Supervisor:
1. Introduction 3
2. Historical sketch 4
3. "Place" of the ODZ when solving equations and inequalities 5-6
4. Features and danger of ODZ 7
5. ODZ - there is a solution 8-9
6. Finding the ODU is extra work. Equivalence of transitions 10-14
7. ODZ in the exam 15-16
8. Conclusion 17
9. Literature 18
1. Introduction
Problem:the equations and inequalities in which one needs to find the ODV did not find a place in the algebra course of a systematic presentation, perhaps that is why I and my peers often make mistakes when solving such examples, having spent a lot of time solving them, while forgetting about the ODV.
Purpose:be able to analyze the situation and draw logically correct conclusions in examples where it is necessary to take into account DHS.
Tasks:
1. Study theoretical material;
2. Solve a lot of equations, inequalities: a) fractional-rational; b) irrational; c) logarithmic; d) containing inverse trigonometric functions;
3. Apply the studied materials in a situation that differs from the standard one;
4. Create a work on the topic "Range of acceptable values: theory and practice"
Project work:i started working on the project by repeating the functions known to me. Many of them are limited in scope.
ODZ occurs:
1. When solving fractional rational equations and inequalities
2. When solving irrational equations and inequalities
3. When solving logarithmic equations and inequalities
4. When solving equations and inequalities containing inverse trigonometric functions
Having solved many examples from various sources (exam manuals, textbooks, reference books), I systematized the solution of examples according to the following principles:
You can solve the example and take into account the LDZ (the most common way)
You can solve the example without taking into account the DHS
· It is possible to come to the right decision only taking into account the DHS
Methods used in the work:1) analysis; 2) statistical analysis; 3) deduction; 4) classification; 5) forecasting.
I studied the analysis of the USE results over the past years. Many mistakes were made in the examples in which DHS should be taken into account. This emphasizes once again relevance my theme.
2. Historical sketch
Like the rest of the concepts of mathematics, the concept of a function did not develop immediately, but went a long way of development. In the work of P. Fermat "Introduction and study of flat and solid places" (1636, publ. 1679) says: "Whenever there are two unknown quantities in the final equation, there is a place." Essentially here we are talking about functional dependence and its graphic representation (Fermat's “place” means a line). The study of lines according to their equations in "Geometry" by R. Descartes (1637) also indicates a clear idea of \u200b\u200bthe mutual dependence of two variables. I. Barrow (Lectures on Geometry, 1670) establishes in a geometric form the reciprocity of the actions of differentiation and integration (of course, without using these terms themselves). This already testifies to a completely clear knowledge of the concept of function. In a geometric and mechanical form, we find this concept in I. Newton. However, the term "function" first appears only in 1692 by G. Leibniz and, moreover, not quite in its modern understanding. G. Leibniz calls a function the various segments associated with any curve (for example, the abscissa of its points). In the first printed course "Analysis of infinitesimal curves for knowledge" by L'Hôpital (1696), the term "function" is not used.
The first definition of a function in a sense close to the modern one is found in I. Bernoulli (1718): "A function is a quantity composed of a variable and a constant." This not quite clear definition is based on the idea of \u200b\u200bdefining a function by an analytical formula. The same idea appears in the definition of L. Euler, given by him in his "Introduction to the Analysis of the Infinite" (1748): "The function of a variable quantity is an analytical expression composed in some way of this variable quantity and numbers or constant quantities." However, L. Euler is no stranger to the modern understanding of a function, which does not connect the concept of a function with any analytical expression of it. In his "Differential Calculus" (1755) says: "When some quantities depend on others in such a way that when the latter change, they themselves are subject to change, the former are called functions of the latter."
Since the beginning of the 19th century, more and more often they define the concept of a function without mentioning its analytical representation. In "Treatise on differential and integral calculus" (1797-1802) S. Lacroix says: "Any quantity, the value of which depends on one or many other quantities, is called a function of these latter." In the "Analytical theory of heat" J. Fourier (1822) there is a phrase: "The function f (x) denotes a completely arbitrary function, that is, a sequence of data values, subject or not to the general law and corresponding to all values xcontained between 0 and any quantity x". NI Lobachevsky's definition is also close to the modern one: “... The general concept of a function requires that a function from x name the number that is given for each x and together with x gradually changes. The value of a function can be given either by an analytical expression, or by a condition that provides a means to test all numbers and choose one of them, or, finally, a dependence can exist and remain unknown. " It also says a little below: "The broad view of the theory admits the existence of dependence only in the sense that the numbers one with the other in connection are understood as if by the data together." Thus, the modern definition of a function, free from references to the analytical task, usually attributed to P. Dirichlet (1837), was repeatedly proposed before him.
The domain of definition (permissible values) of a function y is the set of values \u200b\u200bof the independent variable x, for which this function is defined, i.e., the range of variation of the independent variable (argument).
