How to check the parity of the function. Properties of functions. Sufficient conditions for the existence of extremum

evenif at all \\ (x \\) from its definition area is true: \\ (f (-x) \u003d f (x) \\).

An even function graph is symmetrical with respect to the axis \\ (Y \\):

Example: the function \\ (f (x) \u003d x ^ 2 + \\ cos X \\) is even, because \\ (f (-X) \u003d (- x) ^ 2 + \\ cos ((- x)) \u003d x ^ 2 + \\ cos x \u003d f (x) \\).

\\ (\\ BLACKTRIANGLERIGHT \\) The function \\ (f (x) \\) is called oddif at all \\ (x \\) from its definition area is true: \\ (f (-x) \u003d - f (x) \\).

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

Example: the function \\ (f (x) \u003d x ^ 3 + x \\) is an odd, because \\ (f (-x) \u003d (- x) ^ 3 + (- x) \u003d - x ^ 3 - x \u003d - (x ^ 3 + x) \u003d - f (x) \\).

\\ (\\ blacktriangleright \\) Functions that are neither even nor odd, are called functions general view. Such a function can always be unique in the form of an even and odd function as the sum.

For example, the function \\ (f (x) \u003d x ^ 2-x \\) is the sum of the even function \\ (F_1 \u003d x ^ 2 \\) and the odd \\ (F_2 \u003d -x \\).

\\ (\\ blacktriangleright \\) Some properties:

1) The product and private two functions of the same parity are an even function.

2) the work and private two functions of different parity - other function.

3) The sum and difference of even functions is an even function.

4) The sum and the difference of odd functions is an odd function.

5) If \\ (F (x) \\) is an even function, the equation \\ (f (x) \u003d c \\ (C \\ in \\ mathbb (r) \\)) has the only root then and only when when \\ (x \u003d 0 \\).

6) If \\ (F (x) \\) is an even or odd function, and the equation \\ (F (x) \u003d 0 \\) has a root \\ (x \u003d b \\), then this equation will necessarily have a second root \\ (x \u003d -B \\).

\\ (\\ BLACKTRIANGLERIGHT \\) The function \\ (F (x) \\) is called periodic to \\ (x \\), if for a certain number \\ (t \\ ne 0 \\) is made \\ (F (x) \u003d f (x + t) \\), where \\ (x, x + t \\ in x \\). The smallest \\ (t \\), for which this equality is satisfied, is called the main (basic) function period.

In the periodic function, any number of species \\ (nt \\), where \\ (n \\ in \\ mathbb (z) \\) will also be a period.

Example: anyone trigonometric function is periodic;
functions \\ (f (x) \u003d \\ sin x \\) and \\ (f (x) \u003d \\ cos x \\) the main period is \\ (2 \\ pi \\), for functions \\ (f (x) \u003d \\ mathrm ( TG) \\, x \\) and \\ (f (x) \u003d \\ mathrm (CTG) \\, X \\) The main period is \\ (\\ pi \\).

In order to build a graph of a periodic function, you can construct its schedule on any segment length \\ (T \\) (the main period); Then the schedule of the entire function is completed with a shift of the constructed part by an integer number of periods to the right and left:

\\ (\\ (\\ blacktriangleright \\) The definition area \\ (D (F) \\) functions \\ (F (x) \\) is a set consisting of all the values \u200b\u200bof the argument \\ (x \\), in which the function makes sense (defined).

Example: function \\ (F (X) \u003d \\ SQRT X + 1 \\) Definition area: \\ (X \\ IN

Task 1 # 6364

Task level: equal to ege

Under what values \u200b\u200bof the parameter \\ (a \\) equation

has the only solution?

Note that since \\ (x ^ 2 \\) and \\ (\\ cos x \\) are even functions, if the equation has a root \\ (x_0 \\), it will also have a root \\ (- x_0 \\).
Indeed, let \\ (x_0 \\) - the root, that is, equality \\ (2x_0 ^ 2 + a \\ mathrm (Tg) \\, (\\ cos x_0) + a ^ 2 \u003d 0 \\) right. Substitute \\ (- x_0 \\): \\ (2 (-x_0) ^ 2 + A \\ MathRM (TG) \\, (\\ cos (-x_0)) + a ^ 2 \u003d 2x_0 ^ 2 + A \\ MathRM (TG) \\, (\\ cos x_0) + a ^ 2 \u003d 0 \\).

