even, if for all \(x\) from its domain is true: \(f(-x)=f(x)\) .
The graph of an even function is symmetrical about the \(y\) axis:
Example: the function \(f(x)=x^2+\cos x\) is even, because \(f(-x)=(-x)^2+\cos((-x))=x^2+\cos x=f(x)\).
\(\blacktriangleright\) The function \(f(x)\) is called odd, if for all \(x\) from its domain is true: \(f(-x)=-f(x)\) .
The graph of an odd function is symmetrical with respect to the origin:
Example: the function \(f(x)=x^3+x\) is odd because \(f(-x)=(-x)^3+(-x)=-x^3-x=-(x^3+x)=-f(x)\).
\(\blacktriangleright\) Functions that are neither even nor odd are called functions general view. Such a function can always be uniquely represented as the sum of an even and an odd function.
For example, the function \(f(x)=x^2-x\) is the sum of an even function \(f_1=x^2\) and an odd function \(f_2=-x\) .
\(\blacktriangleright\) Some properties:
1) The product and quotient of two functions of the same parity is an even function.
2) The product and quotient of two functions of different parity is an odd function.
3) The sum and difference of even functions is an even function.
4) The sum and difference of odd functions is an odd function.
5) If \(f(x)\) is an even function, then the equation \(f(x)=c \ (c\in \mathbb(R)\) ) has a unique root if and only if, when \(x =0\) .
6) If \(f(x)\) is an even or odd function, and the equation \(f(x)=0\) has a root \(x=b\) , then this equation will necessarily have a second root \(x =-b\) .
\(\blacktriangleright\) A function \(f(x)\) is called periodic on \(X\) if for some number \(T\ne 0\) we have \(f(x)=f(x+T) \) , where \(x, x+T\in X\) . The smallest \(T\) , for which this equality holds, is called the main (basic) period of the function.
A periodic function has any number of the form \(nT\) , where \(n\in \mathbb(Z)\) will also be a period.
Example: any trigonometric function is periodic;
for the functions \(f(x)=\sin x\) and \(f(x)=\cos x\) the principal period is \(2\pi\) , for the functions \(f(x)=\mathrm( tg)\,x\) and \(f(x)=\mathrm(ctg)\,x\) main period is \(\pi\) .
In order to plot a periodic function, you can plot its graph on any segment of length \(T\) (main period); then the graph of the entire function is completed by shifting the constructed part by an integer number of periods to the right and to the left:
\(\blacktriangleright\) The domain \(D(f)\) of the function \(f(x)\) is the set consisting of all values of the argument \(x\) for which the function makes sense (is defined).
Example: the function \(f(x)=\sqrt x+1\) has a domain of definition: \(x\in
Task 1 #6364
Task level: Equal to the Unified State Examination
For what values of the parameter \(a\) the equation
has a unique solution?
Note that since \(x^2\) and \(\cos x\) - even functions, then if the equation has a root \(x_0\) , it will also have a root \(-x_0\) .
Indeed, let \(x_0\) be a root, that is, the equality \(2x_0^2+a\mathrm(tg)\,(\cos x_0)+a^2=0\) right. Substitute \(-x_0\) : \(2 (-x_0)^2+a\mathrm(tg)\,(\cos(-x_0))+a^2=2x_0^2+a\mathrm(tg)\,(\cos x_0)+a ^2=0\).
Thus, if \(x_0\ne 0\) , then the equation will already have at least two roots. Therefore, \(x_0=0\) . Then:
We got two parameter values \(a\) . Note that we have used the fact that \(x=0\) is exactly the root of the original equation. But we never used the fact that he is the only one. Therefore, it is necessary to substitute the resulting values of the parameter \(a\) into the original equation and check for which exactly \(a\) the root \(x=0\) will indeed be unique.
1) If \(a=0\) , then the equation will take the form \(2x^2=0\) . Obviously, this equation has only one root \(x=0\) . Therefore, the value \(a=0\) suits us.
2) If \(a=-\mathrm(tg)\,1\) , then the equation takes the form \ We rewrite the equation in the form \ Because \(-1\leqslant \cos x\leqslant 1\), then \(-\mathrm(tg)\,1\leqslant \mathrm(tg)\,(\cos x)\leqslant \mathrm(tg)\,1\). Therefore, the values of the right side of the equation (*) belong to the interval \([-\mathrm(tg)^2\,1; \mathrm(tg)^2\,1]\).
Since \(x^2\geqslant 0\) , then the left side of equation (*) is greater than or equal to \(0+ \mathrm(tg)^2\,1\) .
