When 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. The DHS provides significant assistance in terms of monitoring the admissibility of the ongoing transformations. Let's dwell on this in more detail.
The essence of the approach is as follows: the ODZ of variables for the original expression is compared with the ODZ of variables for the expression obtained as a result of performing identical transformations, and based on the results of the comparison, appropriate conclusions are drawn.
In general, identical transformations can
- do not affect the ODZ;
- lead to an expansion of the DHS;
- 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 let's do the following identical transformation with this expression - let's bring like terms , as a result it will take the form x 2 +4 x . Obviously, the ODZ variable x of this expression is also the set R . Thus, the transformation did not change the ODZ.
Let's move on. Take the expression x+3/x−3/x . In this case, the ODZ 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 come to the expression x, for which the ODZ is R. What we see: as a result of the transformation, the ODZ has expanded (the number zero has been 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 after transformations. Take the expression 
. The ODZ of the variable x is determined by the inequality (x−1) (x−3)≥0, suitable for its solution, for example, as a result we have (−∞, 1]∪∪; edited by S. A. Telyakovskii. - 17- e ed. - M.: Education, 2008. - 240 pp.: illustrations - ISBN 978-5-09-019315-3.
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Let's start by finding domain of definition of the sum of functions. It is clear that such a function makes sense for all such values of the variable for which all the functions that make up the sum make sense. Therefore, there is no doubt about the validity of the following statement:
If the function f is the sum of n functions f 1 , f 2 , …, f n , that is, the function f is given by the formula y=f 1 (x)+f 2 (x)+…+f n (x) , then the domain of the function f is the intersection of the domains of the functions f 1 , f 2 , …, f n . Let's write it as .
Let's agree to continue using records like the last one, by which we mean written inside a curly bracket, or the simultaneous fulfillment of any conditions. This is convenient and quite naturally resonates with the meaning of systems.
Example.
Given a function y=x 7 +x+5+tgx , and we need to find its domain.
Decision.
The function f is represented by the sum of four functions: f 1 is a power function with an exponent of 7 , f 2 is a power function with an exponent of 1 , f 3 is a constant function and f 4 is a tangent function.
Looking at the table of domains of definition of the basic elementary functions, we find that D(f 1)=(−∞, +∞) , D(f 2)=(−∞, +∞) , D(f 3)=(−∞, +∞) , and the domain of the tangent is the set of all real numbers, except for the numbers 
.
The domain of the function f is the intersection of the domains of the functions f 1 , f 2 , f 3 and f 4 . It is quite obvious that this is the set of all real numbers, with the exception of the numbers .
Answer:
set of all real numbers except .
Let's move on to finding domains of the product of functions. For this case, a similar rule holds:
If the function f is the product of n functions f 1 , f 2 , …, f n , that is, the function f is given by the formula y=f 1 (x) f 2 (x) ... f n (x), then the domain of the function f is the intersection of the domains of the functions f 1 , f 2 , …, f n . So, .
It is understandable, in the indicated area all the functions of the product are defined, and hence the function f itself.
Example.
Y=3 arctgx lnx .
Decision.
The structure of the right side of the formula that defines the function can be considered as f 1 (x) f 2 (x) f 3 (x) , where f 1 is a constant function, f 2 is the arc tangent function, and f 3 is the logarithmic function with base e.
We know that D(f 1)=(−∞, +∞) , D(f 2)=(−∞, +∞) and D(f 3)=(0, +∞) . Then 
.
Answer:
the domain of the function y=3 arctgx lnx is the set of all real positive numbers.
Let us dwell separately on finding the domain of definition of the function given by the formula y=C·f(x) , where C is some real number. It is easy to show that the domain of this function and the domain of the function f coincide. Indeed, the function y=C f(x) is the product of a constant function and a function f . The domain of a constant function is the set of all real numbers, and the domain of the function f is D(f) . Then the domain of the function y=C f(x) is 
, which was to be shown.
So, the domains of the functions y=f(x) and y=C·f(x) , where С is some real number, coincide. For example, if the domain of the root is , it becomes clear that D(f) is the set of all x from the domain of the function f 2 for which f 2 (x) is included in the domain of the function f 1 .
Thus, domain of a complex function y=f 1 (f 2 (x)) is the intersection of two sets: the set of all x such that x∈D(f 2) and the set of all x such that f 2 (x)∈D(f 1) . That is, in our notation 
(this is essentially a system of inequalities).
Let's take a look at a few examples. In the process, we will not describe in detail, as this is beyond the scope of this article.
Example.
Find the domain of the function y=lnx 2 .
Decision.
The original function can be represented as y=f 1 (f 2 (x)) , where f 1 is a logarithm with base e, and f 2 is a power function with exponent 2.
Turning to the known domains of definition of the basic elementary functions, we have D(f 1)=(0, +∞) and D(f 2)=(−∞, +∞) .
Then 
So we found the domain of definition of the function we needed, it is the set of all real numbers except zero.
Answer:
(−∞, 0)∪(0, +∞) .
Example.
What is the scope of the function 
?
Decision.
This function is complex, it can be considered as y \u003d f 1 (f 2 (x)) , where f 1 is a power function with exponent, and f 2 is the arcsine function, and we need to find its domain.
Let's see what we know: D(f 1)=(0, +∞) and D(f 2)=[−1, 1] . It remains to find the intersection of sets of values x such that x∈D(f 2) and f 2 (x)∈D(f 1) : 
For arcsinx>0, let's recall the properties of the arcsine function. The arcsine increases over the entire domain of definition [−1, 1] and vanishes at x=0 , therefore, arcsinx>0 for any x from the interval (0, 1] .
Let's go back to the system: 
Thus, the desired domain of definition of the function is a half-interval (0, 1] .
Answer:
(0, 1] .
Now let's move on to complex general functions y=f 1 (f 2 (…f n (x)))) . The domain of the function f in this case is found as 
.
Example.
Find the scope of a function 
.
Decision.
The given complex function can be written as y \u003d f 1 (f 2 (f 3 (x))), where f 1 - sin, f 2 - function of the root of the fourth degree, f 3 - lg.
We know that D(f 1)=(−∞, +∞) , D(f 2)=∪∪/ Access mode: Materials of sites www.fipi.ru, www.eg
Appendix 1
Practical work "ODZ: when, why and how?"
|
Option 1 |
Option 2 |
|
│х+14│= 2 - 2х |
|
|
│3-х│=1 - 3х |
Annex 2
Answers to the tasks of practical work "ODZ: when, why and how?"
|
Option 1 |
Option 2 |
|
Answer: no roots |
Answer: x is any number except x=5 |
|
9x+ = +27 ODZ: x≠3 Answer: no roots |
ODZ: x=-3, x=5. Answer: -3;5. |
|
y= -decreases, y= -increases So the equation has at most one root. Answer: x=6. |
ODZ: → →х≥5 Answer: x≥5, x≤-6. |
|
│х+14│=2-2х ODZ:2-2х≥0, х≤1 х=-4, х=16, 16 does not belong to ODZ |
Decreases - increases The equation has at most one root. Answer: no roots. |
|
0, ODZ: x≥3, x≤2 Answer: x≥3, x≤2 |
8x+ = -32, ODZ: x≠-4. Answer: no roots. |
|
x=7, x=1. Answer: no solution |
Increasing - decreasing Answer: x=2. |
|
0 ODZ: x≠15 Answer: x is any number except x=15. |
│3-х│=1-3х, ODZ: 1-3х≥0, х≤ x=-1, x=1 does not belong to the ODZ. Answer: x=-1. |