Thus, if \\ (x_0 \\ ne 0 \\), the equation will already have at least two roots. Consequently, \\ (x_0 \u003d 0 \\). Then:

We obtained two values \u200b\u200bof the parameter \\ (a \\). Note that we used what \\ (x \u003d 0 \\) is precisely the root of the original equation. But we have not used anywhere that it is the only one. Therefore, it is necessary to substitute the resulting values \u200b\u200bof the parameter \\ (a \\) to the initial equation and check at which \\ (a \\) root \\ (x \u003d 0 \\) will indeed be the only one.

1) If \\ (a \u003d 0 \\), the equation will take the form \\ (2x ^ 2 \u003d 0 \\). Obviously, this equation has only one root \\ (x \u003d 0 \\). Consequently, the value \\ (a \u003d 0 \\) is suitable for us.

2) if \\ (a \u003d - \\ mathrm (Tg) \\, 1 \\), the equation will take the form \ Rewrite the equation in the form \ As \\ (- 1 \\ leqslant \\ cos x \\ leqslant 1 \\)T. \\ (- \\ MathRM (TG) \\, 1 \\ Leqslant \\ Mathrm (TG) \\, (\\ COS X) \\ Leqslant \\ Mathrm (TG) \\, 1 \\). Consequently, the values \u200b\u200bof the right part of the equation (*) belong to the segment \\ ([- \\ MathRM (TG) ^ 2 \\, 1; \\ MathRM (TG) ^ 2 \\, 1] \\).

Since \\ (x ^ 2 \\ geqslant 0 \\), the left part of the equation (*) is greater than or equal to \\ (0+ \\ MathRM (TG) ^ 2 \\, 1 \\).

Thus, equality (*) can be performed only when both parts of the equation are \\ (\\ MathRM (TG) ^ 2 \\, 1 \\). And this means that \\ [\\ Begin (Cases) 2x ^ 2 + \\ MathRM (TG) ^ 2 \\, 1 \u003d \\ MathRM (TG) ^ 2 \\, 1 \\\\\\\\\\ Mathrm (TG) \\, 1 \\ Cdot \\ Mathrm (TG) \\ \u003d \\ MathRM (TG) \\, 1 \\ END (CASES) \\ quad \\ leftrightarrow \\ quad x \u003d 0 \\] Consequently, the value \\ (a \u003d - \\ mathrm (TG) \\, 1 \\) is suitable for us.

Answer:

\\ (A \\ in \\ (- \\ MathRM (TG) \\, 1; 0 \\) \\)

Task 2 # 3923

Task level: equal to ege

Find all parameter values \u200b\u200b\\ (a \\), each you are a function graph \

symmetrical on the start of the coordinates.

If the graph of the function is symmetrical relative to the start of the coordinates, this function is odd, that is, it is made \\ (f (-x) \u003d - f (x) \\) for any \\ (x \\) from the function of determining the function. Thus, it is required to find those values \u200b\u200bof the parameter at which it is made \\ (f (-x) \u003d - f (x). \\)

\\ [\\ begin (aligned) & 3 \\ mathrm (tg) \\, \\ left (- \\ dfrac (AX) 5 \\ RIGHT) +2 \\ sin \\ dfrac (8 \\ pi a + 3x) 4 \u003d - \\ left (3 \\ , \\ dfrac (AX) 5 + 2 \\ sin \\ dfrac (8 \\ pi a + 3x) 4 \u003d - \\ left (3 \\ MathRM (TG) \\, \\ left (\\ dfrac (AX) 5 \\ RIGHT) +2 \\ 3X) 4 \u003d 0 \\ quad \\ rightarrow \\ quad2 \\ sin \\ dfrac12 \\ left (\\ dfrac (8 \\ pi a + 3x) 4+ \\ dfrac (8 \\ PI A-3X) 4 \\ Right) \\ CDOT \\ COS \\ DFRAC12 \\ left (\\ dfrac (8 \\ pi a + 3x) 4- \\ dfrac (8 \\ pi a-3x) 4 \\ right) \u003d 0 \\ quad \\ rightarrow \\ quad \\ sin (2 \\ pi a) \\ cdot \\ cos \\ The latter equation must be performed for all \\ (x \\) from the definition area \\ (F (x) \\), therefore,

\\ (\\ sin (2 \\ pi a) \u003d 0 \\ rightarrow a \u003d \\ dfrac n2, n \\ in \\ mathbb (z) \\) \\ (\\ dfrac n2, n \\ in \\ mathbb (z) \\).