Thus, equality (*) can only hold when both sides of the equation are equal to \(\mathrm(tg)^2\,1\) . And this means that \[\begin(cases) 2x^2+\mathrm(tg)^2\,1=\mathrm(tg)^2\,1 \\ \mathrm(tg)\,1\cdot \mathrm(tg)\ ,(\cos x)=\mathrm(tg)^2\,1 \end(cases) \quad\Leftrightarrow\quad \begin(cases) x=0\\ \mathrm(tg)\,(\cos x) =\mathrm(tg)\,1 \end(cases)\quad\Leftrightarrow\quad x=0\] Therefore, the value \(a=-\mathrm(tg)\,1\) suits us.
Answer:
\(a\in \(-\mathrm(tg)\,1;0\)\)
Task 2 #3923
Task level: Equal to the Unified State Examination
Find all values of the parameter \(a\) , for each of which the graph of the function \
symmetrical about the origin.
If the graph of a function is symmetric with respect to the origin, then such a function is odd, that is, \(f(-x)=-f(x)\) is satisfied for any \(x\) from the function's domain. Thus, it is required to find those parameter values for which \(f(-x)=-f(x).\)
\[\begin(aligned) &3\mathrm(tg)\,\left(-\dfrac(ax)5\right)+2\sin \dfrac(8\pi a+3x)4= -\left(3\ mathrm(tg)\,\left(\dfrac(ax)5\right)+2\sin \dfrac(8\pi a-3x)4\right)\quad \Rightarrow\quad -3\mathrm(tg)\ ,\dfrac(ax)5+2\sin \dfrac(8\pi a+3x)4= -\left(3\mathrm(tg)\,\left(\dfrac(ax)5\right)+2\ sin \dfrac(8\pi a-3x)4\right) \quad \Rightarrow\\ \Rightarrow\quad &\sin \dfrac(8\pi a+3x)4+\sin \dfrac(8\pi a- 3x)4=0 \quad \Rightarrow \quad2\sin \dfrac12\left(\dfrac(8\pi a+3x)4+\dfrac(8\pi a-3x)4\right)\cdot \cos \dfrac12 \left(\dfrac(8\pi a+3x)4-\dfrac(8\pi a-3x)4\right)=0 \quad \Rightarrow\quad \sin (2\pi a)\cdot \cos \ frac34 x=0 \end(aligned)\]
The last equation must hold for all \(x\) from the domain \(f(x)\) , hence \(\sin(2\pi a)=0 \Rightarrow a=\dfrac n2, n\in\mathbb(Z)\).
Answer:
\(\dfrac n2, n\in\mathbb(Z)\)
Task 3 #3069
Task level: Equal to the Unified State Examination
Find all values of the parameter \(a\) , for each of which the equation \ has 4 solutions, where \(f\) is an even periodic function with period \(T=\dfrac(16)3\) defined on the entire real line , and \(f(x)=ax^2\) for \(0\leqslant x\leqslant \dfrac83.\)
(Task from subscribers)
Task 4 #3072
Task level: Equal to the Unified State Examination
Find all values \(a\) , for each of which the equation \
has at least one root.
(Task from subscribers)
We rewrite the equation in the form \
and consider two functions: \(g(x)=7\sqrt(2x^2+49)\) and \(f(x)=3|x-7a|-6|x|-a^2+7a\ ) .
The function \(g(x)\) is even, has a minimum point \(x=0\) (and \(g(0)=49\) ).
The function \(f(x)\) for \(x>0\) is decreasing, and for \(x<0\)
– возрастающей, следовательно, \(x=0\)
– точка максимума.
Indeed, for \(x>0\) the second module expands positively (\(|x|=x\) ), therefore, regardless of how the first module expands, \(f(x)\) will be equal to \( kx+A\) , where \(A\) is an expression from \(a\) , and \(k\) is equal to either \(-9\) or \(-3\) . For \(x<0\)
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 the functions \(f\) and \(g\) have at least one intersection point. Therefore, you need: \ Solving this set of systems, we get the answer: \\]
Answer:
\(a\in \(-7\)\cup\)
Task 5 #3912
Task level: Equal to the Unified State Examination
Find all values of the parameter \(a\) , for each of which the equation \
has six different solutions.
