Many problems lead us to search for a set of function values ​​on a certain segment or throughout the entire domain of definition. Such tasks include various evaluations of expressions and solving inequalities.

In this article, we will define the range of values ​​of a function, consider methods for finding it, and analyze in detail the solution of examples from simple to more complex. All material will be provided with graphic illustrations for clarity. So this article is a detailed answer to the question of how to find the range of a function.

Definition.

The set of values ​​of the function y = f(x) on the interval X is the set of all values ​​of a function that it takes when iterating over all .

Definition.

Function range y = f(x) is the set of all values ​​of a function that it takes when iterating over all x from the domain of definition.

The range of the function is denoted as E(f) .

The range of a function and the set of values ​​of a function are not the same thing. We will consider these concepts equivalent if the interval X when finding the set of values ​​of the function y = f(x) coincides with the domain of definition of the function.

Also, do not confuse the range of the function with the variable x for the expression on the right side of the equality y=f(x) . Region acceptable values variable x for the expression f(x) - this is the domain of definition of the function y=f(x) .

The figure shows several examples.

Function graphs are shown with thick blue lines, thin red lines are asymptotes, red dots and lines on the Oy axis show the range of values ​​of the corresponding function.

As you can see, the range of values ​​of a function is obtained by projecting the graph of the function onto the y-axis. It can be one single number (first case), a set of numbers (second case), a segment (third case), an interval (fourth case), an open ray (fifth case), a union (sixth case), etc.

So what do you need to do to find the range of values ​​of a function?

Let's start from the very beginning simple case: we'll show you how to define a set of values continuous function y = f(x) on the segment .

It is known that a function continuous on an interval reaches its maximum and minimum values ​​on it. Thus, the set of values ​​of the original function on the segment will be the segment . Consequently, our task comes down to finding the largest and smallest values ​​of the function on the segment.

For example, let's find the range of values ​​of the arcsine function.

Example.

Specify the range of the function y = arcsinx .

Solution.

The area of ​​definition of the arcsine is the segment [-1; 1]. Let's find the greatest and smallest value functions on this segment.

The derivative is positive for all x from the interval (-1; 1), that is, the arcsine function increases over the entire domain of definition. Consequently, it takes the smallest value at x = -1, and the largest at x = 1.

We have obtained the range of arcsine function .

Example.

Find the set of function values on the segment.

Solution.

Let's find the largest and smallest value of the function on a given segment.

Let us determine the extremum points belonging to the segment:

We calculate the values ​​of the original function at the ends of the segment and at points :

Therefore, the set of values ​​of a function on an interval is the interval .

Now we will show how to find the set of values ​​of a continuous function y = f(x) in the intervals (a; b) , .

First, we determine the extremum points, extrema of the function, intervals of increase and decrease of the function on a given interval. Next, we calculate at the ends of the interval and (or) the limits at infinity (that is, we study the behavior of the function at the boundaries of the interval or at infinity). This information is enough to find the set of function values ​​on such intervals.

Example.

Define the set of function values ​​on the interval (-2; 2) .

Solution.

Let's find the extremum points of the function falling on the interval (-2; 2):

Dot x = 0 is a maximum point, since the derivative changes sign from plus to minus when passing through it, and the graph of the function goes from increasing to decreasing.

there is a corresponding maximum of the function.

Let's find out the behavior of the function as x tends to -2 on the right and as x tends to 2 on the left, that is, we find one-sided limits:

What we got: when the argument changes from -2 to zero, the function values ​​increase from minus infinity to minus one-fourth (the maximum of the function at x = 0), when the argument changes from zero to 2, the function values ​​decrease to minus infinity. Thus, the set of function values ​​on the interval (-2; 2) is .

Example.

Specify the set of values ​​of the tangent function y = tgx on the interval.

Solution.

The derivative of the tangent function on the interval is positive , which indicates an increase in function. Let's study the behavior of the function at the boundaries of the interval:

Thus, when the argument changes from to, the function values ​​increase from minus infinity to plus infinity, that is, the set of tangent values ​​on this interval is the set of all real numbers.

Example.

Find the range of a function natural logarithm y = lnx.

Solution.

The natural logarithm function is defined for positive values argument . On this interval the derivative is positive , this indicates an increase in the function on it. Let's find the one-sided limit of the function as the argument tends to zero on the right, and the limit as x tends to plus infinity:

We see that as x changes from zero to plus infinity, the values ​​of the function increase from minus infinity to plus infinity. Therefore, the range of the natural logarithm function is the entire set of real numbers.

Example.

Solution.

This function is defined for everyone real values x. Let us determine the extremum points, as well as the intervals of increase and decrease of the function.

Consequently, the function decreases at , increases at , x = 0 is the maximum point, the corresponding maximum of the function.

Let's look at the behavior of the function at infinity:

Thus, at infinity the values ​​of the function asymptotically approach zero.

We found that when the argument changes from minus infinity to zero (the maximum point), the function values ​​increase from zero to nine (to the maximum of the function), and when x changes from zero to plus infinity, the function values ​​decrease from nine to zero.

Look at the schematic drawing.