Answer:

Task 3 # 3069

Find all the parameter values \u200b\u200b\\ (a \\), each you each of which the equation \\ has 4 solutions, where \\ (F \\) is an even periodic with a period \\ (T \u003d \\ DFRAC (16) 3 \\) a function defined on the entire numeric direct , moreover, \\ (f (x) \u003d ax ^ 2 \\) when

Task level: equal to ege

\\ (0 \\ leqslant x \\ leqslant \\ dfrac83. \\) (Task from subscribers)

Task 4 # 3072

Find all values \u200b\u200b\\ (a \\), each you are

Task level: equal to ege

It has at least one root. \

rewrite the equation in the form

Task 4 # 3072

and consider two functions: \\ (G (x) \u003d 7 \\ sqrt (2x ^ 2 + 49) \\) and \\ (f (x) \u003d 3 | x-7a | -6 | x | -a ^ 2 + 7a \\ The function \\ (G (x) \\) is even, has a point of minimum \\ (x \u003d 0 \\) (and \\ (G (0) \u003d 49 \\)). \ The function \\ (F (x) \\) with \\ (x\u003e 0 \\) is decreasing, and with \\ (x

<0\) – возрастающей, следовательно, \(x=0\) – точка максимума.
Indeed, with \\ (x\u003e 0 \\), the second module will reveal positively (\\ (| x | \u003d x \\)), therefore, regardless of how the first module is revealed, \\ (F (x) \\) will be equal to \\ ( KX + A \\), where \\ (a \\) is the expression from \\ (a \\), and \\ (k \\) is equal to either \\ (- 9 \\), or \\ (- 3 \\). With \\ (x<0\) наоборот: второй модуль раскроется отрицательно и \(f(x)=kx+A\) , где \(k\) равно либо \(3\) , либо \(9\) .
Find the value \\ (F \\) at the maximum point: \\

In order for the equation to have at least one solution, it is necessary that the graphs of functions \\ (F \\) and \\ (G \\) have at least one point of intersection. Therefore, you need: \ Resolving this set of systems, we will receive the answer: \\]

Answer:

\\ (A \\ In \\ (- 7 \\) \\ CUP \\)

Task 5 # 3912

Task level: equal to ege

Find all the parameter values \u200b\u200b\\ (a \\), each you are \

has six different solutions.

We will replace \\ ((\\ sqrt2) ^ (x ^ 3-3x ^ 2 + 4) \u003d T \\), \\ (T\u003e 0 \\). Then the equation will take the form \ We will gradually write out the conditions under which the initial equation will have six solutions.
notice, that quadratic equation \\ ((*) \\) can maximize two solutions. Any cubic equation \\ (AX ^ 3 + BX ^ 2 + CX + D \u003d 0 \\) may have no more than three solutions. Consequently, if the equation \\ ((*) \\) has two different solutions (positive!, since \\ (t \\) must be greater than zero) \\ (t_1 \\) and \\ (t_2 \\), then by making a replacement, we will get: \\ [\\ left [\\ begin (gathered) \\ begin (aligned) & (\\ sqrt2) ^ (x ^ 3-3x ^ 2 + 4) \u003d t_1 \\\\ \\ (\\ sqrt2) ^ (x ^ 3-3x ^ 2 +4) \u003d T_2 \\ END (Aligned) \\ END (Gathered) \\ Right. \\] Since any positive number can be represented as \\ (\\ sqrt2 \\) to some extent, for example, \\ (t_1 \u003d (\\ sqrt2) ^ (\\ log _ (\\ sqrt2) t_1) \\), the first equation of the aggregate will rewrite in the form of \ As we have already spoken, any cubic equation has no more than three solutions, therefore, each equation from the aggregate will have no more than three solutions. So, the entire totality will have no more than six decisions.
It means that the initial equation has six solutions, the square equation \\ ((*) \\) must have two different solutions, and each obtained cubic equation (from the aggregate) should have three different solutions (no solution of one equation should coincide with what -Lo decision of the second!)
Obviously, if the square equation \\ ((*) \\) will have one solution, we will not get six solutions in the original equation.

Thus, the solution plan becomes clear. Let's repel the conditions that must be performed.