3. "Place" of the region of admissible values \u200b\u200bwhen solving equations and inequalities
1. When solving fractional rational equations and inequalities the denominator must not be zero.
2. Solution of irrational equations and inequalities.
2.1..gif "width \u003d" 212 "height \u003d" 51 "\u003e.
In this case, there is no need to find the ODV: from the first equation it follows that for the obtained values \u200b\u200bof x, the inequality is fulfilled: https://pandia.ru/text/78/083/images/image004_33.gif "width \u003d" 107 "height \u003d" 27 src \u003d "\u003e is system:
Since and are included in the equation with equal rights, instead of inequality, you can include inequality https://pandia.ru/text/78/083/images/image009_18.gif "width \u003d" 220 "height \u003d" 49 "\u003e 
https://pandia.ru/text/78/083/images/image014_11.gif "width \u003d" 239 "height \u003d" 51 "\u003e 
3. Solution of logarithmic equations and inequalities.
3.1. Scheme for solving a logarithmic equation
But it is enough to check only one condition of the DHS.
3.2..gif "width \u003d" 115 "height \u003d" 48 src \u003d "\u003e. Gif" width \u003d "115" height \u003d "48 src \u003d"\u003e
4. Trigonometric equations of the form are equivalent to the system (instead of inequality, the inequality can be included into the system https://pandia.ru/text/78/083/images/image024_5.gif "width \u003d" 377 "height \u003d" 23 "\u003e are equivalent to the equation
4. Features and danger of the range of permissible values
In math lessons, we are required to find the ODV in each example. At the same time, according to the mathematical essence of the matter, finding the ODD is not at all obligatory, often unnecessary, and sometimes even impossible - and all this without any prejudice to solving the example. On the other hand, it often happens that, having solved an example, schoolchildren forget to take into account the ODZ, write it down as the final answer, take into account only some conditions. This circumstance is well known, but the "war" continues every year and, it seems, will go on for a long time.
Consider, for example, this inequality:
Here the ODV is sought and the inequality is resolved. However, when solving this inequality, schoolchildren sometimes believe that it is quite possible to do without the search for DHS, more precisely, it is possible to do without the condition
Indeed, to get the correct answer, it is necessary to take into account both inequality and.
And here, for example, the solution to the equation: https://pandia.ru/text/78/083/images/image032_4.gif "width \u003d" 79 height \u003d 75 "height \u003d" 75 "\u003e
which is tantamount to working with ODZ. However, even in this example, such work is superfluous - it is enough to check the fulfillment of only two of these inequalities, and any two.
Let me remind you that any equation (inequality) can be reduced to the form. ODZ is just a function definition area on the left side. The fact that this area should be monitored follows from the definition of the root as a number from the area of \u200b\u200bdefinition of the given function, thereby from the ODZ. Here is a funny example on this topic..gif "width \u003d" 20 "height \u003d" 21 src \u003d "\u003e has a scope of many positive numbers (this, of course, is an agreement - to consider the function for,, but reasonable), and then -1 is not is the root.
5. Range of valid values \u200b\u200b- there is a solution
And finally, in the mass of examples, finding the ODZ allows you to get the answer without cumbersome calculations, or even verbally.
1. OD3 is an empty set, which means that the original example has no solutions.
1)
2) 3) 
2.InODZ one or more numbers are found, and a simple substitution quickly determines the roots.
1)
, x \u003d 3
2)
Here, the ODZ contains only the number 1, and after the substitution, it is clear that it is not a root.
3) There are two numbers in the ODZ: 2 and 3, and both fit.
4)\u003e In the ODZ there are two numbers 0 and 1, and only 1 is suitable.
The DLT can be effectively used in combination with the analysis of the expression itself.
5) 
< ОДЗ: Но в правой части неравенства могут быть только положительные числа, поэтому оставляем х=2. Тогда в неравенство подставим 2.
6) 
From the ODZ it follows that whence we have ..gif "width \u003d" 143 "height \u003d" 24 "\u003e From the ODZ we have:. But then and. Since, then there are no solutions.
From the ODZ we have: https: //pandia.ru/text/78/083/images/image060_0.gif "width \u003d" 48 "height \u003d" 24 "\u003e\u003e, which means,. Solving the last inequality, we get x<- 4, что не входит в ОДЗ. Поэтому решения нет.
3) ODZ:. Since then
On the other hand, https: //pandia.ru/text/78/083/images/image068_0.gif "width \u003d" 160 "height \u003d" 24 "\u003e
ODZ :. Consider the equation on the interval [-1; 0).
It fulfills the following inequalities https://pandia.ru/text/78/083/images/image071_0.gif "width \u003d" 68 "height \u003d" 24 src \u003d "\u003e. Gif" width \u003d "123" height \u003d "24 src \u003d "\u003e and no solutions. With the function and https://pandia.ru/text/78/083/images/image076_0.gif "width \u003d" 179 "height \u003d" 25 "\u003e. ODZ: x\u003e 2..gif" width \u003d "233" height \u003d "45 src \u003d"\u003e Find the ODZ:
An integer solution is possible only when x \u003d 3 and x \u003d 5. By checking, we find that the root x \u003d 3 does not fit, which means the answer is: x \u003d 5.
