Find out the function is even or odd. Parity and odd functions

The parity and oddness of the function are one of its main properties, and in parity occupies an impressive part of the school course in mathematics. It determines the nature of the behavior of the function and greatly facilitates the construction of an appropriate schedule.

Determine the parity of the function. Generally speaking, the function under study is considered even if for opposite values \u200b\u200bof an independent variable (X), which are in its definition area, the corresponding values \u200b\u200bof y (functions) will be equal.

We will give a more stringent definition. Consider some function f (x), which is set in the region D. It will be even if for any point x in the field of definition:

• -x (opposite point) also lies in this area of \u200b\u200bdefinition,

• f (-x) \u003d f (x).

From the given definition, the condition necessary for the determination of such a function is followed, namely, symmetry relative to the point of the onset of coordinates, since if some point b is contained in the field of determining the even function, the corresponding point is also lies in this area. From the above, therefore, the conclusion implies: the even function has symmetrical to the ordinate (oy) axis.

How to practice the parity of the function?

Let be given using the formula H (x) \u003d 11 ^ x + 11 ^ (- x). Following the algorithm that arises directly from the definition, we examine primarily its definition area. Obviously, it is defined for all the values \u200b\u200bof the argument, that is, the first condition is fulfilled.

The next step is to substitute instead of the argument (x) its opposite value (-X).
We get:
h (-x) \u003d 11 ^ (- x) + 11 ^ x.
Since the addition satisfies the commutative (transitional) law, then obviously h (-x) \u003d h (x) and the specified functional dependence is even.

Check the parity of the function h (x) \u003d 11 ^ x-11 ^ (- x). Following the same algorithm, we obtain that H (-x) \u003d 11 ^ (- x) -11 ^ x. I will make minus, as a result, we have
h (-X) \u003d - (11 ^ x-11 ^ (- x)) \u003d - h (x). Consequently, H (x) is odd.

By the way, you should recall that there are functions that cannot be classified according to these features, they are called neither even or odd.

Even functions have a number of interesting properties:

• as a result of the addition of such functions, it is also obtained;
• as a result, the subtraction of such functions receive even;
• even, also even;
• as a result of multiplying two such functions, it is also obtained;
• as a result of multiplication of odd and even functions, the odd is obtained;
• as a result of the division of odd and even functions, they receive odd;
• the derivative of such a function is odd;
• if we build an odd function in a square, we get even.

The readiness of the function can be used in solving equations.

To solve the equation of type G (x) \u003d 0, where the left part of the equation is an even function, it will be enough to find it solutions for non-negative values \u200b\u200bof the variable. The obtained roots of the equation must be combined with opposite numbers. One of them is subject to verification.

This is successfully used to solve non-standard tasks with a parameter.

For example, is there any value of the parameter A, in which the equation 2x ^ 6-X ^ 4-AX ^ 2 \u003d 1 will have three roots?

If we consider that the variable is included in the equation in even degrees, it is clear that the replacement of X is x the specified equation will not change. It follows that if a certain number is its root, then it is also the opposite number. The conclusion is obvious: the roots of the equation other than zero are included in the many of his "couples" solutions.

It is clear that the number 0 itself is not, that is, the number of roots of a similar equation can only be even even and, naturally, none of the parameter value it cannot have three roots.

But the number of roots of the equation 2 ^ x + 2 ^ (- x) \u003d AX ^ 4 + 2x ^ 2 + 2 may be odd, and for any value of the parameter. Indeed, it is easy to check that many roots of this equation Contains solutions "couples". Check whether 0 is the root. When substituting it into the equation, we obtain 2 \u003d 2. Thus, in addition to the "paired" 0, it is also the root, which proves their odd amount.

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Objectives:

• form the concept of readiness and innergity of the function, learn the ability to determine and use these properties when research Functions, build graphs;
• develop the creative activity of students logical thinking, the ability to compare, summarize;
• educate hard work, mathematical culture; Develop communicative qualities .

Equipment:multimedia Installation, Interactive Board, Distribution Material.

Forms of work:frontal and group with elements of search and research activities.

Information sources:

1. Algebra9 Class A.G Mordkovich. Textbook.
2. Algebra 9 Class A.G Mordkovich. Task.
3. Algebra grade 9. Tasks for learning and student development. Belenkova E.Yu. Lebedinsieva E.A.

DURING THE CLASSES

1. Organizational moment

Setting the goals and objectives of the lesson.