Now it is clearly visible that the range of values ​​of the function is .

Finding the set of values ​​of the function y = f(x) on intervals requires similar research. We will not dwell on these cases in detail now. We will meet them again in the examples below.

Let the domain of definition of the function y = f(x) be the union of several intervals. When finding the range of values ​​of such a function, the sets of values ​​on each interval are determined and their union is taken.

Example.

Find the range of the function.

Solution.

The denominator of our function should not go to zero, that is, .

First, let's find the set of function values ​​on the open ray.

Derivative of a function is negative on this interval, that is, the function decreases on it.

We found that as the argument tends to minus infinity, the function values ​​asymptotically approach unity. When x changes from minus infinity to two, the values ​​of the function decrease from one to minus infinity, that is, on the interval under consideration, the function takes on a set of values. We do not include unity, since the values ​​of the function do not reach it, but only asymptotically tend to it at minus infinity.

We proceed similarly for open beam.

On this interval the function also decreases.

The set of function values ​​on this interval is the set .

Thus, the desired range of values ​​of the function is the union of the sets and .

Graphic illustration.

Special attention should be paid to periodic functions. Range of values periodic functions coincides with the set of values ​​on the interval corresponding to the period of this function.

Example.

Find the range of the sine function y = sinx.

Solution.

This function is periodic with a period of two pi. Let's take a segment and define the set of values ​​​​on it.

The segment contains two extremum points and .

We calculate the values ​​of the function at these points and on the boundaries of the segment, select the smallest and highest value:

Hence, .

Example.

Find the range of a function .

Solution.

We know that the arc cosine range is the segment from zero to pi, that is, or in another post. Function can be obtained from arccosx by shifting and stretching along the abscissa axis. Such transformations do not affect the range of values, therefore, . Function obtained from stretching three times along the Oy axis, that is, . And the last stage of transformation is a shift of four units down along the ordinate. This leads us to double inequality

Thus, the required range of values ​​is .

Let us give the solution to another example, but without explanations (they are not required, since they are completely similar).

Example.

Define Function Range .

Solution.

Let us write the original function in the form . Range of values power function is interval . That is, . Then

Hence, .

To complete the picture, we should talk about finding the range of values ​​of a function that is not continuous on the domain of definition. In this case, we divide the domain of definition into intervals by break points, and find sets of values ​​on each of them. By combining the resulting sets of values, we obtain the range of values ​​of the original function. We recommend you remember

The concept of function and everything connected with it is traditionally complex and not fully understood. A special stumbling block when studying a function and preparing for the Unified State Exam is the domain of definition and the range of values ​​(changes) of the function.
Often students do not see the difference between the domain of a function and the domain of its values.
And if students manage to master the tasks of finding the domain of definition of a function, then the tasks of finding the set of values ​​of a function cause them considerable difficulties.
The purpose of this article: to familiarize yourself with methods for finding function values.
As a result of considering this topic, theoretical material was studied, methods for solving problems of finding sets of function values ​​were considered, didactic material For independent work students.
This article can be used by a teacher in preparing students for graduation and entrance exams, when studying the topic “Value Range of a Function” on extracurricular activities elective courses in mathematics.

I. Determining the range of values ​​of a function.

The domain (set) of values ​​E(y) of the function y = f(x) is the set of such numbers y 0, for each of which there is a number x 0 such that: f(x 0) = y 0.

Let us recall the ranges of values ​​of the main elementary functions.

Let's look at the table.

Note also that the range of value of any polynomial of even degree is the interval , where n is the largest value of this polynomial.

II. Properties of functions used when finding the range of a function

To successfully find the set of values ​​of a function, you must have a good knowledge of the properties of the basic elementary functions, especially their domains of definition, ranges of values, and the nature of monotonicity. Let us present the properties of continuous, monotone differentiable functions that are most often used when finding the set of function values.

Properties 2 and 3, as a rule, are used together with the property of an elementary function to be continuous in its domain of definition. At the same time, the simplest and short solution The task of finding the set of values ​​of a function is achieved on the basis of property 1, if using simple methods it is possible to determine the monotonicity of the function. The solution to the problem is even simpler if the function, in addition, is even or odd, periodic, etc. Thus, when solving problems of finding sets of values ​​of a function, one should, as necessary, check and use the following properties of the function:

• continuity;
• monotone;
• differentiability;
• even, odd, periodicity, etc.

Simple tasks to find the set of values ​​of a function are mostly oriented:

a) to use the simplest estimates and restrictions: (2 x >0, -1≤sinx?1, 0≤cos 2 x?1, etc.);

b) to isolate a complete square: x 2 – 4x + 7 = (x – 2) 2 + 3;

c) for transformation trigonometric expressions: 2sin 2 x – 3cos 2 x + 4 = 5sin 2 x +1;

d) using the monotonicity of the function x 1/3 + 2 x-1 increases by R.

III. Let's consider ways to find the ranges of functions.

a) sequentially finding the values ​​of complex function arguments;
b) estimation method;
c) use of the properties of continuity and monotonicity of a function;
d) use of derivative;
e) using the largest and smallest values ​​of the function;
f) graphical method;
g) parameter input method;
h) inverse function method.