6. Finding the range of acceptable values \u200b\u200bis extra work. Equivalence of transitions.
Examples can be given where the situation is clear even without finding the DHS.
1. 
Equality is impossible, because when subtracting from a smaller expression, a larger one should get a negative number.
2. 
.
The sum of two non-negative functions cannot be negative.
I will also give examples where finding the LDZ is difficult, and sometimes simply impossible.
And, finally, the search for the DHS is very often just extra work, which can be perfectly dispensed with, thereby proving an understanding of what is happening. There are a huge number of examples here, so I will choose only the most typical ones. The main solution is in this case equivalent transformations in the transition from one equation (inequality, system) to another.
1.
... ODZ is not needed, because, having found those values \u200b\u200bof x for which x2 \u003d 1, we cannot get x \u003d 0.
2.. ODZ is not needed, because we find out when the equality of the radical expression to a positive number is fulfilled.
3.. ODU is not needed for the same reasons as in the previous example.
4. 
ODZ is not needed, because the radical expression is equal to the square of some function, and therefore cannot be negative.
5. 
6. ..gif "width \u003d" 271 "height \u003d" 51 "\u003e To solve only one restriction for the radical expression is enough. Indeed, from the written mixed system it follows that the other radical expression is also non-negative.
8. DHS is not needed for the same reasons as in the previous example.
9. 
ODZ is not needed, since it is enough that two of the three expressions under the signs of the logarithm are positive to ensure that the third is positive.
10. 
.gif "width \u003d" 357 "height \u003d" 51 "\u003e ODZ is not needed for the same reasons as in the previous example.
It is worth noting, however, that when solving by the method of equivalent transformations, knowledge of the ODV (and properties of functions) helps.
Here are some examples.
1. . OD3, whence the positiveness of the expression on the right side follows, and it is possible to write an equation equivalent to this one in this form https://pandia.ru/text/78/083/images/image101_0.gif "width \u003d" 112 "height \u003d" 27 "\u003e ODZ:. But then, and when solving this inequality, it is not necessary to consider the case when the right side is less than 0.
3.. From the ODZ it follows that, and therefore the case when https://pandia.ru/text/78/083/images/image106_0.gif "width \u003d" 303 "height \u003d" 48 "\u003e The transition in general looks like this:
https://pandia.ru/text/78/083/images/image108_0.gif "width \u003d" 303 "height \u003d" 24 "\u003e
Two cases are possible: 0
Hence, the original inequality is equivalent to the following set of systems of inequalities:
The first system has no solutions, and from the second we get: x<-1 – решение неравенства.
Understanding the conditions of equivalence requires knowledge of some subtleties. For example, why are such equations equivalent:
Or 
And finally, perhaps the most essential. The fact is that equivalence guarantees the correctness of the answer if some transformations of the equation itself are performed, but is not used for transformations in only one of the parts. Abbreviation, the use of different formulas in one of the parts do not fall under the theorems of equivalence. I have already given some examples of this kind. Let's look at some more examples.
1. This decision is natural. On the left, by the property of the logarithmic function, go to the expression ..gif "width \u003d" 111 "height \u003d" 48 "\u003e
Solving this system, we get the result (-2 and 2), which, however, is not an answer, since the number -2 is not included in the DHS. So what, do we need to establish a DHS? Of course not. But since we used a certain property of the logarithmic function in the solution, then we must ensure the conditions under which it is fulfilled. Such a condition is the positiveness of expressions under the sign of the logarithm..gif "width \u003d" 65 "height \u003d" 48 "\u003e.
2. 
..gif "width \u003d" 143 "height \u003d" 27 src \u003d "\u003e numbers must be substituted in this way 
... Who wants to do such boring calculations? .Gif "width \u003d" 12 "height \u003d" 23 src \u003d "\u003e add a condition, and you can immediately see that only the number meets this condition https://pandia.ru/text/78/083/ images / image128_0.gif "width \u003d" 117 "height \u003d" 27 src \u003d "\u003e) showed 52% of the applicants. One of the reasons for such low rates is the fact that many graduates did not select the roots obtained from the equation after it was squared.
3) Consider, for example, the solution to one of the tasks C1: "Find all the values \u200b\u200bof x for which the points of the graph of the function 
lie above the corresponding points of the function graph ". :
X< 10. Они отмечают, что в первом случае решений нет, а во втором – корнями являются числа –1 и . При этом выпускники не учитывают условие x < 10.
8. Conclusion
To summarize, we can say that there is no universal method for solving equations and inequalities. Every time, if you want to understand what you are doing, and not to act mechanically, a dilemma arises: what solution to choose, in particular, to look for an ODZ or not? I think that my experience will help me solve this dilemma. I will stop making mistakes by learning how to use the DLO correctly. Time will tell whether I can do it, or rather the Unified State Exam.