Let's make the substitution \((\sqrt2)^(x^3-3x^2+4)=t\) , \(t>0\) . Then the equation will take the form \
We will gradually write out the conditions under which the original equation will have six solutions.
notice, that quadratic equation\((*)\) can have at most two solutions. Any cubic equation \(Ax^3+Bx^2+Cx+D=0\) can have no more than three solutions. Therefore, if the equation \((*)\) has two different solutions (positive!, since \(t\) must be greater than zero) \(t_1\) and \(t_2\) , then, having made the reverse substitution, we we get: \[\left[\begin(gathered)\begin(aligned) &(\sqrt2)^(x^3-3x^2+4)=t_1\\ &(\sqrt2)^(x^3-3x^2 +4)=t_2\end(aligned)\end(gathered)\right.\] Since any positive number can be represented as \(\sqrt2\) to some extent, for example, \(t_1=(\sqrt2)^(\log_(\sqrt2) t_1)\), then the first equation of the set will be rewritten in the form \
As we have already said, any cubic equation has no more than three solutions, therefore, each equation from the set will have no more than three solutions. This means that the whole set will have no more than six solutions.
This means that in order for the original equation to have six solutions, the quadratic equation \((*)\) must have two different solutions, and each resulting cubic equation (from the set) must have three different solutions (and not a single solution of one equation should coincide with which - or by the decision of the second!)
Obviously, if the quadratic equation \((*)\) has one solution, then we will not get six solutions for the original equation.
Thus, the solution plan becomes clear. Let's write out the conditions that must be met point by point.
1) For the equation \((*)\) to have two different solutions, its discriminant must be positive: \
2) We also need both roots to be positive (because \(t>0\) ). If the product of two roots is positive and their sum is positive, then the roots themselves will be positive. Therefore, you need: \[\begin(cases) 12-a>0\\-(a-10)>0\end(cases)\quad\Leftrightarrow\quad a<10\]
Thus, we have already provided ourselves with two distinct positive roots \(t_1\) and \(t_2\) .
3)
Let's look at this equation \
For what \(t\) will it have three different solutions? Thus, we have determined that both roots of the equation \((*)\) must lie in the interval \((1;4)\) . How to write this condition? Thus, we need to intersect the \(a\) parameter values found in the 1st, 2nd and 3rd paragraphs, and we will get the answer: \[\begin(cases) a\in (-\infty;8-2\sqrt3)\cup(8+2\sqrt3;+\infty)\\ a<10\\
4 Definition 1. The function is called even
(odd
) if together with each value of the variable Thus, a function can be even or odd only when its domain of definition is symmetrical with respect to the origin of coordinates on the real line (numbers X and - X simultaneously belong Function Function Function The graph of an even function is symmetrical about the axis OU, since if the point When proving whether a function is even or odd, the following statements are useful. Theorem 1. a) The sum of two even (odd) functions is an even (odd) function. b) The product of two even (odd) functions is an even function. c) The product of an even and an odd function is an odd function. d) If f is an even function on the set X, and the function g
defined on the set e) If f is an odd function on the set X, and the function g
defined on the set Proof. Let us prove, for example, b) and d). b) Let d) Let f
is an even function. Then. The other assertions of the theorem are proved similarly. The theorem has been proven. Theorem 2. Any function Proof. Function Function Definition 2. Function Such a number T called period
functions Definition 1 implies that if T– function period Definition 3. The smallest of the positive periods of a function is called its main
period. Theorem 3. If T is the main period of the function f, then the remaining periods are multiples of it. Proof. Assume the opposite, that is, that there is a period that is It is well known that trigonometric functions are periodic. Main period (because Meaning T, determined from the first equality, cannot be a period, since it depends on X, i.e. is a function of X, not a constant number. The period is determined from the second equality: An example of a more complex periodic function is the Dirichlet function Note that if T is a rational number, then for any rational number T. Therefore, any rational number T is the period of the Dirichlet function. It is clear that this function has no main period, since there are positive rational numbers arbitrarily close to zero (for example, a rational number can be made by choosing n arbitrarily close to zero). Theorem 4. If function f
set on the set X and has a period T, and the function g
set on the set Proof. We have therefore that is, the assertion of the theorem is proved. For example, since cos
x
has a period Definition 4. Functions that are not periodic are called non-periodic
. Attention! The slide preview is for informational purposes only and may not represent the full extent of the presentation. If you are interested in this work, please download the full version. Goals: Equipment: multimedia installation, interactive whiteboard, handouts. Forms of work: frontal and group with elements of search and research activities. Information sources: DURING THE CLASSES 1. Organizational moment Setting goals and objectives of the lesson. 2.
Checking homework No. 10.17 (Problem book 9th grade A.G. Mordkovich). a) at = f(X), f(X) = b) f (–2) = –3; f (0) = –1; f(5) = 69; c) 1. D( f) = [– 2; + ∞) (Did you use the feature exploration algorithm?) Slide.