Let us reveal the essence of these methods using specific examples.

Example 1: Find the range E(y) functions y = log 0.5 (4 – 2 3 x – 9 x).

Let's solve this example using the method sequential finding values ​​of complex function arguments. Having highlighted perfect square under the logarithm, we transform the function

y = log 0.5 (5 – (1 + 2 3 x – 3 2x)) = log 0.5 (5 – (3 x + 1) 2)

And we will sequentially find the sets of values ​​of its complex arguments:

E(3 x) = (0;+∞), E(3 x + 1) = (1;+∞), E(-(3 x + 1) 2 = (-∞;-1), E(5 – (3 x +1) 2) = (-∞;4)

Let's denote t= 5 – (3 x +1) 2, where -∞≤ t≤4. Thus, the problem is reduced to finding the set of values ​​of the function y = log 0.5 t on the ray (-∞;4) . Since the function y = log 0.5 t is defined only for, its set of values ​​on the ray (-∞;4) coincides with the set of function values ​​on the interval (0;4), which is the intersection of the ray (-∞;4) with domain of definition (0;+∞) logarithmic function. On the interval (0;4) this function is continuous and decreasing. At t> 0 it tends to +∞, and when t = 4 takes the value -2, so E(y) =(-2, +∞).

Example 2: Find the range of a function

y = cos7x + 5cosx

Let's solve this example using the estimation method, the essence of which is to estimate a continuous function from below and above and to prove that the function reaches the lower and upper bounds of the estimates. In this case, the coincidence of the set of function values ​​with the interval from the lower bound of the estimate to the upper one is determined by the continuity of the function and the absence of other values ​​for it.

From the inequalities -1≤cos7x?1, -5≤5cosx?5 we obtain the estimate -6≤y?6. At x = p and x = 0, the function takes values ​​-6 and 6, i.e. reaches the lower and upper bounds of the estimate. As a linear combination of the continuous functions cos7x and cosx, the function y is continuous throughout number axis, therefore, by the property of a continuous function, it takes all values ​​from -6 to 6 inclusive, and only them, since due to the inequalities -6≤y?6, other values ​​are impossible for it. Hence, E(y)= [-6;6].

Example 3: Find the range E(f) functions f(x)= cos2x + 2cosx.

Using the cosine formula double angle transform the function f(x)= 2cos 2 x + 2cosx – 1 and denote t=cosx. Then f(x)= 2t 2 + 2t – 1. Since E(cosx) =

[-1;1], then the range of values ​​of the function f(x) coincides with the set of values ​​of the function g (t)= 2t 2 + 2t – 1 on the segment [-1;1], which we will find graphical method. Having plotted the function y = 2t 2 + 2t – 1 = 2(t + 0.5) 2 – 1.5 on the interval [-1;1], we find E(f) = [-1,5; 3].

Note: many problems with a parameter are reduced to finding the set of values ​​of a function, mainly related to the solvability and number of solutions to equations and inequalities. For example, the equation f(x)= a is solvable if and only if

a E(f) Likewise, Eq. f(x)= a has at least one root located on some interval X, or does not have a single root on this interval if and only if a belongs or does not belong to the set of values ​​of the function f(x) on the interval X. Also studied using a set of function values ​​and inequalities f(x)≠ A, f(x)> a, etc. In particular, f(x)≠ and for all admissible values ​​of x, if a E(f)

Example 4. For what values ​​of the parameter a does the equation (x + 5) 1/2 = a(x 2 + 4) have a single root on the interval [-4;-1].

Let's write the equation in the form (x + 5) 1/2 / (x 2 + 4) = a. The last equation has at least one root on the interval [-4;-1] if and only if a belongs to the set of values ​​of the function f(x) =(x + 5) 1/2 / (x 2 + 4) on the segment [-4;-1]. Let's find this set using the property of continuity and monotonicity of the function.

On the interval [-4;-1] the function y = xІ + 4 is continuous, decreasing and positive, therefore the function g(x) = 1/(x 2 + 4) is continuous and increases on this segment, since when divided by positive function the nature of the monotonicity of the function changes to the opposite. Function h(x) =(x + 5) 1/2 is continuous and increasing in its domain of definition D(h) =[-5;+∞) and, in particular, on the segment [-4;-1], where it is, in addition, positive. Then the function f(x)=g(x) h(x), as the product of two continuous, increasing and positive functions, is also continuous and increasing on the segment [-4;-1], therefore its set of values ​​on [-4;-1] is the segment [ f(-4); f(-1)] = . Consequently, the equation has a solution on the interval [-4;-1], and the only one (by the property of the continuous monotonic function), at 0.05 ≤ a ≤ 0.4

Comment. Solvability of the equation f(x) = a on a certain interval X is equivalent to belonging to the values ​​of the parameter A set of function values f(x) on X. Consequently, the set of values ​​of the function f(x) on the interval X coincides with the set of parameter values A, for which the equation f(x) = a has at least one root on the interval X. In particular, the range of values E(f) functions f(x) matches the set of parameter values A, for which the equation f(x) = a has at least one root.