9. Literature
Etc. "Algebra and the beginning of analysis 10-11" problem book and textbook, M .: "Education", 2002. "Handbook of elementary mathematics." M .: "Science", 1966. Newspaper "Mathematics" №46, Newspaper "Mathematics" № Newspaper "Mathematics" № "History of mathematics in school VII-VIII grades". M .: "Education", 1982. and others. "The most complete edition of options for real tasks of the Unified State Exam: 2009 / FIPI" - M .: "Astrel", 2009. and others. "Unified State Exam. Maths. Universal materials for training students / FIPI "- M .:" Intellect-Center ", 2009. and others." Algebra and the beginning of analysis 10-11 ". M .: "Education", 2007., "Workshop on solving problems in school mathematics (workshop on algebra)." M .: Education, 1976. "25000 lessons of mathematics". M .: "Education", 1993. "Preparing for the Olympiads in mathematics." M .: "Exam", 2006. "Encyclopedia for children" MATHEMATICS "" volume 11, M .: Avanta +; 2002. Materials of sites www. *****, www. *****.
Solving various problems, we very often have to carry out identical transformations of expressions. But it happens that some kind of transformation is permissible in some cases, but not in others. ODZ provides significant assistance in terms of control of the admissibility of the ongoing transformations. Let's dwell on this in more detail.
The essence of the approach is as follows: the ODV of variables for the original expression are compared with the ODV of variables for the expression obtained as a result of performing identical transformations, and the corresponding conclusions are drawn based on the comparison results.
In general, identical transformations can
- do not affect LDU;
- lead to the expansion of DLU;
- lead to a narrowing of the ODZ.
Let's explain each case with an example.
Consider the expression x 2 + x + 3 x, the ODZ of the variable x for this expression is the set R. Now we will do the following identical transformation with this expression - we will give similar terms, as a result it will take the form x 2 + 4 · x. Obviously, the LDZ variable x of this expression is also the set R. Thus, the transformation carried out did not change the DHS.
Let's move on. Take the expression x + 3 / x − 3 / x. In this case, the GDV is determined by the condition x ≠ 0, which corresponds to the set (−∞, 0) ∪ (0, + ∞). This expression also contains similar terms, after reduction of which we arrive at the expression x for which the ODV is R. What we see: as a result of the transformation performed, an expansion of the ODZ occurred (the number zero was added to the ODZ of the variable x for the original expression).
It remains to consider an example of narrowing the range of admissible values \u200b\u200bafter the transformations. Let's take the expression 
... The ODZ of the variable x is defined by the inequality (x − 1) · (x − 3) ≥0, for its solution it is suitable, for example, as a result we have (−∞, 1] ∪∪; edited by S. A. Telyakovsky. - 17- ed. - M.: Education, 2008 .-- 240 p.: ill. - ISBN 978-5-09-019315-3.
In equations and inequalities of the form,,,,, the intersection of the domains of definition of functions and is called the range of permissible values \u200b\u200b(ODV) of the variable, as well as the ODV of the equation or inequality, respectively.
When solving equations (inequalities) with one variable, when the question arises - whether to find the ODV, you can often hear a categorical "yes" and an equally categorical "no". “First you need to find the ODZ, and then proceed to solve the equation (inequality),” some say. “There is no need to waste time on ODZ, in the course of solving we will go to an equivalent equation (inequality) or to an equivalent system of equations and inequalities or only inequalities. In the end, if it's an equation, then you can check, ”others say.
So whether to find ODZ?
Of course, there is no single answer to this question. Finding the DHS equation or inequality is not a necessary element of the solution. In each specific example, this issue is resolved individually.
In some cases, finding the ODV simplifies the solution of an equation or inequality (examples 1-5), and in some cases it is even a necessary step in the solution (examples 1, 2, 4).
In other cases (examples 6, 7), it is worth abandoning the preliminary finding of the ODZ, since it makes the solution more cumbersome.
Example 1. Solve the equation.
Squaring both sides of the equation will not simplify, but complicate it and will not allow you to get rid of radicals. We need to look for another solution.
Let's find the ODZ equation:
Thus, the ODZ contains only one value, and, therefore, only the number 4 can serve as the root of the original equation. By direct substitution, we make sure that is the only root of the equation.
Example 2. Solve the equation.
The presence in the equation of radicals of various degrees - the second, third and sixth - makes the solution difficult. Therefore, first of all, we will find the ODZ of the equation:
By direct substitution, we make sure that is the root of the original equation.
Example 3. Solve inequality.
Of course, it is possible to solve this inequality by considering the cases:,, but finding the ODZ immediately simplifies this solution.
ODZ: 
Substituting this single value into the original inequality, we get a false numerical inequality. Therefore, the original inequality has no solution.
Answer: there is no solution.
Example 4. Solve the equation.
Let's write the equation in the form.
An equation of the form is equivalent to a mixed system 
those. 
Of course, there is no need to find the ODZ.
In our case, we get an equivalent system 
those. 
