Function. The domain of definition and the range of values of a function. Function graphs. Solving typical problems What does a set of function values mean?
Function y=f(x) is such a dependence of the variable y on the variable x, when each valid value of the variable x corresponds to a single value of the variable y.
Function definition domain D(f) is the set of all possible values of the variable x.
Function Range E(f) is the set of all admissible values of the variable y.
Graph of a function y=f(x) is a set of points on the plane whose coordinates satisfy a given functional dependence, that is, points of the form M (x; f(x)). The graph of a function is a certain line on a plane.
If b=0 , then the function will take the form y=kx and will be called direct proportionality.
D(f) : x \in R;\enspace E(f) : y \in R
The graph of a linear function is a straight line.
The slope k of the straight line y=kx+b is calculated using the following formula:
k= tan \alpha, where \alpha is the angle of inclination of the straight line to the positive direction of the Ox axis.
1) The function increases monotonically for k > 0.
For example: y=x+1
2) The function decreases monotonically as k< 0 .
For example: y=-x+1
3) If k=0, then giving b arbitrary values, we obtain a family of straight lines parallel to the Ox axis.
For example: y=-1
Inverse proportionality
Inverse proportionality called a function of the form y=\frac (k)(x), where k is a non-zero real number
D(f) : x \in \left \( R/x \neq 0 \right \); \: E(f) : y \in \left \(R/y \neq 0 \right \).
Function graph y=\frac (k)(x) is a hyperbole.
1) If k > 0, then the graph of the function will be located in the first and third quarters of the coordinate plane.
For example: y=\frac(1)(x)
2) If k< 0 , то график функции будет располагаться во второй и четвертой координатной плоскости.
For example: y=-\frac(1)(x)
Power function
Power function is a function of the form y=x^n, where n is a non-zero real number
1) If n=2, then y=x^2. D(f) : x \in R; \: E(f) : y \in; main period of the function T=2 \pi
Often, as part of solving problems, we have to look for many values of a function on a domain of definition or a segment. For example, this should be done when solving different types inequalities, evaluations of expressions, etc.
In this material, we will tell you what the range of values of a function is, give the main methods by which it can be calculated, and analyze problems of varying degrees of complexity. For clarity, individual provisions are illustrated with graphs. After reading this article, you will get a comprehensive understanding of the range of a function.
Let's start with basic definitions.
Definition 1
The set of values of the function y = f (x) on some interval x is the set of all values that this function takes when iterating over all values x ∈ X .
Definition 2
The range of values of a function y = f (x) is the set of all its values that it can take when searching through the values of x from the range x ∈ (f).
The range of values of a certain function is usually denoted by E (f).
Please note that the concept of the set of values of a function is not always identical to its range of values. These concepts will be equivalent only if the interval of values of x when finding a set of values coincides with the domain of definition of the function.
It is also important to distinguish between the range of values and the range of acceptable values of the variable x for the expression on the right side y = f (x). The range of permissible values x for the expression f (x) will be the domain of definition of this function.
Below is an illustration showing some examples. Blue lines are function graphs, red lines are asymptotes, red points and lines on the ordinate axis are function ranges.
Obviously, the range of values of a function can be obtained by projecting the graph of the function onto the O y axis. Moreover, it can represent either a single number or a set of numbers, a segment, an interval, an open ray, a union of numerical intervals, etc.
Let's look at the main ways to find the range of values of a function.
Let's start by defining the set of values of the continuous function y = f (x) on a certain segment denoted [ a ; b ] . We know that a function that is continuous on a certain segment reaches its minimum and maximum on it, that is, the largest m a x x ∈ a ; b f (x) and the smallest value m i n x ∈ a ; b f (x) . This means that we get a segment m i n x ∈ a ; bf(x); m a x x ∈ a ; b f (x) , which will contain the sets of values of the original function. Then all we need to do is to find the indicated minimum and maximum points on this segment.
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 l and 4 x 2 - 15 x + 12 = 0 D = - 15 2 - 4 4 12 = 33 x 2 = 15 - 33 8 ≈ 1. 16 ∈ 1 ; 4 ; x 3 = 15 + 33 8 ≈ 2 . 59 ∈ 1 ; 4
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. Consequently, 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 considerations, 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. 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 with values of the argument 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 when the value of the argument is - 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 ], the result is a constant function y = - 1. Consequently, the entire set of its values 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 that are 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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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 equation y=f(x) . The range of permissible values of the variable x for the expression f(x) is the domain of definition of the function y=f(x) .
The figure shows several examples.
Graphs of functions 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 with the simplest case: we will show how to determine the set of values of a 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 largest and smallest value of the function 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 the natural logarithm function y = lnx.
Solution.