Example 5: Find the range E(f) functions

Let us solve the example by introducing a parameter according to which E(f) matches the set of parameter values A, for which the equation

has at least one root.

When a = 2, the equation is linear - 4x - 5 = 0 with a non-zero coefficient for unknown x, therefore it has a solution. For a≠2, the equation is quadratic, so it is solvable if and only if its discriminant

Since the point a = 2 belongs to the segment

then the desired set of parameter values A, therefore, the range of values E(f) will be the entire segment.

How direct development method of introducing a parameter when finding a set of function values, you can consider the method of the inverse function, to find which you need to solve the equation for x f(x)=y, considering y to be a parameter. If this equation has the only solution x =g(y), then the range of values E(f) original function f(x) coincides with the domain of definition D(g) inverse function g(y). If the equation f(x)=y has several solutions x =g 1 (y), x =g 2 (y) etc., then E(f) is equal to the union of the domains of the function g 1 (y), g 2 (y) etc.

Example 6: Find the range E(y) functions y = 5 2/(1-3x).

From Eq.

we'll find inverse function x = log 3 ((log 5 y – 2)/(log 5 y)) and its domain of definition D(x):

Since the equation for x has a unique solution, then

E(y) = D(x) = (0; 1)(25;+∞ ).

If the domain of definition of a function consists of several intervals or the function is defined on different intervals different formulas, then to find the range of values ​​of a function, you need to find the sets of function values ​​on each interval and take their union.

Example 7: Find Ranges f(x) And f(f(x)), Where

f(x) on the ray (-∞;1], where it coincides with the expression 4 x + 9 4 -x + 3. Let us denote t = 4 x. Then f(x) = t + 9/t + 3, where 0< t ≤ 4 , так как показательная функция непрерывно возрастает на луче (-∞;1] и стремится к нулю при х → -∞. Тем самым множество значений функции f(x) on the ray (-∞;1] coincides with the set of values ​​of the function g(t) = t + 9/t + 3, on the interval (0;4], which we find using the derivative g’(t) = 1 – 9/t 2. On the interval (0;4] derivative g'(t) is defined and vanishes there at t = 3. At 0<t<3 она отрицательна, а при 3<t<4 положительна. Следовательно, в интервале (0;3) функция g(t) decreases, and in the interval (3;4) it increases, remaining continuous throughout the entire interval (0;4), so g (3)= 9 – the smallest value of this function on the interval (0;4], while its greatest value does not exist, so when t→0 function on the right g(t)→+∞. Then, by the property of a continuous function, the set of values ​​of the function g(t) on the interval (0;4], and therefore a set of values f(x) on (-∞;-1], there will be a ray .

Now, combining the intervals - the sets of function values f(f(x)), denote t = f(x). Then f(f(x)) = f(t), where For the specified t function f(t)= 2cos( x-1) 1/2+ 7 and it again takes all values ​​from 5 to 9 inclusive, i.e. range E(fІ) = E(f(f(x))) =.

Similarly, denoting z = f(f(x)), you can find the range of values E(f 3) functions f(f(f(x))) = f(z), where 5 ≤ z ≤ 9, etc. Make sure E(f 3) = .

The most universal method of finding a set of function values ​​is to use the largest and smallest values ​​of the function on a given interval.

Example 8. At what parameter values r inequality 8 x - р ≠ 2 x+1 – 2 x holds for all -1 ≤ x< 2.

Having designated t = 2x, we write the inequality in the form р ≠ t 3 – 2t 2 + t. Because t = 2x– continuous increasing function on R, then for -1 ≤ x< 2 переменная

2 -1 ≤ t<2 2 ↔

0.5 ≤ t< 4, и исходное неравенство выполняется для всех -1 ≤ x < 2 тогда и только тогда, когда r different from the function values f(t) = t 3 – 2t 2 + t at 0.5 ≤ t< 4.

Let's first find the set of values ​​of the function f(t) on the segment where it has a derivative everywhere f’(t) =3t 2 – 4t + 1. Hence, f(t) is differentiable, and therefore continuous on the interval. From Eq. f’(t) = 0 find the critical points of the function t = 1/3, t = 1, the first of which does not belong to the segment, and the second belongs to it. Because f(0.5) = 1/8, f(1) = 0, f(4) = 36, then, according to the property of a differentiable function, 0 is the smallest, and 36 is the largest value of the function f(t) on the segment. Then f(t), as a continuous function, it takes on the interval all values ​​from 0 to 36 inclusive, and the value 36 takes only when t=4, therefore, for 0.5 ≤ t< 4, она принимает все значения из промежутка . Мы знаем, что функция, непрерывная на некотором отрезке, достигает на нем своего минимума и максимума, то есть наибольшего m a x x ∈ a ; b f (x) и наименьшего значения m i n x ∈ a ; b f (x) . Значит, у нас получится отрезок m i n x ∈ a ; b f (x) ; m a x x ∈ a ; b f (x) , в котором и будут находиться множества значений исходной функции. Тогда все, что нам нужно сделать, – это найти на этом отрезке указанные точки минимума и максимума.