Equation is equivalent to a set 
The equation has no rational roots, but it can have irrational roots, which will cause difficulties for students to find. Therefore, we will look for another solution.
Let's return to the original equation, write it in the form.
Find ODZ:.
When the right side of the equation, and the left side 
... Consequently, the original equation in the range of admissible values \u200b\u200bof the variable x is equivalent to the system of equations 
the solution of which is only one value.
Thus, in this example, it is precisely the finding of the ODZ that made it possible to solve the original equation.
Example 5. Solve the equation.
Since, a, then when solving the original equation it will be necessary to get rid of the modules (to open them).
Therefore, at first it makes sense to find the ODZ equations:
So, ODZ:
Let's simplify the original equation using the properties of logarithms.
Since in the range of permissible values \u200b\u200bof the variable x and then 
, a, then we get an equivalent equation:
Taking into account that in the ODZ, we pass to the equivalent equation 
and solve it by dividing both parts by 3.
Answer: - 4.75.
Comment.
If you do not find the ODZ, then when solving the equation it would be necessary to consider four cases:,,,. At each of these intervals of constancy of the expressions under the modulus sign, it would be necessary to open the modules and solve the resulting equation. In addition, also check. We see that finding the ODV of the original equation greatly simplifies its solution.
Example 7. Solve inequality 
.
Since the variable x is included in the base of the logarithm, then when solving this inequality it will be necessary to consider two cases: and. Therefore, it is impractical to find the ODZ separately.
So, we represent the original inequality in the form 
and it will be tantamount to a combination of two systems:
Answer: 
.
To find the domains of common functions, in this lesson we will solve equations and inequalities in one variable.
There will also be tasks for an independent solution, to which you can see the answers.
What is the scope of a function? Let's take a look at the function graph in the figure. Each point of the function graph corresponds to a certain value of the "x" - the argument of the function and a certain value of the "game" - the function itself. From the argument - "x" - the "game" is calculated - the value of the function. The domain of a function is a set of all values \u200b\u200bof the "x" for which there exists, that is, the "game" - the value of the function, can be calculated. In other words, the set of argument values \u200b\u200bon which the "function works". Most of the functions are defined by formulas. Therefore, the domain of a function is also the largest set on which the formula makes sense.
The figure shows the graph of the function. The denominator of a fraction cannot be zero, since you cannot divide by zero. Therefore, equating the denominator to zero, we get a value that is not included in the domain of the function: 1. And the domain of the function is all the values \u200b\u200bof "x" from minus infinity to one and from one to plus infinity. This is clearly seen on the graph.
Example 0. How to find the domain of definition of the function y is equal to the square root of x minus five (radical expression x minus five) ()? You just need to solve the inequality
x - 5 ≥ 0 ,
since in order for us to get the real value of the game, the radical expression must be greater than or equal to zero. We get the solution: the domain of the function - all x values \u200b\u200bare greater than or equal to five (or x belongs to the interval from five inclusive to plus infinity).
In the drawing above, there is a fragment of the numerical axis. On it, the domain of definition of the considered function is shaded, while in the "plus" direction, the shading continues indefinitely along with the axis itself.
Constant domain
Constant (constant) defined for any real values x R real numbers. It can be written like this: the domain of this function is the entire number line] - ∞; + ∞ [.
Example 1. Find the domain of a function y = 2 .
Decision. The domain of the function is not indicated, which means that by virtue of the above definition, we mean the natural domain of definition. Expression f(x) \u003d 2 is defined for any real values x , therefore, this function is defined on the entire set R real numbers.
Therefore, in the drawing above, the number line is shaded throughout from minus infinity to plus infinity.
Root scope n-th degree
In the case when the function is given by the formula and n - natural number:
Example 2. Find the domain of a function .
Decision. As follows from the definition, an even root makes sense if the radical expression is non-negative, that is, if - 1 ≤ x ≤ 1. Therefore, the domain of this function is [- 1; 1] .
The shaded area of \u200b\u200bthe number line in the drawing above is the definition area of \u200b\u200bthis function.
Domain of power function
Domain of a power function with an integer exponent
if a a - positive, then the domain of the function is the set of all real numbers, that is] - ∞; + ∞ [;
if a a - negative, then the domain of the function is the set] - ∞; 0 [∪] 0; + ∞ [, that is, the whole number line except zero.
In the corresponding drawing, from above, the entire number line is shaded, and the point corresponding to zero is punctured (it is not included in the domain of the function definition).
Example 3. Find the domain of a function .
Decision. The first term is an integer degree of x equal to 3, and the degree of x in the second term can be represented as one - also an integer. Therefore, the domain of this function is the whole number line, that is] - ∞; + ∞ [.
Domain of a power function with fractional exponent
In the case when the function is given by the formula:
if - positive, then the domain of the function is the set 0; + ∞ [.
Example 4. Find the domain of a function .
Decision. Both terms in the function expression are power functions with positive fractional exponents. Consequently, the domain of this function is the set - ∞; + ∞ [.