1) To the equation \\ ((*) \\) had two different solutions, its discriminant must be positive: \

2) It is also necessary that both roots are positive (since \\ (T\u003e 0 \\)). If the product of the two roots is positive and the amount is positive, then the roots themselves will be positive. Therefore, you need: \\ [\\ Begin (Cases) 12-A\u003e 0 \\\\ - (A-10)\u003e 0 \\ End (Cases) \\ Quad \\ Leftrightarrow \\ Quad A<10\]

Thus, we have already provided two different positive roots \\ (t_1 \\) and \\ (t_2 \\).

3) Let's look at such an equation \ At what \\ (t \\) will it have three different solutions?
Consider the function \\ (f (x) \u003d x ^ 3-3x ^ 2 + 4 \\).
You can decompose on multipliers: \ Consequently, its zeros: \\ (x \u003d -1; 2 \\).
If you find the derivative \\ (f "(x) \u003d 3x ^ 2-6x \\), then we obtain two extremum points \\ (x_ (max) \u003d 0, x_ (min) \u003d 2 \\).
Therefore, the schedule looks like this:


We see that any horizontal straight line \\ (y \u003d k \\), where \\ (0 \\ (x ^ 3-3x ^ 2 + 4 \u003d \\ log _ (\\ sqrt2) t \\) had three different solutions, you need to \\ (0<\log_ {\sqrt2}t<4\) .
Thus, you need: \\ [\\ Begin (Cases) 0<\log_{\sqrt2}t_1<4\\ 0<\log_{\sqrt2}t_2<4\end{cases}\qquad (**)\] Let us immediately note that if the numbers \\ (t_1 \\) and \\ (t_2 \\) are different, then the numbers \\ (\\ log _ (\\ sqrt2) t_1 \\) and \\ (\\ log _ (\\ sqrt2) t_2 \\) will be different, So the equations \\ (x ^ 3-3x ^ 2 + 4 \u003d \\ log _ (\\ sqrt2) t_1 \\) and \\ (x ^ 3-3x ^ 2 + 4 \u003d \\ log _ (\\ sqrt2) t_2 \\) Will have the roots at all.
The system \\ ((**) \\) can be rewritten so: \\ [\\ Begin (Cases) 1

Thus, we determined that both roots of the equation \\ ((*) \\) must lie in the interval \\ ((1; 4) \\). How to write this condition?
In explicit form, write the roots we will not.
Consider the function \\ (G (T) \u003d T ^ 2 + (A-10) T + 12-A \\). Its graph is a parabola with branches up, which has two intersection points with the abscissa axis (we recorded this condition in paragraph 1)). How should its schedule look like that the intersection points with the abscissa axis were in the interval \\ ((1; 4) \\)? So:


First, the values \u200b\u200b\\ (g (1) \\) and \\ (g (g) \\) functions at points \\ (1 \\) and \\ (4 \\) should be positive, secondly, the pearabol vertex \\ (T_0 \\ Therefore, you can write the system: \\ [\\ begin (Cases) 1 + A-10 + 12-A\u003e 0 \\\\ 4 ^ 2 + (A-10) \\ CDOT 4 + 12-A\u003e 0 \\\\ 1<\dfrac{-(a-10)}2<4\end{cases}\quad\Leftrightarrow\quad 4

Thus, we need to cross the value of the parameter \\ (a \\), found in the 1st, 2nd and 3rd paragraphs, and we will get the answer: \\ [\\ begin (Cases) A \u200b\u200b\\ in (- \\ infty; 8-2 \\ SQRT3) \\ CUP (8 + 2 \\ SQRT3; + \\ INFTY) \\\\ A<10\\ 4

Definition1. Functioning even (odd ) if together with each value of the variable
value - h.also belong
and equality is performed

Thus, the function can be even or odd only when its area of \u200b\u200bdetermining is symmetrical relative to the origin on the numeric line (the number h.and - h.at the same time belong
). For example, a function
is not even and odd, since its area of \u200b\u200bdefinition
not symmetrical about the start of coordinates.

Function
even because
symmetrical relative to the start of coordinates and.

Function
odd because
and
.

Function
is not even and odd, because though
and symmetrical on the origin of the coordinates, equality (11.1) is not performed. For example,.

An even function graph is symmetrical about the axis OUsince if the point

also belongs to graphics. Schedule of an odd function is symmetrical relative to the start of coordinates, since
belongs graphics and point
also belongs to schedule.

In proof of parity or oddness, the following statements are useful.