Let's take a problem in which we need to determine the range of arcsine values.

Example 1

Condition: find the range of values ​​y = a r c sin x .

Solution

In the general case, the domain of definition of the arcsine is located on the segment [ - 1 ; 1]. We need to determine the largest and smallest value of the specified function on it.

y " = a r c sin x " = 1 1 - x 2

We know that the derivative of the function will be positive for all values ​​of x located in the interval [ - 1 ; 1 ], that is, throughout the entire domain of definition, the arcsine function will increase. This means that it will take the smallest value when x is equal to - 1, and the largest value is when x is equal to 1.

m i n x ∈ - 1 ; 1 a r c sin x = a r c sin - 1 = - π 2 m a x x ∈ - 1 ; 1 a r c sin x = a r c sin 1 = π 2

Thus, the range of values ​​of the arcsine function will be equal to E (a r c sin x) = - π 2; π 2.

Answer: E (a r c sin x) = - π 2 ; π 2

Example 2

Condition: calculate the range of values ​​y = x 4 - 5 x 3 + 6 x 2 on the given interval [ 1 ; 4].

Solution

All we need to do is calculate the largest and smallest value of the function in a given interval.

To determine extremum points, the following calculations must be made:

y " = x 4 - 5 x 3 + 6 x 2 " = 4 x 3 + 15 x 2 + 12 x = x 4 x 2 - 15 x + 12 y " = 0 ⇔ x (4 x 2 - 15 x + 12 ) = 0 x 1 = 0 ∉ 1 ; 4 and 4 x 2 - 15 x + 12 = 0 D = - 15 2 - 4 · 4 · 12 = 33 x 2 = 15 - 33 8 ≈ 1 . ; 4 ; x 3 = 15 + 33 8 ≈ 2 .

Now let's find the values ​​of the given function at the ends of the segment and points x 2 = 15 - 33 8; x 3 = 15 + 33 8:

y (1) = 1 4 - 5 1 3 + 6 1 2 = 2 y 15 - 33 8 = 15 - 33 8 4 - 5 15 - 33 8 3 + 6 15 - 33 8 2 = = 117 + 165 33 512 ≈ 2. 08 y 15 + 33 8 = 15 + 33 8 4 - 5 · 15 + 33 8 3 + 6 · 15 + 33 8 2 = = 117 - 165 33 512 ≈ - 1 . 62 y (4) = 4 4 - 5 4 3 + 6 4 2 = 32

This means that the set of function values ​​will be determined by the segment 117 - 165 33 512; 32.

Answer: 117 - 165 33 512 ; 32 .

Let's move on to finding the set of values ​​of the continuous function y = f (x) in the intervals (a ; b), and a ; + ∞ , - ∞ ; b , - ∞ ; + ∞ .

Let's start by determining the largest and smallest points, as well as the intervals of increasing and decreasing on a given interval. After this, we will need to calculate one-sided limits at the ends of the interval and/or limits at infinity. In other words, we need to determine the behavior of the function under given conditions. We have all the necessary data for this.

Example 3

Condition: calculate the range of the function y = 1 x 2 - 4 on the interval (- 2 ; 2) .

Solution

Determine the largest and smallest value of a function on a given segment

y " = 1 x 2 - 4 " = - 2 x (x 2 - 4) 2 y " = 0 ⇔ - 2 x (x 2 - 4) 2 = 0 ⇔ x = 0 ∈ (- 2 ; 2)

We got a maximum value equal to 0, since it is at this point that the sign of the function changes and the graph begins to decrease. See illustration:

That is, y (0) = 1 0 2 - 4 = - 1 4 will be the maximum value of the function.

Now let’s determine the behavior of the function for an x ​​that tends to - 2 s right side and k + 2 on the left side. In other words, we find one-sided limits:

lim x → - 2 + 0 1 x 2 - 4 = lim x → - 2 + 0 1 (x - 2) (x + 2) = = 1 - 2 + 0 - 2 - 2 + 0 + 2 = - 1 4 · 1 + 0 = - ∞ lim x → 2 + 0 1 x 2 - 4 = lim x → 2 + 0 1 (x - 2) (x + 2) = = 1 2 - 0 - 2 2 - 0 + 2 = 1 4 1 - 0 = - ∞

It turns out that the function values ​​will increase from minus infinity to - 1 4 when the argument changes from - 2 to 0. And when the argument changes from 0 to 2, the function values ​​decrease towards minus infinity. Therefore, the set of values ​​of a given function on the interval we need will be (- ∞ ; - 1 4 ] .

Answer: (- ∞ ; - 1 4 ] .

Example 4

Condition: indicate the set of values ​​y = t g x on a given interval - π 2; π 2.

Solution

We know that in the general case the derivative of the tangent is - π 2; π 2 will be positive, that is, the function will increase. Now let’s determine how the function behaves within the given boundaries:

lim x → π 2 + 0 t g x = t g - π 2 + 0 = - ∞ lim x → π 2 - 0 t g x = t g π 2 - 0 = + ∞

We have obtained an increase in the values ​​of the function from minus infinity to plus infinity when the argument changes from - π 2 to π 2, and we can say that the set of solutions to this function will be the set of all real numbers.