Domain of exponential and logarithmic functions
Domain of exponential function
In the case when the function is given by a formula, the domain of the function is the whole number line, that is] - ∞; + ∞ [.
Domain of the logarithmic function
The logarithmic function is defined under the condition that its argument is positive, that is, the domain of its definition is the set] 0; + ∞ [.
Find the scope of the function yourself and then see the solution
Domain of trigonometric functions
Function scope y \u003d cos ( x) is also the set R real numbers.
Function scope y \u003d tg ( x) - a bunch of R
real numbers other than numbers 
.
Function scope y \u003d ctg ( x) - a bunch of R real numbers other than numbers.
Example 8. Find the domain of a function .
Decision. The external function is a decimal logarithm and the conditions of the domain of definition of the logarithmic function in general apply to its domain of definition. That is, her argument must be positive. The argument here is the x sine. Turning an imaginary compass in a circle, we see that the condition sin x \u003e 0 is violated when "x" is equal to zero, "pi", two, multiplied by "pi" and generally equal to the product of "pi" and any even or odd integer.
Thus, the domain of this function is given by the expression
,
where k is an integer.
Domain of inverse trigonometric functions
Function scope y \u003d arcsin ( x) - the set [-1; 1] .
Function scope y \u003d arccos ( x) - also the set [-1; 1] .
Function scope y \u003d arctg ( x) - a bunch of R real numbers.
Function scope y \u003d arcctg ( x) is also the set R real numbers.
Example 9. Find the domain of a function .
Decision. Let's solve the inequality:
Thus, we obtain the domain of definition of this function - the segment [- 4; 4] .
Example 10. Find the domain of a function
.
Decision. Let's solve two inequalities:
Solution of the first inequality:
Solution of the second inequality:
Thus, we get the domain of this function - a segment.
Fraction definition area
If the function is given by a fractional expression in which the variable is in the denominator of the fraction, then the domain of the function is the set R real numbers, except for such x at which the denominator of the fraction vanishes.
Example 11. Find the domain of a function .
Decision. Solving the equality to zero of the denominator of the fraction, we find the domain of this function - the set] - ∞; - 2 [∪] - 2; + ∞ [.
Example 12. Find the domain of a function .
Decision. Let's solve the equation:
Thus, we obtain the domain of this function - ]- ∞; - 1[ ∪ ]- 1 ; 1[ ∪ ]1 ;+ ∞[ .
How ?
Solution examples
If something is missing somewhere, then there is something somewhere
We continue to study the section "Functions and schedules", and the next station of our journey is. An active discussion of this concept began in the article on sets and continued in the first lesson on function graphswhere I looked at elementary functions, and in particular their domains. Therefore, I recommend for dummies to start with the basics of the topic, since I will not dwell on some basic points again.
It is assumed that the reader knows the domain of the following functions: linear, quadratic, cubic function, polynomials, exponential, sine, cosine. They are defined at (set of all real numbers)... For tangents, arcsines, so be it, forgive \u003d) - more rare graphs are not immediately remembered.
The domain of definition seems to be a simple thing, and a natural question arises, what will the article be about? In this lesson, I will look at common tasks for finding the scope of a function. Moreover, we will repeat one variable inequality, the skills of solving which will be required in other problems of higher mathematics. The material, by the way, is all school-based, so it will be useful not only for students, but also for students. The information, of course, does not pretend to be encyclopedic, but here are not contrived "dead" examples, but roasted chestnuts, which are taken from real practical works.
Let's start with a quick cut into the topic. Briefly about the main thing: we are talking about a function of one variable. Its scope is many values \u200b\u200b"x"for which exist the meaning of the "gamers". Let's consider a conditional example: 
The scope of this function is the union of spans:
(for those who have forgotten: - the unification icon). In other words, if you take any value of "x" from the interval, or from, or from, then for each such "x" there will be a value "y".
Roughly speaking, where the domain is, there is a graph of the function. But the half-interval and the "tse" point are not included in the definition area and the graph is not there.
How do I find the scope of a function? Many people remember the children's counting-book: “stone, scissors, paper”, and in this case it can be safely rephrased: “root, fraction and logarithm”. Thus, if you encounter a fraction, root or logarithm on your life path, then you should immediately be very, very alert! Tangent, cotangent, arcsine, arccosine are much less common, and we will also talk about them. But first, sketches from the life of ants:
Domain of a function containing a fraction
Suppose you are given a function containing some fraction. As you know, you cannot divide by zero:, so those x values \u200b\u200bthat turn the denominator to zero are not included in the scope of this function.
I will not dwell on the simplest functions like 
and so on, since everyone can see perfectly the points that are not included in their definition area. Consider more meaningful fractions:
Example 1
Find the domain of a function
Decision: There is nothing special in the numerator, but the denominator must be non-zero. Let's set it to zero and try to find the "bad" points:
The resulting equation has two roots: 
... These values outside the scope of the function... Indeed, plug or into the function and you will see that the denominator vanishes.