2. Let's check the table that you were asked on the slide. Domain Function zeros Constancy intervals x = -5, х € (–5;3) U х € (–∞;–5) U x ∞ -5, х € (–5;3) U х € (–∞;–5) U x ≠ -5, x € (–∞; –5) U x € (–5; 2) 3.
Knowledge update – Functions are given. X ≠ 0 and not defined. 4. New material - While doing this work, guys, we have revealed one more property of the function, unfamiliar to you, but no less important than the others - this is the evenness and oddness of the function. Write down the topic of the lesson: “Even and odd functions”, our task is to learn how to determine the even and odd functions, find out the significance of this property in the study of functions and plotting. Def. one Function at = f (X) defined on the set X is called even, if for any value XЄ X in progress equality f (–x) = f (x). Give examples. Def. 2 Function y = f(x), defined on the set X is called odd, if for any value XЄ X the equality f(–х)= –f(х) is fulfilled. Give examples. Where did we meet the terms "even" and "odd"? 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 at x and - x, thus it is assumed that the function is also defined at the value X, and at - X. ODA 3. If a number set together with each of its elements x contains the opposite element x, then the set X is called a symmetric set. Examples: (–2;2), [–5;5]; (∞;∞) are symmetric sets, and , [–5;4] are nonsymmetric. - Do even functions have a domain of definition - a symmetric set? The odd ones? Slide
Algorithm for examining a function for parity 1. Determine whether the domain of the function is symmetrical. 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): Examples:
Investigate the function for parity a) at= x 5 +; b) at= ; in) at= . Solution. a) h (x) \u003d x 5 +, 1) D(h) = (–∞; 0) U (0; +∞), symmetric set. 2) h (- x) \u003d (-x) 5 + - x5 - \u003d - (x 5 +), 3) h (- x) \u003d - h (x) \u003d\u003e function h(x)= x 5 + odd. b) y =, at = f(X), D(f) = (–∞; –9)? (–9; +∞), asymmetric set, so the function is neither even nor odd. in) 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) y \u003d x 2 (2x - x 3), b) y \u003d Mutual check on slide. 6. Homework: №11.11, 11.21,11.22; Proof of the geometric meaning of the parity property. *** (Assignment of the USE option). 1. The odd function y \u003d f (x) is defined on the entire real 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) = at X = 3. 7. Summing up Hide Show Let the function be given by the formula: y=2x^(2)-3 . By assigning any value to the independent variable x , you can use this formula to calculate the corresponding values of the dependent variable y . For example, if x=-0.5 , then using the formula, we get that the corresponding value of y is y=2 \cdot (-0.5)^(2)-3=-2.5 . Given any value taken by the x argument in the formula y=2x^(2)-3 , only one function value can be calculated that corresponds to it. The function can be represented as a table: Using this table, you can figure out that for the value of the argument -1, the value of the function -3 will correspond; and the value x=2 will correspond to y=0, and so on. It is also important to know that each argument value in the table corresponds to only one function value. More functions can be set using graphs. With the help of the graph, it is established which value of the function correlates with a certain value of x. Most often, this will be an approximate value of the function. The function is even function, when f(-x)=f(x) for any x from the domain. Such a function will be symmetrical about the Oy axis. The function is odd function when f(-x)=-f(x) for any x in the domain. Such a function will be symmetrical about the origin O (0;0) . The function is not even, nor odd and called general function when it does not have symmetry about the axis or origin. We examine the following function for parity: f(x)=3x^(3)-7x^(7) D(f)=(-\infty ; +\infty) with a symmetrical domain of definition about the origin. f(-x)= 3 \cdot (-x)^(3)-7 \cdot (-x)^(7)= -3x^(3)+7x^(7)= -(3x^(3)-7x^(7))= -f(x). Hence, the function f(x)=3x^(3)-7x^(7) is odd. The function y=f(x) , in the domain of which f(x+T)=f(x-T)=f(x) is true for any x, is called periodic function with period T \neq 0 . Repetition of the graph of the function on any segment of the abscissa axis, which has length T . Intervals where the function is positive, that is, f (x) > 0 - segments of the abscissa axis, which correspond to the points of the graph of the function that lie above the abscissa axis. f(x) > 0 on (x_(1); x_(2)) \cup (x_(3); +\infty) Gaps where the function is negative, i.e. 