2. Checking homework

№10.17 (Problem 9kl. A.G. Mordkovich).

but) w. = f.(h.), f.(h.) =

b) f. (–2) = –3; f. (0) = –1; f.(5) = 69;

c) 1. D ( f.) = [– 2; + ∞)
2. E ( f.) = [– 3; + ∞)
3. f.(h.) \u003d 0 when h. ~ 0,4
4. f.(h.)\u003e 0 when h. > 0,4 ; f.(h.) < 0 при – 2 < h. < 0,4.
5. The function increases when h. € [– 2; + ∞)
6. The function is limited to below.
7. w. Nym \u003d - 3, w. Naib does not exist
8. The function is continuous.

(You used the function research algorithm?) Slide.

2. The table that you wondered, check by the slide.

Fill the table
	Domain	Zero function	Intervals of the sign		Coordinates of points intersection graphics with OU
		x \u003d -5 x \u003d 2.	x € (-5; 3) u U (2; ∞)	x € (-∞; -5) u U (-3; 2)
	x ∞ -5 x ≠ 2.		x € (-5; 3) u U (2; ∞)	x € (-∞; -5) u U (-3; 2)
	x ≠ -5 x ≠ 2.		x € (-∞; -5) u U (2; ∞)	x € (-5; 2)

3. Actualization of knowledge

- Dana functions.
- Specify the definition area for each function.
- Compare the value of each function for each pair of the value of the argument: 1 and - 1; 2 and - 2.
- For some of these functions in the field of definition, equality are performed f.(– h.) = f.(h.), f.(– h.) = – f.(h.)? (the obtained data is in the table) Slide

		f.(1) and f.(– 1)	f.(2) and f.(– 2)	graphics	f.(– h.) = –f.(h.)	f.(– h.) = f.(h.)
1. f.(h.) =
2. f.(h.) = h. 3
3. f.(h.) = | h. |
4. F.(h.) = 2h. – 3
5. f.(h.) =	h. ≠ 0
6. f.(h.)=	h. > –1		and not definitely

4. New Material

- Performing this workGuys, we revealed another property of a function unfamiliar to you, but no less important than the rest - this is the readiness and oddness of the function. Write down theme of the lesson: "Even and odd functions", our task is to learn how to determine the readiness and oddness of the function, to find out the significance of this property in the study of functions and the construction of graphs.
So, find definitions in the textbook and read (p. 110) . Slide

ORD. oneFunction w. = f. (h.) specified on the set x called thoughtIf for any value h. Є X is performed equality F (-m) \u003d F (x). Give examples.

ORD. 2.Function y \u003d f (x)specified on the set x called oddIf for any value h. Є x equality f (s) \u003d -f (x) is performed. Give examples.

Where did we meet with the terms "even" and "odd"?
Which of these functions will be read, what do you think? Why? What are odd? Why?
For any type of type w.= x N.where n. - An integer can be argued that the function is odd n. - odd and functions are black n. - younger.
- Functions of type w. \u003d I. w. = 2h. - 3 are not anything, neither internally, because Equality are not performed f.(– h.) = – f.(h.), f.(– h.) = f.(h.)

Studying the question of whether the function is even or internally called the study of functions to readiness.Slide

In definitions 1 and 2, it was discussed about the values \u200b\u200bof the function at x and there, thereby it is assumed that the function is determined and when h.and when - h..

OPR 3. If the numerical set together with each element x contains the opposite element, then the set H.called symmetric set.

Examples:

(-2; 2), [-5; 5]; (∞; ∞) - symmetric sets, a, [-5; 4] - asymmetrical.

- Intelligent functions, the definition area is a symmetrical set? In odd?
- if D ( f.) - Asymmetric set, then what is the function?
- Thus, if the function w. = f.(h.) - something or odd, then its area of \u200b\u200bdefinition D ( f.) - symmetrical set. Is it true that the reverse statement is true if the definition area is a symmetric set, then it is black, or internally?
- Therefore, the presence of a symmetric set of the definition area is a necessary condition, but insufficient.
- So how to investigate the function for parity? Let's try to make an algorithm.

Slide

Algorithm Research Functions for Ready

1. Install whether the function definition is symmetrical. If not, the function is not aware or intense. If so, then go to step 2 algorithm.

2. Make an expression for f.(– H.).

3. Compare f.(– H.).and f.(h.):

• if a f.(– H.).= f.(h.), then the function is even;
• if a f.(– H.).= – f.(h.), then the function is odd;
• if a f.(– H.) ≠ f.(h.) I. f.(– H.) ≠ –f.(h.), the function is not aware or intense.

Examples:

Explore the function a) w. \u003d x 5 +; b) w. \u003d; in) w.= .

Decision.

a) h (x) \u003d x 5 +,

1) D (H) \u003d (-∞; 0) U (0; + ∞), a symmetrical set.