Theorem1. a) the sum of two even (odd) functions have an even function (odd).

b) The product of two even (odd) functions have an even function.

c) The product of even and odd functions has an odd function.

d) if f.- even function on the set H., and function g. defined on a set
, then function
- even.

e) if f.- odd feature on the set H., and function g. defined on a set
and even (odd), then the function
- Even (odd).

Evidence. We prove, for example, b) and d).

b) Let
and
- even functions. Then, therefore. Similarly, the case of odd functions is considered.
and
.

d) Let f. - even function. Then.

The remaining statements of the theorem are proved similarly. Theorem is proved.

Theorem2. Any function
set H., symmetrical relative to the start of coordinates, can be represented as the sum of even and odd functions.

Evidence. Function
can be written in the form

.

Function
- even, since
, and function
- odd, because. In this way,
where
- even, and
- odd functions. Theorem is proved.

Definition2. Function
called periodic if there is a number
, such that in any
numbers
and
also belong areas of definition
and equality are performed

Such a number T.called period functions
.

From Definition 1 it follows that if T.- Period of function
, then the number - T.also is a function period
(since when replacing T.on the - T.equality is preserved). With the help of the method of mathematical induction, you can show that if T.- Period of function f., that I.
is also a period. It follows that if the function has a period, then it has infinitely many periods.

Definition3. The smallest of the positive periods of the function is called it basic period.

Theorem3. If T.- the main period of the function f., the remaining periods are painted to him.

Evidence. Suppose nasty, that is, there is a period functions f. (\u003e 0), not multiple T.. Then, share on the T.with the remnant, we get
where
. therefore

i.e - Period of function f., and
, and this contradicts what T.- the main period of the function f.. The assertion of the theorem follows from the resulting contradiction. Theorem is proved.

It is well known that trigonometric functions are periodic. The main period
and
raven
,
and
. Find a function of the function
. Let be
- The period of this function. Then

(as
.

elijah
.

Value T.defined from the first equality can not be a period because it depends on h.. is a function OT. h., not a constant number. The period is determined from the second equality:
. Periods are infinitely a lot, when
the smallest positive period is obtained at
:
. This is the main period of the function.
.

An example of a more complex periodic function is the Dirichlet function

Note that if T.- rational number, then
and
are rational numbers with rational h.and irrational with irrational h.. therefore

with any rational number T.. Therefore, any rational number T.is a period of Dirichlet function. It is clear that there is no main period in this function, since there are positive rational numbers, how much are close to zero (for example, a rational settlement to choose n.how much is close to zero).

Theorem4. If the function f. set on the set H.and has a period T., and function g. set on the set
, then a complex function
also has a period T..

Evidence. We have, therefore

that is, the statement of the theorem is proved.

For example, since cos. x. has a period
, then functions
have a period
.

Definition4. Functions that are not periodic are called non-periodic .

An even function.

Even called function, the sign of which does not change when the sign changes x..

x. Equality is performed f.(–x.) = f.(x.). Sign x. does not affect the sign y..

An even function graph is symmetrical with respect to the coordinate axis (Fig. 1).

Examples of even functions:

y. \u003d COS. x.

y. = x. 2

y. = –x. 2

y. = x. 4

y. = x. 6

y. = x. 2 + x.

Explanation:
Take the function y. = x. 2 or y. = –x. 2 .
With any meaning x. Positive function. Sign x. does not affect the sign y.. The graph is symmetrical relative to the coordinate axis. This is an even function.

An odd feature.

Odd called function, the sign of which changes when the sign changes x..

Speaking otherwise, for any value x. Equality is performed f.(–x.) = –f.(x.).

The schedule of an odd function is symmetrical relative to the start of coordinates (Fig. 2).

Examples of odd function:

y. \u003d SIN x.

y. = x. 3

y. = –x. 3

Explanation:

Take the function y \u003d - x. 3 .
All values w. It will be with a minus sign. That is the sign x. affects the sign y.. If an independent variable is a positive number, then the function is positive, if an independent variable is a negative number, then the function is negative: f.(–x.) = –f.(x.).
The graph of the function is symmetrical on the start of the coordinates. This is an odd feature.

Properties of even and odd functions:

NOTE:

Not all functions are even or odd. There are functions that do not obey such graduation. For example, the function of the root w. = √h. Does not apply to any other, nor to odd functions (Fig. 3). When listed the properties of such functions, a corresponding description should be given: neither even or odd.

