14.0pt;line-height: 150%"> Thus, the fifth inequality of the system includes both cases of the previous method.
The topic has been proven.
Basic provisions of the theory of the rationalization method.
The rationalization method is to replace a complex expression F(x ) to a simpler expression G(x ), at which the inequality G(x )EN-US" style="font-size:14.0pt;line-height:150%;font-family:Calibri">F(x )0 in the expression definition area F(x).
Let's highlight some expressions F and their corresponding rationalizing expressions G, where u, v, , p, q - expressions with two variables ( u > 0; u ≠ 1; v > 0, > 0), a - fixed number (a > 0, a ≠ 1).
Expression F | Expression G |
|
(a –1)( v – φ) |
||
|
||
1 b |
|
|
|
||
|
||
2 b |
|
|
|
||
|
)
Proof
1. Let logav - logaφ > 0, that is logav > logaφ, and a > 0, a ≠ 1, v > 0,
φ > 0.
If 0< a < 1, то по свойству убывающей логарифмической функции имеем v < φ . This means that the system of inequalities holds
a -1<0
v – φ < 0
Whence follows the inequality (a – 1)( v – φ ) > 0 true in the domain of expressionF = logav - logaφ.
If a > 1, That v > φ . Therefore, there is an inequality ( a – 1)( v – φ )> 0. Conversely, if the inequality holds ( a – 1)( v – φ )> 0 on the range of acceptable values ( a > 0, a ≠ 1, v> 0, φ > 0),then in this region it is equivalent to the combination of two systems.
a – 1<0 a – 1 > 0
v – φ < 0 v – φ > 0
Every system implies the inequalitylogav > logaφ, that is logav - logaφ > 0.
Similarly, we consider the inequalities F< 0, F ≤ 0, F ≥ 0.
2. Let some number A> 0 and A≠ 1, then we have
logu v- loguφ = EN-US" style="font-size:14.0pt;line-height:150%">v - 1)( u- 1)(φ –u).
4.From inequality uv- uφ > 0 should uv > uφ. Let the number a > 1, thenloga uv > logauφ or
( u – φ) loga u > 0.
Hence, taking into account replacement 1b and the conditiona > 1 we get
( v – φ)( a – 1)( u – 1) > 0, ( v – φ)( u – 1) > 0. Similarly, the inequalities are proved F< 0,
F ≤ 0, F ≥ 0.
5. The proof is similar to Proof 4.
6. The proof of substitution 6 follows from the equivalence of the inequalities | p | > | q | and p 2 > q 2
(|p|< | q | и p 2 < q 2 ).
Let us compare the volume of solutions to inequalities containing a variable in the base of the logarithm classical method and the rationalization method
3. Conclusion
I believe that the tasks that I set for myself while completing the work have been achieved. The project has practical significance, since the method proposed in the work can significantly simplify the solution of logarithmic inequalities. As a result, the number of calculations leading to the answer is reduced by approximately half, which saves not only time, but also allows you to potentially make fewer arithmetic and careless errors. Now when solving C3 problems I use this method.
4. List of used literature
1. , – Methods for solving inequalities with one variable. – 2011.
2. – Mathematics manual. – 1972.
3. - Mathematics for applicants. Moscow: MTsNMO, 2008.