Answer: - ∞ ; + ∞ .

Example 5

Condition: determine the range of the natural logarithm function y = ln x.

Solution

We know that this function is defined for positive values ​​of the argument D (y) = 0 ; + ∞ . The derivative on a given interval will be positive: y " = ln x " = 1 x . This means that the function increases on it. Next we need to define a one-sided limit for the case when the argument tends to 0 (on the right side) and when x goes to infinity:

lim x → 0 + 0 ln x = ln (0 + 0) = - ∞ lim x → ∞ ln x = ln + ∞ = + ∞

We found that the values ​​of the function will increase from minus infinity to plus infinity as the values ​​of x change from zero to plus infinity. This means that the set of all real numbers is the range of values ​​of the natural logarithm function.

Answer: the set of all real numbers is the range of values ​​of the natural logarithm function.

Example 6

Condition: determine the range of the function y = 9 x 2 + 1 .

Solution

This function is defined provided that x is a real number. Let us calculate the largest and smallest values ​​of the function, as well as the intervals of its increase and decrease:

y " = 9 x 2 + 1 " = - 18 x (x 2 + 1) 2 y " = 0 ⇔ x = 0 y " ≤ 0 ⇔ x ≥ 0 y " ≥ 0 ⇔ x ≤ 0

As a result, we determined that this function will decrease if x ≥ 0; increase if x ≤ 0 ; it has a maximum point y (0) = 9 0 2 + 1 = 9 with a variable equal to 0.

Let's see how the function behaves at infinity:

lim x → - ∞ 9 x 2 + 1 = 9 - ∞ 2 + 1 = 9 1 + ∞ = + 0 lim x → + ∞ 9 x 2 + 1 = 9 + ∞ 2 + 1 = 9 1 + ∞ = + 0

It is clear from the record that the function values ​​in this case will asymptotically approach 0.

To summarize: when the argument changes from minus infinity to zero, the function values ​​increase from 0 to 9. When the argument values ​​change from 0 to plus infinity, the corresponding function values ​​will decrease from 9 to 0. We have shown this in the figure:

It shows that the range of values ​​of the function will be the interval E (y) = (0 ; 9 ]

Answer: E (y) = (0 ; 9 ]

If we need to determine the set of values ​​of the function y = f (x) on the intervals [ a ; b) , (a ; b ] , [ a ; + ∞) , (- ∞ ; b ] , then we will need to carry out exactly the same studies. We will not analyze these cases for now: we will encounter them later in problems.

But what if the domain of definition of a certain function is a union of several intervals? Then we need to calculate the sets of values ​​​​on each of these intervals and combine them.

Example 7

Condition: determine what the range of values ​​will be y = x x - 2 .

Solution

Since the denominator of the function should not be turned to 0, then D (y) = - ∞; 2 ∪ 2 ; + ∞ .

Let's start by defining the set of function values ​​on the first segment - ∞; 2, which is an open beam. We know that the function on it will decrease, that is, the derivative of this function will be negative.

lim x → 2 - 0 x x - 2 = 2 - 0 2 - 0 - 2 = 2 - 0 = - ∞ lim x → - ∞ x x - 2 = lim x → - ∞ x - 2 + 2 x - 2 = lim x → - ∞ 1 + 2 x - 2 = 1 + 2 - ∞ - 2 = 1 - 0

Then, in cases where the argument changes towards minus infinity, the function values ​​will asymptotically approach 1. If the values ​​of x change from minus infinity to 2, then the values ​​will decrease from 1 to minus infinity, i.e. the function on this segment will take values ​​from the interval - ∞; 1. We exclude unity from our reasoning, since the values ​​of the function do not reach it, but only asymptotically approach it.

For open beam 2; + ∞ we perform exactly the same actions. The function on it is also decreasing:

lim x → 2 + 0 x x - 2 = 2 + 0 2 + 0 - 2 = 2 + 0 = + ∞ lim x → + ∞ x x - 2 = lim x → + ∞ x - 2 + 2 x - 2 = lim x → + ∞ 1 + 2 x - 2 = 1 + 2 + ∞ - 2 = 1 + 0

The values ​​of the function on a given segment are determined by the set 1; + ∞ . This means that the range of values ​​we need for the function specified in the condition will be the union of sets - ∞ ; 1 and 1; + ∞ .

Answer: E (y) = - ∞ ; 1 ∪ 1 ; + ∞ .

This can be seen on the graph:

A special case is periodic functions. Their range of values ​​coincides with the set of values ​​on the interval that corresponds to the period of this function.

Example 8

Condition: determine the range of values ​​of sine y = sin x.