Periodic functions.

As you know, frequency is the repeatability of certain processes at a certain interval. Functions describing these processes are called periodic functions. That is, these features, in whose charts there are elements that are repeated with certain numeric intervals.

In July 2020, NASA launches an expedition to Mars. The spacecraft will deliver an electronic media to Mars with the names of all registered expedition participants.

If this post decided your problem or just liked you, share a link to it with my friends on social networks.

One of these code options must be copied and insert into the code of your webpage, preferably between the tags and or immediately after the tag . According to the first version, MathJax is loaded faster and slows down the page. But the second option automatically tracks and loads the latest MathJax versions. If you insert the first code, it will need to be periodically updated. If you insert the second code, the pages will be loaded slower, but you will not need to constantly monitor MathJax updates.

Connect MathJax is the easiest way to Blogger or WordPress: Add a widget for inserting a third-party JavaScript code to insert the first or second version of the download code presented above and place the widget closer to the beginning of the template (by the way, it is not at all necessary Since the MathJax script is loaded asynchronously). That's all. Now read Mathml, Latex and Asciimathml markup syntax, and you are ready to insert mathematical formulas on the web pages of your site.

Another New Year's Eve ... Frosty Weather and Snowflakes on the window glass ... All this prompted me to again write about ... Fractals, and about what he knows about this Alpha tungshes. On this occasion, there is an interesting article in which there are examples of two-dimensional fractal structures. Here we will consider more complex examples of three-dimensional fractals.

The fractal can be clearly imagined (describe), as a geometric shape or body (having in mind that both there are many, in this case, a set of points), the details of which have the same shape as the original figure itself. That is, it is a self-like structure, considering the details of which with an increase, we will see the same form as without increasing. Whereas in the case of a conventional geometric shape (not fractal), with an increase we will see parts that have a simpler form than the original figure itself. For example, with a sufficiently large increase, the part of the ellipse looks like a straight line. With fractals, this does not happen: with any increases, we will again see the same complex shape, which will repeat again and again.

Benoit Mandelbrot (Benoit Mandelbrot), the founder of the science of fractals, in his article Fractals and art in the name of science wrote: "Fractals are geometric forms that are equally complex in their details, as in its general form. That is, if Part of the fractal will be increased to the size of the whole, it will look like an integer, or exactly, or, possibly, with a small deformation. "

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Ways to set the function

Suppose the function is defined by the formula: y \u003d 2x ^ (2) -3. Assigning any values \u200b\u200bof an independent variable x can be calculated using this formula corresponding values \u200b\u200bof the dependent variable y. For example, if X \u003d -0.5, then, using the formula, we obtain that the corresponding value of Y is equal to Y \u003d 2 \\ Cdot (-0.5) ^ (2) -3 \u003d -2.5.

Taking any value taken by the x argument in the formula y \u003d 2x ^ (2) -3, you can only calculate the value of the function that corresponds to it. The function can be represented as a table:

Using this table, it can be disassembled that for the value of the argument -1 it will correspond to the value of the -3 function; And the value x \u003d 2 will correspond to y \u003d 0, etc. It is also important to know that each value of the argument in the table corresponds to only one function value.

More functions can be set using graphs. Using the graph, it is set to what function value is correlated with a specific value x. Most often, it will be the approximate value of the function.

Even and odd function

Function is even functionWhen f (-x) \u003d f (x) for any x from the definition area. This function will be symmetrical with respect to the OY axis.

Function is an odd featureWhen f (-x) \u003d - f (x) for any x from the definition area. This function will be symmetrical relative to the start of the coordinates O (0; 0).

Function is nor even, neither odd and called function of common typeWhen it does not have symmetry relative to the axis or the start of coordinates.

We study the parity below:

f (x) \u003d 3x ^ (3) -7x ^ (7)

D (f) \u003d (- \\ infty; + \\ infty) with a symmetric area of \u200b\u200bdefinition relative to the start of coordinates. f (-X) \u003d 3 \\ Cdot (-X) ^ (3) -7 \\ CDOT (-X) ^ (7) \u003d -3x ^ (3) + 7x ^ (7) \u003d - (3x ^ (3) -7x ^ (7)) \u003d -f (x).

So, the function f (x) \u003d 3x ^ (3) -7x ^ (7) is odd.