The natural logarithm function is defined for positive values of the 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 all real values of 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 the 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. The range of values of 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 
. The range of values of the power function is the 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 remembering 3 on the left, the values of the function tend to minus one, and as x tends to 3 on the right, the values of the function tend to plus infinity.
Thus, we divide the domain of definition of the function into three intervals.
On the interval we have the function 
. Since then 
Thus, the set of values of the original function on the interval is [-6;2] .
On the half-interval we have a constant function y = -1. That is, the set of values of the original function on the interval consists of a single element .
The function is defined for all valid argument values. Let us find out the intervals of increase and decrease of the function.
The derivative vanishes at x=-1 and x=3. Let's mark these points on the number line and determine the signs of the derivative on the resulting intervals. 
The function decreases by 
, increases by [-1; 3] , x=-1 minimum point, x=3 maximum point.
Let's calculate the corresponding minimum and maximum of the function: 
Let's check the behavior of the function at infinity: 
The second limit was calculated using .
Let's make a schematic drawing.
When the argument changes from minus infinity to -1, the function values decrease from plus infinity to -2e, when the argument changes from -1 to 3, the function values increase from -2e to, when the argument changes from 3 to plus infinity, the function values decrease from to zero, but they don't reach zero.
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 consideration of this topic, theoretical material was studied, methods for solving problems of finding sets of function values were considered, didactic material was selected for independent work students.
This article can be used by a teacher in preparing students for final and entrance exams, when studying the topic “Value domain 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.
| Function | Multiple meanings |
| y = kx+ b | E(y) = (-∞;+∞) |
| y = x2n | E(y) = |
| y = cos x | E(y) = [-1;1] |
| y = tan x | E(y) = (-∞;+∞) |
| y = ctg x | E(y) = (-∞;+∞) |
| y = arcsin x | E(y) = [-π/2 ; π/2] |
| y = arcos x | E(y) = |
| y = arctan x | E(y) = (-π/2 ; π/2) |
| y = arcctg x | E(y) = (0; π) |
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. In this case, the simplest and most concise solution to the problem of finding the set of values of a function is achieved based on property 1, if it is possible to determine the monotonicity of the function using simple methods. 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) to transform 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 by sequentially finding the 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;+∞) of the 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 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 continuous functions cos7x and cosx, the function y is continuous on the entire 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 her. Hence, E(y)= [-6;6].
Example 3: Find the range E(f) functions f(x)= cos2x + 2cosx.
Using the double angle cosine formula, we 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 find graphically. 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 a 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 a unique one (by the property of a continuous monotonic function), for 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.
As a direct development of the method of introducing a parameter when finding a set of function values, we can consider the method of the inverse function, to find which it is necessary to solve the equation for x f(x)=y, considering y to be a parameter. If this equation has only decision 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.
let's find the 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 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 values of the function 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 that 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
E( y) = (– ∞, + ∞)
E( y) = (– ∞, + ∞)
E( y) = (– ∞, + ∞)
E( y) = (0, + ∞)
Using this knowledge, can we immediately find the sets of values of the functions written on the board? (see table 2).
What can help answer this question? (Graphs of these functions).
How to graph the first function? (Lower the parabola 4 units down).
|
Function |
Multiple meanings |
|
y = x 2 – 4 |
E( y) = [-4, + ∞) |
|
y = + 5 |
E( y) = |
|
y = – 5 cos x |
E( y) = [- 5, 5] |
|
y = tg ( x+ / 6) – 1 |
E( y) = (– ∞, + ∞) |
|
y = sin( x+ / 3) – 2 |
E( y) = [- 3, - 1] |
|
y =| x – 1 | + 3 |
E( y) = |
|
y =| ctg x| |
E( y) = |
|
y =  = | cos(x + /4) | |
E( y) = |
|
y =(x – 5) 2 + 3 |
E( y) = . Find the set of function values:   . 
Introduction of an algorithm for solving problems of finding a set of values of trigonometric functions. Let's see how we can apply our existing experience to the various tasks included in the Unified Exam options. 1. Finding the values of functions for a given argument value. Example. Find the value of the function y = 2 cos(π/2+ π/4 ) – 1, If x = -π/2. Solution. y(-π/2) = 2 cos(- π/2 – π/4 )- 1= 2 cos(π/2 + π/4 )- 1 = - 2 sinπ/4 – 1 = - 2  – 1 =
= – 2. Finding the range of values of trigonometric functions
1≤ sinX≤ 1 2 ≤ 2 sinX≤ 2 9 ≤ 11+2sinX≤ 13 3 ≤ Let's write down the integer values of the function on the interval. This is the number 3. Answer: 3.
at= sin 2 X- 2 3 sinx + 3 2 - 3 2 + 8, at= (sinX- 3) 2 -1. E ( sinX) = [-1;1]; E ( sinX -3) = [-4;-2]; E ( sinX -3) 2 = ; E ( at) = . Answer: .
Can we find the set of values of this function? (No.) What should be done? (Reduce to one function.) How to do it? (Use cos 2 formula x= 1-sin 2 x.) So, at= 1-sin 2 x+ 2sin x –2, y= -sin 2 x+ 2sin x –1, at= -(sin x –1) 2 . Well, now we can find a set of values and choose the smallest one. 1 ≤ sin x ≤ 1, 2 ≤ sin x – 1 ≤ 0, 0 ≤ (sin x – 1) 2 ≤ 4, 4 ≤ -(sin x -1) 2 ≤ 0. This means that the smallest value of the function at name= –4. Answer: -4.
Solution. at= 1-cos 2 x+cos x + 1,5, at= -cos 2 x+ 2∙0.5∙cos x - 0,25 + 2,75, at= -(cos x- 0,5) 2 + 2,75. E(cos x) = [-1;1], E(cos x – 0,5) = [-1,5;0,5], E(cos x – 0,5) 2 = , E(-(cos x-0,5) 2) = [-2,25;0], E( at) = . Largest function value at naib= 2.75; smallest value at name= 0.5. Let's find the product of the largest and smallest value of the function: at naib ∙at name = 0,5∙2,75 = 1,375. Answer: 1.375. Solution. Let's rewrite the function in the form at =, at = Let us now find the set of values of the function. E(sin x) = [-1, 1], E(6sin x) = [-6, 6], E(6sin x + 1) = [-5, 7], E((6sin x + 1) 2) = , E(– (6sin x + 1) 2) = [-49, 0], E(– (6sin x + 1) 2 + 64) = , E( y) = [ Let's find the sum of the integer values of the function: 4 + 5 + 6 + 7 + 8 = 30. Answer: 30.  Solution. 1) 2) Therefore 2 X belong to the second quarter. 3) In the second quarter, the sine function decreases and is continuous. This means that this function 4) Let's calculate these values:
Answer :
 Solution. 1) Since a sine takes values from -1 to 1, then the set of difference values 2) Arc cosine is a monotonically decreasing and continuous function. This means that the set of values of the expression is a segment 3) When multiplying this segment by Answer:  Solution. Since arctangent is an increasing function, then 2) When increasing X from 3) When increasing from 4) Using the formula expressing the sine through the tangent of a half angle, we find that
This means that the desired set of values is the union of segments Answer: at= a sin x + b cos x or at= a sin (Rx) + b cos (Rx).
Solution. Let's find the value Let's transform the expression 15 sin 2x + 20 cos 2x = 25 ( 25 sin (2x + The set of function values y = sin (2x + Then the set of values of the original function is -25 Answer:
[-25; 25].
Function at= сtg X is decreasing on the interval [π/4; π/2], therefore, the function will take the smallest value when x =π/2, that is at(π/2) = сtg π/2 = 0; and the greatest value is at x=π/4, that is at(π/4) = сtg π/4 = 1. Answer: 1, 0.  . Solution. Let us select in equality It follows that the graph of the function f(x) is either a hyperbola (a≠ 0) or a straight line without a point. Moreover, if a; 2a) and (2a; If a = 0, then f(x) = -2 throughout the entire domain of definition x ≠ 0. Therefore, it is obvious that the required values of the parameter are not equal to zero. Since we are interested in the function values only on the interval [-1; 1], then the classification of situations is determined by the fact that the asymptote x = 2a of the hyperbola (a≠0) is located relative to this segment. Case 1. All points in the interval [-1; 1] are to the right of the vertical asymptote x = 2a, that is, when 2a Case 2. The vertical asymptote crosses the interval [-1; 1], and the function decreases (as in case 1), that is, when Case 3. The vertical asymptote crosses the interval [-1; 1] and the function increases, that is -1 Case 4. All points in the interval [-1; 1] are to the left of the vertical asymptote, that is, 1 a > . and second Reception 5. Simplification of the formula defining a fractional-rational function Reception 6. Finding a set of values of quadratic functions (by finding the vertex of a parabola and establishing the nature of the behavior of its branches). Reception 7. Introduction of an auxiliary angle for finding the set of values of some trigonometric functions. Share with friends or save for yourself:
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– 1.
, i.e. E (y) = .
,
, 8].
that is X belongs to the first quarter.
before 
.
. When multiplied by 
this segment will go into the segment 
.
.
we get 
.
.
.
before 
argument 2 X increases from 
before 
. Since the sine increases over such an interval, the function 
to 1.
argument 2 X increases from 
. Since the sine decreases on such an interval, then the function 
to 1.
.
And 
, that is, the segment 
= 
= 25.
) = 25 () =
), where cos 
sin (2x + 
) and, if a > 0, increases monotonically on these rays.
.