Answer: domain: 
The record reads like this: "the domain of definition is all real numbers except for a set consisting of values 
". As a reminder, the backslash symbol in mathematics denotes logical subtraction, and curly braces denote a set. The answer can be equivalently written as a union of three intervals:
As you like.
At points 
function suffers endless breaks, and the straight lines given by the equations 
are vertical asymptotes for the graph of this function. However, this is a slightly different topic, and I will not focus on this further.
Example 2
Find the domain of a function
The assignment is essentially oral, and many of you will find the scope almost immediately. The answer is at the end of the lesson.
Will a fraction always be "bad"? Not. For example, a function is defined on the entire number axis. Whatever value "x" we take, the denominator will not vanish, moreover, it will always be positive:. Thus, the scope of this function is:.
All functions like 
identified and continuous on the .
Slightly more complicated is the situation when the denominator has occupied a square trinomial:
Example 3
Find the domain of a function 
Decision: try to find the points at which the denominator vanishes. For this we decide quadratic equation:
The discriminant turned out to be negative, which means that there are no real roots, and our function is defined on the entire number axis.
Answer: domain:
Example 4
Find the domain of a function 
This is an example for a stand alone solution. Solution and answer at the end of the lesson. I advise you not to be lazy with simple tasks, as misunderstandings will accumulate for further examples.
Function scope with root
The square root function is defined only for those values \u200b\u200bof "x" when the radical expression is non-negative:. If the root is located in the denominator, then the condition is obviously tougher:. Similar calculations are valid for any root of a positive even degree: 
, however, the root is already the 4th degree in research functions I don't remember.
Example 5
Find the domain of a function 
Decision: the radical expression must be non-negative:
Before continuing with the solution, let me remind you of the basic rules for working with inequalities, known from school.
Pay special attention! Inequalities are now being considered with one variable - that is, for us there is only one dimension along the axis... Please do not confuse with inequalities of two variableswhere the entire coordinate plane is geometrically involved. However, there are also pleasant coincidences! So, for inequality, the following transformations are equivalent:
1) The terms can be transferred from part to part, changing their (terms) signs.
2) Both sides of the inequality can be multiplied by a positive number.
3) If both sides of the inequality are multiplied by negative number, then you need to change the sign of the inequality itself... For example, if there was “more”, then it will become “less”; if it was "less than or equal", it will become "greater than or equal".
In the inequality, we transfer the "three" to the right side with a change in sign (rule # 1):
Multiply both sides of the inequality by -1 (rule # 3):
Let's multiply both sides of the inequality by (rule # 2):
Answer: domain: 
The answer can also be written with the equivalent phrase: "the function is defined at".
Geometrically, the region of definition is depicted by hatching the corresponding intervals on the abscissa axis. In this case:
I remind once again the geometric meaning of the domain of definition - the graph of the function 
exists only in the shaded area and is absent at.
In most cases, a purely analytical finding of the domain of definition is suitable, but when the function is very confused, you should draw an axis and make notes.
Example 6
Find the domain of a function
This is an example for a stand alone solution.
When there is a square binomial or trinomial under the square root, the situation becomes a little more complicated, and now we will analyze the solution technique in detail:
Example 7
Find the domain of a function 
Decision: the radical expression must be strictly positive, that is, we need to solve the inequality. In the first step, we try to factor out the square trinomial: 
The discriminant is positive, we are looking for roots: 
Thus the parabola 
intersects the abscissa axis at two points, which means that part of the parabola is located below the axis (inequality), and part of the parabola is above the axis (the inequality we need).
Since the coefficient, the branches of the parabola look up. It follows from the above that the inequality is fulfilled on the intervals (the branches of the parabola go up to infinity), and the vertex of the parabola is located on the interval below the abscissa axis, which corresponds to the inequality:
! Note:
if you do not fully understand the explanations, please draw the second axis and the whole parabola! It is advisable to return to the article and the manual Hot Formulas School Mathematics Course.
Please note that the points themselves are punctured (not included in the solution), since our inequality is strict.
Answer: domain:
In general, many inequalities (including the one considered) are solved by the universal interval method, known again from the school curriculum. But in the cases of square two- and three-term, in my opinion, it is much more convenient and faster to analyze the position of the parabola relative to the axis. And the main method - the method of intervals, we will analyze in detail in the article Function zeros. Intervals of constancy.
Example 8
Find the domain of a function
This is an example for a stand alone solution. The sample contains detailed comments on the logic of reasoning + the second way of solving and another important transformation of inequality, without knowing which the student will limp on one leg ..., ... hmm ... at the expense of the leg, perhaps, he got excited, rather - on one toe. Thumb.
Can a square root function be defined on the whole number line? Sure. All familiar faces:. Or a similar amount with an exhibitor:. Indeed, for any values \u200b\u200bof "x" and "ka":, therefore, it is true and.