Periodic function

The function y \u003d f (x), in the definition area of \u200b\u200bwhich for any x, the equality f (x + t) \u003d f (x - t) \u003d f (x) is called periodic function With the period T \\ NEQ 0.

Repeating the graph of the function on any cut of the abscissa axis, which has a length t.

The gaps where the function is positive, i.e. F (x)\u003e 0 - cuts of the abscissa axis, which correspond to the points of the graph of the function lying above the abscissa axis.

f (x)\u003e 0 on (x_ (1); x_ (2)) \\ CUP (X_ (3); + \\ INFTY)

Gaps where the function is negative, that is, f (x)< 0 - отрезки оси абсцисс, которые отвечают точкам графика функции, лежащих ниже оси абсцисс.

f (X)< 0 на (- \\ infty; x_ (1)) \\ Cup (x_ (2); x_ (3))

Limit functions

Limited below It is customary to call the function y \u003d f (x), x \\ in x when there is such a number A for which the inequality f (x) \\ GEQ A is performed for any x \\ in x.

An example of a limited bottom of the function: y \u003d \\ sqrt (1 + x ^ (2)) as Y \u003d \\ SQRT (1 + x ^ (2)) \\ GEQ 1 for any x.

Limited from above The function y \u003d f (x), x \\ in x is called when there is such a number b for which the inequality F (x) \\ NEQ B is performed for any x \\ in x.

An example of a limited bottom function: y \u003d \\ sqrt (1-x ^ (2)), x \\ in [-1; 1] Since Y \u003d \\ SQRT (1 + x ^ (2)) \\ NEQ 1 for any X \\ in [-1; 1].

Limited It is customary to call the function y \u003d f (x), x \\ in x when there is such a number k\u003e 0 for which the inequality \\ left is performed | F (X) \\ Right | \\ NEQ K for any x \\ in x.

Example of limited function: y \u003d \\ sin x is limited on the entire numeric axis, since \\ left | \\ Sin X \\ Right | \\ NEQ 1..

Increasing and decreasing function

About the function that increases on the interval under consideration it is customary to talk about increasing function Then, when the greater value of X corresponds to the greater value of the function y \u003d f (x). From here it turns out that taking two arbitrary values \u200b\u200bof the argument x_ (1) and x_ (2) from the interval under consideration, with x_ (1)\u003e x_ (2), it will be y (x_ (1))\u003e y (x_ (2)).

Function that decreases on the interval under consideration is called decreasing function Then, when a greater value of X corresponds to the smaller value of the function y (x). From here it turns out that the two arbitrary values \u200b\u200bof the argument x_ (1) and x_ (2), with x_ (1)\u003e x_ (2), will be y (x_ (2), will be out of this< y(x_{2}) .

Roots function It is customary to call the points in which the function f \u003d y (x) crosses the abscissa axis (they are obtained as a result of solving the equation y (x) \u003d 0).

a) if at x\u003e 0 an even function increases, then it decreases with x< 0

b) when with x\u003e 0 an even function decreases, then it increases with x< 0

c) when with x\u003e 0 an odd function increases, then it increases with x< 0

d) when an odd function will decrease with x\u003e 0, then it will decrease with x< 0

Extreme function

Point of minimum function Y \u003d F (x) It is customary to call this point x \u003d x_ (0), in which its neighborhood will have other points (except for the point x \u003d x_ (0)), and the inequality f (x)\u003e F will be performed (x_ (0)). y_ (min) - designation function at MIN point.

Point of maximum function Y \u003d F (x) It is customary to call such a point x \u003d x_ (0), in which its neighborhood will have other points (except for the point x \u003d x_ (0)), and then the inequality F (x) will be performed< f(x^{0}) . y_{max} - обозначение функции в точке max.

Prerequisite

According to the farm theorem: F "(x) \u003d 0 when the function f (x), which is differentiated at the point x_ (0), the extremum will appear at this point.

Sufficient condition

1. When the derivative value changes from the plus to minus, then x_ (0) will be a minimum point;
2. x_ (0) - It will be a maximum point only when the derivative changes a sign from a minus plus when switching through a stationary point x_ (0).

The greatest and smallest value of the function on the interval

Computing steps:

1. It is searched for a derivative F "(x);
2. There are stationary and critical points of the function and choose the segment belonging;
3. There are values \u200b\u200bof the function f (x) in stationary and critical points and segments. Less from the results obtained will be the smallest value of the function, and greater - great.