Solution

Sine is a periodic function and its period is 2 pi. We take the segment 0; 2 π and see what the set of values ​​​​on it will be.

y " = (sin x) " = cos x y " = 0 ⇔ cos x = 0 ⇔ x = π 2 + πk , k ∈ Z

Within 0 ; 2 π the function will have extremum points π 2 and x = 3 π 2 . Let's calculate what the function values ​​will be equal to in them, as well as on the boundaries of the segment, and then choose the largest and smallest value.

y (0) = sin 0 = 0 y π 2 = sin π 2 = 1 y 3 π 2 = sin 3 π 2 = - 1 y (2 π) = sin (2 π) = 0 ⇔ min x ∈ 0 ; 2 π sin x = sin 3 π 2 = - 1 , max x ∈ 0 ; 2 π sin x = sin π 2 = 1

Answer: E (sin x) = - 1 ; 1.

If you need to know the ranges of functions such as power, exponential, logarithmic, trigonometric, inverse trigonometric, then we advise you to re-read the article on basic elementary functions. The theory we present here allows us to verify the values ​​stated there. It is advisable to learn them because they are often required when solving problems. If you know the ranges of basic functions, you can easily find the ranges of functions that are obtained from elementary ones using a geometric transformation.

Example 9

Condition: determine the range of values ​​y = 3 a r c cos x 3 + 5 π 7 - 4 .

Solution

We know that the segment from 0 to pi is the arc cosine range. In other words, E (a r c cos x) = 0; π or 0 ≤ a r c cos x ≤ π . We can get the function a r c cos x 3 + 5 π 7 from the arc cosine by shifting and stretching it along the O x axis, but such transformations will not give us anything. This means 0 ≤ a r c cos x 3 + 5 π 7 ≤ π .

The function 3 a r c cos x 3 + 5 π 7 can be obtained from the arc cosine a r c cos x 3 + 5 π 7 by stretching along the ordinate axis, i.e. 0 ≤ 3 a r c cos x 3 + 5 π 7 ≤ 3 π . The final transformation is a shift along the O y axis by 4 values. As a result, we get a double inequality:

0 - 4 ≤ 3 a r c cos x 3 + 5 π 7 - 4 ≤ 3 π - 4 ⇔ - 4 ≤ 3 arccos x 3 + 5 π 7 - 4 ≤ 3 π - 4

We found that the range of values ​​we need will be equal to E (y) = - 4; 3 π - 4 .

Answer: E (y) = - 4 ; 3 π - 4 .

We will write down another example without explanation, because it is completely similar to the previous one.

Example 10

Condition: calculate what the range of the function y = 2 2 x - 1 + 3 will be.

Solution

Let's rewrite the function specified in the condition as y = 2 · (2 ​​x - 1) - 1 2 + 3. For a power function y = x - 1 2 the range of values ​​will be defined on the interval 0; + ∞, i.e. x - 1 2 > 0 . In this case:

2 x - 1 - 1 2 > 0 ⇒ 2 (2 x - 1) - 1 2 > 0 ⇒ 2 (2 x - 1) - 1 2 + 3 > 3

So E(y) = 3; + ∞ .

Answer: E(y) = 3; + ∞ .

Now let's look at how to find the range of values ​​of a function that is not continuous. To do this, we need to divide the entire area into intervals and find sets of values ​​​​in each of them, and then combine what we get. To better understand this, we advise you to review the main types of function breakpoints.

Example 11

Condition: given the function y = 2 sin x 2 - 4 , x ≤ - 3 - 1 , - 3< x ≤ 3 1 x - 3 , x >3. Calculate its range of values.

Solution

This function is defined for all values ​​of x. Let us analyze it for continuity for argument values ​​equal to - 3 and 3:

lim x → - 3 - 0 f (x) = lim x → - 3 2 sin x 2 - 4 = 2 sin - 3 2 - 4 = - 2 sin 3 2 - 4 lim x → - 3 + 0 f (x) = lim x → - 3 (1) = - 1 ⇒ lim x → - 3 - 0 f (x) ≠ lim x → - 3 + 0 f (x)

We have an irremovable discontinuity of the first kind with the value of the argument - 3. As we approach it, the values ​​of the function tend to - 2 sin 3 2 - 4 , and as x tends to - 3 on the right side, the values ​​will tend to - 1 .

lim x → 3 - 0 f (x) = lim x → 3 - 0 (- 1) = 1 lim x → 3 + 0 f (x) = lim x → 3 + 0 1 x - 3 = + ∞

We have an irremovable discontinuity of the second kind at point 3. When a function tends to it, its values ​​approach - 1, when tending to the same point on the right - to minus infinity.

This means that the entire domain of definition of this function is divided into 3 intervals (- ∞ ; - 3 ], (- 3 ; 3 ], (3 ; + ∞).

In the first of them, we got the function y = 2 sin x 2 - 4. Since - 1 ≤ sin x ≤ 1, we get:

1 ≤ sin x 2< 1 ⇒ - 2 ≤ 2 sin x 2 ≤ 2 ⇒ - 6 ≤ 2 sin x 2 - 4 ≤ - 2

This means that on a given interval (- ∞ ; - 3 ] the set of function values ​​is [ - 6 ; 2 ] .

On the half-interval (- 3 ; 3 ] it turned out constant function y = - 1 . Consequently, the entire set of its values ​​in in this case will be reduced to one number - 1.

