Find out if the function is even or odd. Even and odd functions
Evenness and oddness of a function are one of its main properties, and evenness occupies an impressive part of the school mathematics course. It largely determines the nature of the behavior of the function and greatly facilitates the construction of the corresponding graph.
Let us define the parity of the function. Generally speaking, the function under study is considered even if for opposite values of the independent variable (x) located in its domain of definition, the corresponding values of y (function) turn out to be equal.
Let us give a more rigorous definition. Consider some function f (x), which is given in the domain D. It will be even if for any point x located in the domain of definition:
- -x (opposite point) is also in this scope,
- f (-x) = f (x).
The above definition implies a condition necessary for the domain of definition of such a function, namely, symmetry with respect to the point O, which is the origin, since if some point b is contained in the domain of an even function, then the corresponding point - b also lies in this domain. Thus, the conclusion follows from the above: the even function has a form symmetric with respect to the ordinate axis (Oy).
How to determine the parity of a function in practice?
Let it be given using the formula h (x) = 11 ^ x + 11 ^ (- x). Following the algorithm that follows directly from the definition, we first investigate its domain of definition. Obviously, it is defined for all values of the argument, that is, the first condition is satisfied.
The next step is to substitute its opposite value (-x) for argument (x).
We get:
h (-x) = 11 ^ (- x) + 11 ^ x.
Since addition satisfies the commutative (transposable) law, it is obvious that h (-x) = h (x) and the given functional dependence is even.
Let us check the evenness of the function h (x) = 11 ^ x-11 ^ (- x). Following the same algorithm, we get that h (-x) = 11 ^ (- x) -11 ^ x. Taking out the minus, in the end, we have
h (-x) = - (11 ^ x-11 ^ (- x)) = - h (x). Therefore, h (x) is odd.
By the way, it should be recalled that there are functions that cannot be classified according to these criteria, they are called neither even nor odd.
Even functions have a number of interesting properties:
- as a result of the addition of such functions, an even one is obtained;
- as a result of the subtraction of such functions, an even one is obtained;
- even, also even;
- as a result of multiplication of two such functions, an even one is obtained;
- as a result of multiplying the odd and even functions, an odd one is obtained;
- as a result of dividing the odd and even functions, an odd one is obtained;
- the derivative of such a function is odd;
- if we square an odd function, we get an even one.
The parity function can be used when solving equations.
To solve an equation of the type g (x) = 0, where the left-hand side of the equation is an even function, it will be enough to find its solution for non-negative values of the variable. The resulting roots of the equation must be combined with opposite numbers. One of them is subject to verification.
This is also successfully used to solve non-standard problems with a parameter.
For example, is there any value for the parameter a for which the equation 2x ^ 6-x ^ 4-ax ^ 2 = 1 will have three roots?
If we take into account that the variable enters the equation in even powers, then it is clear that replacing x with - x given equation will not change. It follows that if some number is its root, then the opposite number is also the same. The conclusion is obvious: the nonzero roots of the equation are included in the set of its solutions in “pairs”.
It is clear that the number 0 itself is not, that is, the number of roots of such an equation can only be even and, naturally, at no value of the parameter it cannot have three roots.
But the number of roots of the equation 2 ^ x + 2 ^ (- x) = ax ^ 4 + 2x ^ 2 + 2 can be odd, and for any value of the parameter. Indeed, it is easy to check that the set of roots this equation contains solutions in pairs. Let's check if 0 is a root. When substituting it into the equation, we get 2 = 2. Thus, besides the "paired" ones, 0 is also a root, which proves their odd number.
Back forward
Attention! Slide previews are for informational purposes only and may not represent all the presentation options. If you are interested in this work, please download the full version.
Goals:
- to form the concept of evenness and oddness of a function, to teach the ability to define and use these properties when exploration of functions, charting;
- develop the creative activity of students, logical thinking, the ability to compare, generalize;
- to educate hard work, mathematical culture; develop communication skills .
Equipment: multimedia installation, interactive whiteboard, handouts.
Forms of work: frontal and group with elements of search and research activities.
Information sources:
1.Algebra9class A.G. Mordkovich. Textbook.
2.Algebra grade 9 A.G. Mordkovich. Problem book.
3.Algebra grade 9. Assignments for student learning and development. Belenkova E.Yu. Lebedintseva E.A.
DURING THE CLASSES
1. Organizational moment
Setting the goals and objectives of the lesson.
2. Homework check
No. 10.17 (Problem book 9kl. A. G. Mordkovich).
a) at = f(X), f(X) = 
b) f (–2) = –3; f (0) = –1; f(5) = 69;
c) 1.D ( f) = [– 2; + ∞)
2. E ( f) = [– 3; + ∞)
3. f(X) = 0 for X ~ 0,4
4. f(X)> 0 for X > 0,4 ; f(X)
< 0 при – 2 <
X <
0,4.
5. The function increases with X € [– 2; + ∞)
6. The function is limited from below.
7. at naim = - 3, at naib does not exist
8. The function is continuous.
(Did you use the function research algorithm?) Slide.
2. Let's check the table that you were asked on the slide.
| Fill the table | |||||
Domain |
Function zeros |
Intervals of constancy |
Coordinates of points of intersection of the graph with Oy | ||
 |
x = –5, |
х € (–5; 3) U |
х € (–∞; –5) U |
||
 |
x ∞ –5, |
х € (–5; 3) U |
х € (–∞; –5) U |
||
x ≠ –5, |
х € (–∞; –5) U |
x € (–5; 2) |
|||
3. Knowledge update
- Given functions.
- Specify the scope for each function.
- Compare the value of each function for each pair of argument values: 1 and - 1; 2 and - 2.
- For which of these functions in the domain of definition the equalities are fulfilled f(– X)
= f(X), f(– X) = – f(X)? (enter the obtained data into the table) Slide
| f(1) and f(– 1) | f(2) and f(– 2) | charts | f(– X) = –f(X) | f(– X) = f(X) | ||
| 1. f(X) = | ||||||
| 2. f(X) = X 3 | ||||||
| 3. f(X) = | X | | ||||||
| 4.f(X) = 2X – 3 | ||||||
| 5. f(X) = | X ≠ 0 |
|||||
| 6. f(X)= | X > –1 | and not defined. |
4. New material
- Performing this work, guys, we have identified another property of a function that is unfamiliar to you, but no less important than the others - this is the even and odd function. Write down the topic of the lesson: "Even and odd functions", our task is to learn how to determine the evenness and oddness of a function, to find out the significance of this property in the study of functions and plotting.
So, let's find the definitions in the textbook and read (p. 110) ... Slide
Def. one Function at = f (X) given on the set X is called even if for any value XЄ X is executed equality f (–x) = f (x). Give examples.
Def. 2 Function y = f (x) given on the set X is called odd if for any value XЄ X the equality f (–x) = –f (x) holds. Give examples.
Where have we encountered the terms "even" and "odd"?
Which of these functions do you think will be even? Why? What are odd? Why?
For any function of the form at= x n, where n- an integer it can be argued that the function is odd for n- odd and the function is even for n- even.
- View functions at= and at = 2X- 3 are neither even nor odd, since equalities are not satisfied f(– X) = – f(X), f(–
X) = f(X)
The study of the question of whether a function is even or odd is called the study of a function for parity. Slide
Definitions 1 and 2 dealt with the values of the function for x and - x, thus it is assumed that the function is also defined for the value X, and at - X.
Def 3. If a numerical set, together with each of its elements x, also contains the opposite element -x, then the set X called a symmetric set.
Examples:
(–2; 2), [–5; 5]; (∞; ∞) are symmetric sets, and [–5; 4] are asymmetric.
- Is the domain of definition of even functions a symmetric set? The odd ones?
- If D ( f) Is an asymmetric set, then what function?
- Thus, if the function at = f(X) Is even or odd, then its domain of definition is D ( f) Is a symmetric set. Is the converse true, if the domain of a function is a symmetric set, then it is even or odd?
- So the presence of a symmetric set of domains of definition is a necessary condition, but not sufficient.
- So how do you investigate a function for parity? Let's try to compose an algorithm.
Slide
Algorithm for analyzing a function for parity
1. Determine whether the function domain is symmetric. If not, then the function is neither even nor odd. If yes, then go to step 2 of the algorithm.
2. Write an expression for f(–X).
3. Compare f(–X).and f(X):
- if f(–X).= f(X), then the function is even;
- if f(–X).= – f(X), then the function is odd;
- if f(–X) ≠ f(X) and f(–X) ≠ –f(X), then the function is neither even nor odd.
Examples:
Investigate the function for parity a) at= x 5 +; b) at=; v) at= .
Solution.
a) h (x) = x 5 +,
1) D (h) = (–∞; 0) U (0; + ∞), symmetric set.
2) h (- x) = (–x) 5 + - x5 - = - (x 5 +),
3) h (- x) = - h (x) => function h (x)= x 5 + odd.
b) y =,
at = f(X), D (f) = (–∞; –9)? (–9; + ∞), an asymmetric set, so the function is neither even nor odd.
v) f(X) =, y = f (x),
1) D ( f) = (–∞; 3] ≠; b) (∞; –2), (–4; 4]?
Option 2
1. Is the given set symmetric: a) [–2; 2]; b) (∞; 0], (0; 7)?
a); b) y = x · (5 - x 2).
a) y = x 2 (2x - x 3), b) y =
Plot a function graph at = f(X), if at = f(X) Is an even function.
Plot a function graph at = f(X), if at = f(X) Is an odd function.
Mutual verification of slide.
6. Assignment at home: №11.11, 11.21,11.22;
Proof of the geometric meaning of the parity property.
*** (Setting the USE option).
1. The odd function y = f (x) is defined on the whole number line. For any non-negative value of the variable x, the value of this function coincides with the value of the function g ( X) = X(X + 1)(X + 3)(X- 7). Find the value of the function h ( X) = for X = 3.
7. Summing up
A function is called even (odd) if for any and the equality 
.
The graph of an even function is symmetrical about the axis 
.
The graph of an odd function is symmetric about the origin.
Example 6.2. Investigate for evenness or oddness of a function
1)
;
2)
;
3)
.
Solution.
1) The function is defined at 
... Find 
.
Those. 
... This means that this function is even.
2) The function is defined at 
Those. 
... Thus, this function is odd.
3) the function is defined for, i.e. for
,
... Therefore, the function is neither even nor odd. Let's call it a general function.
3. Study of the function for monotonicity.
Function 
is called increasing (decreasing) on some interval if in this interval each more meaning the argument corresponds to the larger (smaller) value of the function.
Functions increasing (decreasing) on a certain interval are called monotone.
If the function 
differentiable on the interval 
and has a positive (negative) derivative 
, then the function 
increases (decreases) in this interval.
Example 6.3... Find the intervals of monotonicity of functions
1)
;
3)
.
Solution.
1) This function is defined on the entire number axis. Let's find the derivative.
The derivative is zero if 
and 
... Definition area - numeric axis, split by dots 
,
at intervals. Let us determine the sign of the derivative in each interval.
In the interval 
the derivative is negative, the function decreases on this interval.
In the interval 
the derivative is positive, therefore, the function increases on this interval.
2) This function is defined if 
or
.
Determine the sign of the square trinomial in each interval.
Thus, the domain of the function
Find the derivative 
,
, if 
, i.e. 
, but 
... Let us determine the sign of the derivative in the intervals 
.
In the interval 
the derivative is negative, therefore, the function decreases on the interval 
... In the interval 
the derivative is positive, the function increases on the interval 
.
4. Investigation of the function for an extremum.
Dot 
is called the maximum (minimum) point of the function 
if there is such a neighborhood of the point 
that for everyone 
from this neighborhood the inequality 
.
The maximum and minimum points of a function are called extreme points.
If the function 
at the point 
has an extremum, then the derivative of the function at this point is zero or does not exist (a necessary condition for the existence of an extremum).
The points at which the derivative is zero or does not exist are called critical.
5. Sufficient conditions for the existence of an extremum.
Rule 1... If, when passing (from left to right) through the critical point 
derivative 
changes the sign from "+" to "-", then at the point 
function 
has a maximum; if from "-" to "+", then the minimum; if 
does not change sign, then there is no extremum.
Rule 2... Let at the point 
first derivative of a function 
is zero 
, and the second derivative exists and is nonzero. If 
, then 
Is the maximum point if 
, then 
Is the minimum point of the function.
Example 6.4 ... Explore the maximum and minimum functions:
1)
;
2)
;
3)
;
4)
.
Solution.
1) The function is defined and continuous on the interval 
.
Find the derivative 
and solve the equation 
, i.e. 
.From here 
- critical points.
Let us determine the sign of the derivative in the intervals, 
.
When crossing points 
and 
the derivative changes sign from "-" to "+", therefore, according to rule 1 
- points of minimum.
When crossing a point 
the derivative changes sign from "+" to "-", therefore 
Is the maximum point.
,
.
2) The function is defined and continuous in the interval 
... Find the derivative 
.
Solving the equation 
, find 
and 
- critical points. If the denominator 
, i.e. 
, then the derivative does not exist. So, 
- the third critical point. Let us determine the sign of the derivative in the intervals.
Consequently, the function has a minimum at the point 
, maximum in points 
and 
.
3) The function is defined and continuous if 
, i.e. at 
.
Find the derivative 
.
Let's find the critical points: 
Point neighborhood 
do not belong to the domain of definition, so they are not so extreme. So, let's explore the critical points 
and 
.
4) The function is defined and continuous on the interval 
... We use rule 2. Find the derivative 
.
Let's find the critical points:
Find the second derivative 
and define its sign at the points 
At points 
function has a minimum.
At points 
the function has a maximum.
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Any fractal is built according to a certain rule, which is consistently applied an unlimited number of times. Each such time is called an iteration.
The iterative algorithm for constructing the Menger sponge is quite simple: the original cube with side 1 is divided by planes parallel to its faces into 27 equal cubes. One central cube and 6 adjacent cubes are removed from it. The result is a set consisting of the remaining 20 smaller cubes. Doing the same with each of these cubes, we get a set, already consisting of 400 smaller cubes. Continuing this process endlessly, we get a Menger sponge.