2) h (x) \u003d (s) 5 + - x5 - \u003d - (x 5 +),

3) h (x) \u003d - h (x) \u003d\u003e function h (x) \u003d x 5 + odd.

b) y \u003d,

w. = f.(h.), D (f) \u003d (-∞; -9)? (-9; + ∞), asymmetrical set, means the function is neither even internally.

in) f.(h.) \u003d, y \u003d f (x),

1) D ( f.) \u003d (-∞; 3] ≠; b) (∞; -2), (-4; 4]?

Option 2.

1. Is the symmetric set set: a) [-2; 2]; b) (∞; 0], (0; 7)?

but); b) y \u003d x · (5 - x 2). 2. Explore the function:

a) y \u003d x 2 · (2x - x 3), b) y \u003d

3. In fig. Built schedule w. = f.(h.), for all h.satisfying the condition h.? 0.
Build a function graph w. = f.(h.), if a w. = f.(h.) - aware of the function.

3. In fig. Built schedule w. = f.(h.), For all x, satisfying the condition x? 0.
Build a function graph w. = f.(h.), if a w. = f.(h.) - odd function.

Multi-test slide.

6. Task for the house: №11.11, 11.21,11.22;

Proof of the geometrical meaning of the properties of attention.

*** (Setting the USE option).

1. The odd function y \u003d f (x) is defined on the entire numeric line. For any non-negative value of the variable x, the value of this function coincides with the value of the G function ( h.) = h.(h. + 1)(h. + 3)(h. - 7). Find the value of the H function ( h.) \u003d O. h. = 3.

7. Summing up

The function is called even (odd) if equality is performed for anyone

.

An even function graph is symmetrical about the axis
.

The schedule of an odd function is symmetrical on the start of the coordinates.

Example 6.2. Explore the parity or oddness of the function

1)
; 2)
; 3)
.

Decision.

1) the function is determined when
. Find
.

Those.
. So this function is even.

2) the function is defined when

Those.
. Thus, this function is odd.

3) The function is defined for, i.e. for

,
. Therefore, the function is neither even or odd. We call it a common type function.

3. Investigation of the function on monotony.

Function
called increasing (decreasing) at some interval, if in this interval each greater value The argument corresponds to a larger (smaller) value of the function.

The functions of increasing (decreasing) are called monotonous at some interval.

If the function
differential on the interval
and has a positive (negative) derivative
, then function
increases (decreases) at this interval.

Example 6.3.. Find intervals of functions monotony

1)
; 3)
.

Decision.

1) This function is determined on the entire numeric axis. Find a derivative.

The derivative is zero if
and
. The definition area is the numerical axis, divided by points
,
at intervals. Determine the sign of the derivative in each interval.

In the interval
the derivative is negative, the function at this interval decreases.

In the interval
the derivative is positive, therefore, the function at this interval increases.

2) This function is defined if
or

.

Determine the sign of square three declections in each interval.

Thus, the field definition area

Find a derivative
,
, if a
.
, but
. 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. Study of the function to extremum.

Point
called a maximum point (minimum) functions
if there is such a neighborhood point that for all
inequality is performed from this neighborhood

.

Maximum points and a minimum of functions are called extremum points.

If the function
at point it has an extremum, the derivative function at this point is zero or does not exist (the necessary condition for the existence of extremum).

The points in which the derivative is equal to zero or is not called critical.

5. Sufficient conditions for the existence of extremum.

Rule 1.. If during the transition (from left to right) through a critical point derivative
changes the sign from "+" to "-", then at the point function
has a maximum; If with "-" on "+", then a minimum; if a
does not change the sign, then the extremum is not.

Rule 2.. Let in the point
first derivative function
equal to zero.
, and the second derivative exists and is different from zero. If a
T. - Maximum point if
T. - point minimum function.

Example 6.4 . Explore the maximum and minimum function:

1)
; 2)
; 3)
;

4)
.

Decision.

1) The function is determined and continuous on the interval
.

Find a derivative
and solve equation
.
.OTSyud
- Critical points.

Determine the sign of the derivative in the intervals,
.

When moving through points
and
the derivative changes the sign from "-" to "+", so according to rule 1
- Minimum points.

When switching through the point
the derivative changes the sign from "+" to "-", therefore
- Maximum point.

,
.

2) The function is defined and continuous in the interval
. Find a derivative
.

Deciding equation
We find
and
- Critical points. If the denominator
.
, the derivative does not exist. So,
- Third critical point. Determine the sign of the derivative in the intervals.

Consequently, the function has a minimum at point
, maximum points
and
.

3) the function is defined and continuous if
. for
.

Find a derivative

.

We will find critical points:

Neighborhoods
do not belong to the definition area, so they are not t. Extremum. So, we investigate critical points
and
.

4) The function is defined and continuous on the interval
. We use Rule 2. Find a derivative
.

We will find critical points:

We find the second derivative
and define her sign at points

At points
the function has a minimum.

At points
the function has a maximum.

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