At the second interval 3 ; + ∞ we have the function y = 1 x - 3 . It is decreasing because y " = - 1 (x - 3) 2< 0 . Она будет убывать от плюс бесконечности до 0 , но самого 0 не достигнет, потому что:

lim x → 3 + 0 1 x - 3 = 1 3 + 0 - 3 = 1 + 0 = + ∞ lim x → + ∞ 1 x - 3 = 1 + ∞ - 3 = 1 + ∞ + 0

This means that the set of values ​​of the original function for x > 3 is the set 0; + ∞ . Now let's combine the results: E (y) = - 6 ; - 2 ∪ - 1 ∪ 0 ; + ∞ .

Answer: E (y) = - 6 ; - 2 ∪ - 1 ∪ 0 ; + ∞ .

The solution is shown in the graph:

Example 12

Condition: there is a function y = x 2 - 3 e x. Determine the set of its values.

Solution

It is defined for all argument values ​​representing real numbers. Let us determine in which intervals this function will increase and in which it will decrease:

y " = x 2 - 3 e x " = 2 x e x - e x (x 2 - 3) e 2 x = - x 2 + 2 x + 3 e x = - (x + 1) (x - 3) e x

We know that the derivative will become 0 if x = - 1 and x = 3. Let's place these two points on the axis and find out what signs the derivative will have on the resulting intervals.

The function will decrease by (- ∞ ; - 1 ] ∪ [ 3 ; + ∞) and increase by [ - 1 ; 3]. The minimum point will be - 1, the maximum - 3.

Now let's find the corresponding function values:

y (- 1) = - 1 2 - 3 e - 1 = - 2 e y (3) = 3 2 - 3 e 3 = 6 e - 3

Let's look at the behavior of the function at infinity:

lim x → - ∞ x 2 - 3 e x = - ∞ 2 - 3 e - ∞ = + ∞ + 0 = + ∞ lim x → + ∞ x 2 - 3 e x = + ∞ 2 - 3 e + ∞ = + ∞ + ∞ = = lim x → + ∞ x 2 - 3 " e x " = lim x → + ∞ 2 x e x = + ∞ + ∞ = = lim x → + ∞ 2 x " (e x) " = 2 lim x → + ∞ 1 e x = 2 1 + ∞ = + 0

L'Hopital's rule was used to calculate the second limit. Let's depict the progress of our solution on a graph.

It shows that the function values ​​will decrease from plus infinity to - 2 e when the argument changes from minus infinity to - 1. If it changes from 3 to plus infinity, then the values ​​will decrease from 6 e - 3 to 0, but 0 will not be reached.

Thus, E(y) = [ - 2 e ; + ∞) .

Answer: E(y) = [ - 2 e ; + ∞)

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Lecture 19. Function. Domain and set of values ​​of a function.

Function is one of the most important mathematical concepts.

Definition: If each number from a certain set x is associated singular y, then we say that the function y(x) is given on this set. In this case, x is called the independent variable or argument, and y is called the dependent variable or the value of a function or simply a function.

The variable y is also said to be a function of the variable x.

Having designated a match by a letter, for example f, it is convenient to write: y=f (x), that is, the value y is obtained from the argument x using the match f. (Read: y equals f of x.) The symbol f (x) denotes the value of the function corresponding to the value of the argument equal to x.

Example 1 Let the function be given by the formula y=2x 2 –6. Then we can write that f(x)=2x 2 –6. Let's find the values ​​of the function for values ​​of x equal to, for example, 1; 2.5;–3; i.e., we find f(1), f(2.5), f(–3):

f(1)=2 1 2 –6=–4;
f(2.5)=2 2.5 2 –6=6.5;
f(–3)=2 (–3) 2 –6= 12.

Note that in the notation of the form y=f (x) other letters are used instead of f: g, etc.

Definition: The domain of a function is all the values ​​of x for which the function exists.

If a function is specified by a formula and its domain of definition is not specified, then the domain of definition of the function is considered to consist of all values ​​of the argument for which the formula makes sense.

In other words, the domain of definition of a function, given formula, is all the values ​​of the argument except those that result in actions we cannot perform. On at the moment we know only two such actions. We can't divide by zero and we can't extract square root from a negative number.

Definition: All values ​​that the dependent variable takes form the range of the function.

The domain of definition of a function describing a real process depends on the specific conditions of its occurrence. For example, the dependence of the length l of an iron rod on the heating temperature t is expressed by the formula, where l 0 is the initial length of the rod, and is the coefficient linear expansion. This formula makes sense for any values ​​of t. However, the domain of definition of the function l=g(t) is an interval of several tens of degrees, for which the law of linear expansion is valid.

Example.

Specify the function range y = arcsinx.

Solution.

The domain of definition of the arcsine is the segment [-1; 1] . Let's find the largest and smallest value of the function on this segment.

The derivative is positive for everyone x from the interval (-1; 1) , that is, the arcsine function increases over the entire domain of definition. Therefore, it takes the smallest value when x = -1, and the greatest at x = 1.

We have obtained the range of arcsine function .

Find the set of function values on the segment .

Solution.

Let's find the largest and smallest value of the function on a given segment.

Let us determine the extremum points belonging to the segment :