Acceptable value range (APV): theory, examples, solutions. How to find the domain of a function? Examples of solutions

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First, let's learn how to find 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 definition of the function f is the intersection of the domains of definition of the functions f 1, f 2, ..., f n. Let's write this as .

Let's agree to continue to use entries similar to the last one, by which we mean written inside a curly brace, or the simultaneous fulfillment of any conditions. This is convenient and quite naturally resonates with the meaning of the systems.

Example.

The function y=x 7 +x+5+tgx is given, and we need to find its domain of definition.

Solution.

The function f is represented by the sum of four functions: f 1 - power function with exponent 7, f 2 - power function with exponent 1, f 3 - constant function and f 4 - tangent function.

Looking at the table of areas for defining the main elementary functions, we find that D(f 1)=(−∞, +∞) , D(f 2)=(−∞, +∞) , D(f 3)=(−∞, +∞) , and the domain of definition of the tangent is the set of all real numbers except numbers .

The domain of definition of the function f is the intersection of the domains of definition 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:

the set of all real numbers except .

Let's move on to finding domain of definition of a product of functions. For this case, a similar rule applies:

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 definition of the function f is the intersection of the domains of definition of the functions f 1, f 2, ..., f n. So, .

This is understandable, in the indicated area all product functions are defined, and hence the function f itself.

Example.

Y=3·arctgx·lnx .

Solution.

The structure of the right-hand side of the formula defining 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 arctangent function, and f 3 is a logarithmic function with base e.

We know that D(f 1)=(−∞, +∞) , D(f 2)=(−∞, +∞) and D(f 3)=(0, +∞) . Then .

Answer:

The domain of definition of the function y=3·arctgx·lnx is the set of all real positive numbers.

Let us separately focus on finding the domain of definition of a function given by the formula y=C·f(x), where C is some real number. It is easy to show that the domain of definition of this function and the domain of definition 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 a function f is D(f) . Then the domain of definition of the function y=C f(x) is , which is what needed to be shown.

So, the domains of definition of the functions y=f(x) and y=C·f(x), where C is some real number, coincide. For example, 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 definition of a complex function y=f 1 (f 2 (x)) is the intersection of two sets: the set of all such x that x∈D(f 2) and the set of all such x for which f 2 (x)∈D(f 1) . That is, in the notation we have adopted (this is essentially a system of inequalities).

Let's look at some example solutions. We will not describe the process in detail, as this is beyond the scope of this article.

Example.

Find the domain of definition of the function y=lnx 2 .

Solution.

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 known areas definitions 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 domain of a function ?

Solution.

This function is complex, it can be considered as y=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 of definition.

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) :

To arcsinx>0, remember the properties of the arcsine function. The arcsine increases throughout the entire domain of definition [−1, 1] and goes to zero at x=0, therefore, arcsinx>0 for any x from the interval (0, 1] .

Let's return to the system:

Thus, the required domain of definition of the function is the half-interval (0, 1].

Answer:

(0, 1] .

Now let's move on to complex functions of the general form y=f 1 (f 2 (...f n (x)))). The domain of definition of the function f in this case is found as .

Example.

Find the domain of a function .

Solution.

A given complex function can be written as y=f 1 (f 2 (f 3 (x))), where f 1 – sin, f 2 – fourth-degree root function, f 3 – log.

We know that D(f 1)=(−∞, +∞) , D(f 2)=- ∞; + ∞[ .

Example 1. Find the domain of a function y = 2 .

Solution. The domain of definition of the function is not indicated, which means that by virtue of the above definition, the natural domain of definition is meant. Expression f(x) = 2 defined for any real values x, hence, this function defined on the entire set R real numbers.

Therefore, in the drawing above, the number line is shaded all the way from minus infinity to plus infinity.

Root definition area 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 .

Solution. As follows from the definition, a root of an even degree makes sense if the radical expression is non-negative, that is, if - 1 ≤ x≤ 1. Therefore, the domain of definition of this function is [- 1; 1] .

The shaded area of ​​the number line in the drawing above is the domain of definition of this function.

Domain of power function

Domain of a power function with an integer exponent

If a- positive, then the domain of definition of the function is the set of all real numbers, that is ]- ∞; + ∞[ ;

If a- negative, then the domain of definition of the function is the set ]- ∞; 0[ ∪ ]0 ;+ ∞[ , that is, the entire number line except zero.

In the corresponding drawing above, the entire number line is shaded, and the point corresponding to zero is punched out (it is not included in the domain of definition of the function).

Example 3. Find the domain of a function .

Solution. The first term is an integer power of x equal to 3, and the power of x in the second term can be represented as one - also an integer. Consequently, the domain of definition of this function is the entire number line, that is ]- ∞; + ∞[ .

Domain of a power function with a fractional exponent

In the case when the function is given by the formula:

if is positive, then the domain of definition of the function is the set 0; + ∞[ .

Example 4. Find the domain of a function .

Solution. Both terms in the function expression are power functions with positive fractional exponents. Consequently, the domain of definition of this function is the set - ∞; + ∞[ .

Domain of exponential and logarithmic functions

Domain of the exponential function

In the case when a function is given by a formula, the domain of definition of the function is the entire number line, that is ] - ∞; + ∞[ .

Domain of the logarithmic function

The logarithmic function is defined provided that its argument is positive, that is, its domain of definition is the set ]0; + ∞[ .

Find the domain of the function yourself and then look at the solution

Domain of trigonometric functions

Function Domain y= cos( x) - also many R real numbers.

Function Domain y= tg( x) - a bunch of R real numbers other than numbers .

Function Domain y= ctg( x) - a bunch of R real numbers, except numbers.

Example 8. Find the domain of a function .

Solution. The external function is a decimal logarithm and its domain of definition is subject to the conditions of the domain of definition logarithmic function at all. That is, her argument must be positive. The argument here is the sine of "x". Turning an imaginary compass around a circle, we see that the condition sin x> 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 definition of this function is given by the expression

,

Where k- an integer.

Domain of definition of inverse trigonometric functions

Function Domain y= arcsin( x) - set [-1; 1] .

Function Domain y= arccos( x) - also the set [-1; 1] .

Function Domain y= arctan( x) - a bunch of R real numbers.

Function Domain y= arcctg( x) - also many R real numbers.

Example 9. Find the domain of a function .

Solution. 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 .

Solution. Let's solve two inequalities:

Solution to the first inequality:

Solution to the second inequality:

Thus, we obtain the domain of definition of this function - the segment.

Fraction scope

If a function is given by a fractional expression in which the variable is in the denominator of the fraction, then the domain of definition of the function is the set R real numbers, except these x, at which the denominator of the fraction becomes zero.

Example 11. Find the domain of a function .

Solution. By solving the equality of the denominator of the fraction to zero, we find the domain of definition of this function - the set ]- ∞; - 2[ ∪ ]- 2 ;+ ∞[ .