Here's a less obvious example: 
... Here the discriminant is negative (the parabola does not intersect the abscissa axis), while the branches of the parabola are directed upwards, therefore, the domain of definition:.
The question is the opposite: can the domain of a function be empty? Yes, and a primitive example immediately suggests itself 
, where the radical expression is negative for any value of "x", and the scope is: (empty set icon). Such a function is not defined at all (of course, the graph is also illusory).
Odd roots 
etc. everything is much better - here the radical expression can be negative... For example, the function is defined on the whole number line. However, the function has a single point still not included in the definition domain, since the denominator is turned to zero. For the same reason for the function 
dots are excluded.
Function domain with logarithm
The third common function is the logarithm. As an example, I will draw the natural logarithm, which comes across in about 99 examples out of 100. If some function contains a logarithm, then its domain of definition should include only those "x" values \u200b\u200bthat satisfy the inequality. If the logarithm is in the denominator:, then additionally the condition is imposed (since).
Example 9
Find the domain of a function
Decision: in accordance with the above, we compose and solve the system: 
A graphical solution for dummies:
Answer: domain:
I will dwell on one more technical point - after all, I have not specified a scale and have not given divisions along the axis. The question arises: how to carry out such drawings in a notebook on checkered paper? Whether to measure the distance between points by cells strictly according to scale? It is more canonical and stricter, of course, to scale, but a schematic drawing that principally reflects the situation is also quite acceptable.
Example 10
Find the domain of a function 
To solve the problem, you can use the method of the previous paragraph - to analyze how the parabola is located relative to the abscissa axis. The answer is at the end of the lesson.
As you can see, in the realm of logarithms, everything is very similar to the situation with a square root: the function 
(square trinomial from Example No. 7) is defined on intervals, and the function 
(square binomial from Example # 6) on the interval. It’s embarrassing to say that type functions are defined on the whole number line.
Useful information
: interesting typical function, it is defined on the whole number line except for the point. According to the property of the logarithm, the "two" can be taken out by a factor outside the logarithm, but so that the function does not change, "x" must be enclosed under the modulus sign: 
... Here is another "practical application" of the module \u003d). This should be done in most cases when you carry even degree, for example: 
... If the base of the degree is obviously positive, for example, then there is no need for the modulus sign and it is enough to get by with parentheses:.
In order not to repeat ourselves, let's complicate the task:
Example 11
Find the domain of a function 
Decision: in this function we have both the root and the logarithm.
The radical expression must be non-negative:, and the expression under the logarithm sign must be strictly positive:. Thus, it is necessary to solve the system:
Many of you know perfectly well or intuitively guess that the solution of the system should satisfy to each condition.
Examining the position of the parabola relative to the axis, we come to the conclusion that the interval (blue shading) satisfies the inequality:
The “red” half-interval obviously corresponds to inequality.
Since both conditions must be met at the same time, then the solution of the system is the intersection of these intervals. "Common interests" are observed at the half-interval.
Answer: domain:
Typical inequality, as demonstrated in Example No. 8, is not difficult to resolve analytically.
The found scope will not change for "similar functions", for example, for 
or 
... You can also add some continuous functions, for example:, or like this: 
, or even like this:. As they say, the root and the logarithm are stubborn things. The only thing is that if one of the functions is "reset" to the denominator, then the scope will change (although in the general case this is not always true). Well, in the theory of matan about this verbal ... oh ... there are theorems.
Example 12
Find the domain of a function 
This is an example for a stand alone solution. Using a drawing is quite appropriate, since the function is not the easiest one.
A couple more examples for securing the material:
Example 13
Find the domain of a function
Decision: compose and solve the system: 
All actions have already been discussed in the course of the article. Let us draw on the number line the interval corresponding to the inequality and, according to the second condition, exclude two points:
The meaning turned out to be completely out of work.
Answer: domain
A little mathematical pun on a variation of the 13th example:
Example 14
Find the domain of a function 
This is an example for a stand alone solution. Whoever missed is in flight ;-)
The final section of the lesson is devoted to more rare, but also "working" functions:
Function domains
with tangents, cotangents, arcsines, arccosines
If it enters into some function, then from its domain of definition excluded points 
where Z - a set of integers. In particular, as noted in the article Graphs and properties of elementary functions, the function has the following values:
That is, the domain of definition of the tangent: 
.
We will not kill ourselves much:
Example 15
Find the domain of a function
Decision: in this case, the following points will not be included in the definition area:
Let's drop the "two" of the left side into the denominator of the right side:
As a result 
:
Answer: domain: 
.
In principle, the answer can be written as a union of an infinite number of intervals, but the construction will turn out to be very cumbersome:
The analytical solution is fully consistent with geometric transformation of the graph: if the function argument is multiplied by 2, then its graph will shrink to the axis twice. Notice how the function has halved its period, and break points doubled in frequency. Tachycardia.
A similar story with cotangent. If it enters into some function, then points are excluded from its domain of definition. In particular, we shoot the following values \u200b\u200bfor the function with an automatic burst:
In other words: