Practical work in mathematics section: “Functions, their properties and graphs” topic: Functions. Domain and set of values of a function
MINISTRY OF EDUCATION OF THE SAKHALIN REGION
GBPOU "CONSTRUCTION TECHNIQUE"
Practical work
In the discipline "Mathematics"
Chapter: " Functions, their properties and graphs.”
Subject: Functions. Domain and set of values of a function. Even and odd functions.
(didactic material)
Compiled by:
Teacher
Kazantseva N.A.
Yuzhno-Sakhalinsk-2017
Practical work in mathematicsby section« and methodologicalinstructions for their implementation are intended for studentsGBPOU "Sakhalin Construction College"
Compiled by : Kazantseva N. A., mathematics teacher
The material contains practical work in mathematics« Functions, their properties and graphs" And instructions for their implementation. The guidelines are compiled in accordance with the work program in mathematics and are intended for students of the Sakhalin Construction College, students studying general education programs.
1) Practical lesson No. 1. Functions. Domain of definition and set of values of a function.……………………………………………………………...4
2)Practical lesson No. 2 . Even and odd functions……………….6
Practical lesson No. 1
Functions. Domain and set of values of a function.
Goals: consolidate skills and problem solving skills on the topic: “The domain of definition and the set of values of a function.
Equipment:
Note. First, you should repeat the theoretical material on the topic: “The domain of definition and the set of values of a function,” after which you can begin to carry out the practical part.
Guidelines:
Definition: Function Domain– this is the set of all values of the argument x on which the function is specified (or the set of x for which the function makes sense).
Designation:D(y),D( f)- domain of definition of a function.
Rule: To find oblastiTo determine a function from a graph, it is necessary to design the graph on OX.
Definition:Function Rangeis the set of y for which the function makes sense.
Designation: E(y), E(f)- the range of the function.
Rule: To find oblastifunction values according to the graph, the graph must be projected onto the op-amp.
№ 1.Find the function values:
a) f(x) = 4 x+ at points 2;20 ;
b) f(x) = 2 · cos(x) at points; 0;
V) f(x) = at points 1;0; 2;
G) f(x) = 6 sin 4 x at points; 0;
e) f(x) = 2 9 x+ 10 at points 2; 0; 5.
№ 2. Find the domain of the function:
a) f(x) = ; b ) f(x) = ; V ) f(x) = ;
G) f(x) = ; d) f(x) = ; e) f (x) = 6 x +1;
and) f(x) = ; h) f(x) = .
№3. Find the range of the function:
A) f(x) = 2+3 x; b) f(x) = 2 7 x + 3.
№ 4. Find the domain of definition and the domain of value of the function whose graph is shown in the figure:
Practical lesson No. 2
Even and odd functions.
Goals: consolidate skills and problem solving skills on the topic: “Even and odd functions.”
Equipment: notebook for practical work, pen, guidelines for completing the work
Note. First, you should repeat the theoretical material on the topic: “Even and odd functions”, after which you can begin to perform the practical part.
Do not forget about the correct formatting of the solution.
Guidelines:
The most important properties of functions include evenness and oddness.
Definition: The function is calledodd changes its meaning to its opposite,
those. f (x)= f (x).
The graph of an odd function is symmetrical about the origin (0;0).
Examples : odd functions are y=x, y=, y= sin x and others
For example, the y= graph is indeed symmetrical about the origin (see Fig. 1):
Fig.1. G graph y= (cubic parabola)
Definition: The function is calledeven , if when changing the sign of the argument, itdoesn't change its meaning, i.e. f (x)= f (x).
The graph of an even function is symmetrical about the op-amp axis.
Examples : even functions are functions y=, y= ,
y= cosx and etc.
For example, let’s show the symmetry of the graph y= relative to the op-amp axis:
Fig.2. Graph =
Tasks for practical work:
№1. Investigate the function for even or odd analytically:
1) f (x) = 2 x 3 – 3; 2) f (x) = 5 x 2 + 3;
3) g (x) = – +; 4) g (x) = –2 x 3 + 3;
5) y(x)= 7xc tgx; 6) y(x)= + cosx;
7) t(x)= tgx 3; 8) t(x)= + sinx.
№2. Investigate the function for even or odd analytically:
1) f (x) = ; 2) f (x) = 6 + · sin 2 x· cosx;
3) f (x) = ; 4) f (x) = 2 + · cos 2 x· sinx;
5) f (x) = ; 6) f (x) = 3 + · sin 4 x· cosx;
7) f (x) = ; 8) f (x) = 3 + · cos 4 x· sinx.
№3. Examine the function for even or odd according to the graph:
№4. Check whether the function is even or odd?
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 needs to be done when solving various types of inequalities, evaluating expressions, etc.
Yandex.RTB R-A-339285-1
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 a function y = f (x) on a certain 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 on the right side and + 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 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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Definition: A numerical function is a correspondence that associates each number x from some given set with a single number y.
Designation:
where x is the independent variable (argument), y is the dependent variable (function). The set of values of x is called the domain of the function (denoted D(f)). The set of values of y is called the range of values of the function (denoted E(f)). The graph of a function is the set of points in the plane with coordinates (x, f(x))
Methods for specifying a function.
- analytical method (using a mathematical formula);
- tabular method (using a table);
- descriptive method (using verbal description);
- graphical method (using a graph).
Basic properties of the function.
1. Even and odd
A function is called even if
– the domain of definition of the function is symmetrical about zero
f(-x) = f(x)
The graph of an even function is symmetrical about the axis 0y
A function is called odd if
– the domain of definition of the function is symmetrical about zero
– for any x from the domain of definition f(-x) = –f(x)
The graph of an odd function is symmetrical about the origin.
2. Frequency
A function f(x) is called periodic with period if for any x from the domain of definition f(x) = f(x+T) = f(x-T) .
The graph of a periodic function consists of unlimitedly repeating identical fragments.
3. Monotony (increasing, decreasing)
The function f(x) is increasing on the set P if for any x 1 and x 2 from this set such that x 1
The function f(x) decreases on the set P if for any x 1 and x 2 from this set, such that x 1 f(x 2) .
4. Extremes
The point X max is called the maximum point of the function f(x) if for all x from some neighborhood of X max the inequality f(x) f(X max) is satisfied.
The value Y max =f(X max) is called the maximum of this function.
X max – maximum point
At max - maximum
A point X min is called a minimum point of the function f(x) if for all x from some neighborhood of X min, the inequality f(x) f(X min) is satisfied.
The value Y min =f(X min) is called the minimum of this function.
X min – minimum point
Y min – minimum
X min , X max – extremum points
Y min , Y max – extrema.
5. Zeros of the function
The zero of a function y = f(x) is the value of the argument x at which the function becomes zero: f(x) = 0.
X 1, X 2, X 3 – zeros of the function y = f(x).
Tasks and tests on the topic "Basic properties of a function"
- Function Properties - Numerical functions 9th grade
Lessons: 2 Assignments: 11 Tests: 1
- Properties of logarithms - Exponential and logarithmic functions grade 11
Lessons: 2 Assignments: 14 Tests: 1
- Square root function, its properties and graph - Square root function. Properties of square root grade 8
Lessons: 1 Assignments: 9 Tests: 1
- Power functions, their properties and graphs - Degrees and roots. Power functions grade 11
Lessons: 4 Assignments: 14 Tests: 1
- Functions - Important topics for reviewing the Unified State Examination in mathematics
Tasks: 24
Having studied this topic, you should be able to find the domain of definition of various functions, determine the monotonicity intervals of a function using graphs, and examine functions for evenness and oddness. Let's consider solving similar problems using the following examples.
Examples.
1. Find the domain of definition of the function.
Solution: the domain of definition of the function is found from the condition
1) Function domain and function range.
The domain of a function is the set of all valid valid argument values x(variable x), for which the function y = f(x) determined. The range of a function is the set of all real values y, which the function accepts.
In elementary mathematics, functions are studied only on the set of real numbers.
2) Function zeros.
Function zero is the value of the argument at which the value of the function is equal to zero.
3) Intervals of constant sign of a function.
Intervals of constant sign of a function are sets of argument values on which the function values are only positive or only negative.
4) Monotonicity of the function.
An increasing function (in a certain interval) is a function in which a larger value of the argument from this interval corresponds to a larger value of the function.
A decreasing function (in a certain interval) is a function in which a larger value of the argument from this interval corresponds to a smaller value of the function.
5) Even (odd) function.
An even function is a function whose domain of definition is symmetrical with respect to the origin and for any X from the domain of definition the equality f(-x) = f(x). The graph of an even function is symmetrical about the ordinate.
An odd function is a function whose domain of definition is symmetrical with respect to the origin and for any X from the domain of definition the equality is true f(-x) = - f(x). The graph of an odd function is symmetrical about the origin.
6) Limited and unlimited functions.
A function is called bounded if there is a positive number M such that |f(x)| ≤ M for all values of x. If such a number does not exist, then the function is unlimited.
7) Periodicity of the function.
A function f(x) is periodic if there is a non-zero number T such that for any x from the domain of definition of the function the following holds: f(x+T) = f(x). This smallest number is called the period of the function. All trigonometric functions are periodic. (Trigonometric formulas).
19. Basic elementary functions, their properties and graphs. Application of functions in economics.
Basic elementary functions. Their properties and graphs
1. Linear function.
Linear function is called a function of the form , where x is a variable, a and b are real numbers.
Number A called the slope of the line, it is equal to the tangent of the angle of inclination of this line to the positive direction of the x-axis. The graph of a linear function is a straight line. It is defined by two points.
Properties of a Linear Function
1. Domain of definition - the set of all real numbers: D(y)=R
2. The set of values is the set of all real numbers: E(y)=R
3. The function takes a zero value when or.
4. The function increases (decreases) over the entire domain of definition.
5. A linear function is continuous over the entire domain of definition, differentiable and .
2. Quadratic function.
A function of the form, where x is a variable, coefficients a, b, c are real numbers, is called quadratic
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