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)) bounded from below it is customary to call a function y=f(x), x \in X when there exists a number A for which the inequality f(x) \geq A holds for any x \in X . An example of a function bounded below: y=\sqrt(1+x^(2)) since y=\sqrt(1+x^(2)) \geq 1 for any x . bounded from above a function y=f(x), x \in X is called if there exists a number B for which the inequality f(x) \neq B holds for any x \in X . An example of a function bounded below: y=\sqrt(1-x^(2)), x \in [-1;1] since y=\sqrt(1+x^(2)) \neq 1 for any x \in [-1;1] . Limited it is customary to call a function y=f(x), x \in X when there exists a number K > 0 for which the inequality \left | f(x) \right | \neq K for any x \in X . Example of a bounded function: y=\sin x is bounded on the whole number line because \left | \sin x \right | \neq 1. It is customary to speak of a function that increases on the interval under consideration as increasing function when a larger value of x will correspond to a larger value of the function y=f(x) . From here it turns out that taking from the considered interval two arbitrary values of the argument x_(1) and x_(2) , and x_(1) > x_(2) , it will be y(x_(1)) > y(x_(2)) . A function that decreases on the interval under consideration is called decreasing function when a larger value of x will correspond to a smaller value of the function y(x) . From here it turns out that taking from the considered interval two arbitrary values of the argument x_(1) and x_(2) , and x_(1) > x_(2) , it will be y(x_(1))< y(x_{2})
. Function roots it is customary to name the points at which the function F=y(x) intersects the abscissa axis (they are obtained as a result of solving the equation y(x)=0 ). a) If an even function increases for x > 0, then it decreases for x< 0
b) When an even function decreases for x > 0, then it increases for x< 0
c) When an odd function increases for x > 0, then it also increases for x< 0
d) When an odd function decreases for x > 0, then it will also decrease for x< 0
Function minimum point y=f(x) it is customary to call such a point x=x_(0) , in which its neighborhood will have other points (except for the point x=x_(0) ), and for them then the inequality f(x) > f (x_(0)) . y_(min) - designation of the function at the point min. Function maximum point y=f(x) it is customary to call such a point x=x_(0) , in which its neighborhood will have other points (except for the point x=x_(0) ), and then the inequality f(x) will be satisfied for them< f(x^{0})
. y_{max}
- обозначение функции в точке max.
According to Fermat's theorem: f"(x)=0, then when the function f(x) , which is differentiable at the point x_(0) , an extremum will appear at this point. Calculation steps: Even function.
Even A function whose sign does not change when the sign is changed is called x. x equality f(–x) = f(x). Sign x does not affect sign y. The graph of an even function is symmetrical about the coordinate axis (Fig. 1). Even function examples: y= cos x y = x 2 y = –x 2 y = x 4 y = x 6 y = x 2 + x Explanation: odd function.
odd is a function whose sign changes when the sign is changed x. In other words, for any value x equality f(–x) = –f(x). The graph of an odd function is symmetrical with respect to the origin (Fig. 2). Examples of an odd function: y= sin x y = x 3 y = –x 3 Explanation: Take the function y = - x 3 . Properties of even and odd functions:
NOTE:
Not all features are even or odd. There are functions that are not subject to such gradation. For example, the root function at = √X does not apply to either even or odd functions (Fig. 3). When listing the properties of such functions, an appropriate description should be given: neither even nor odd. Periodic functions.
As you know, periodicity is the repetition of certain processes at a certain interval. The functions describing these processes are called periodic functions. That is, these are functions in whose graphs there are elements that repeat at certain numerical intervals.
Consider the function \(f(x)=x^3-3x^2+4\) .
Can be multiplied: \
Therefore, its zeros are: \(x=-1;2\) .
If we find the derivative \(f"(x)=3x^2-6x\) , then we get two extreme points \(x_(max)=0, x_(min)=2\) .
Therefore, the graph looks like this: 
We see that any horizontal line \(y=k\) , where \(0
Thus, you need: \[\begin(cases) 0<\log_{\sqrt2}t_1<4\\ 0<\log_{\sqrt2}t_2<4\end{cases}\qquad (**)\]
Let's also note right away 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 \(x^3-3x^2+4=\log_(\sqrt2) t_1\) and \(x^3-3x^2+4=\log_(\sqrt2) t_2\) will have different roots.
The \((**)\) system can be rewritten like this: \[\begin(cases) 1
We will not explicitly write out the roots.
Consider the function \(g(t)=t^2+(a-10)t+12-a\) . Its graph is a parabola with upward branches, which has two points of intersection with the abscissa axis (we wrote this condition in paragraph 1)). How should its graph look like so that the points of intersection with the abscissa axis are in the interval \((1;4)\) ? So: 
Firstly, the values \(g(1)\) and \(g(4)\) of the function at the points \(1\) and \(4\) must be positive, and secondly, the vertex of the parabola \(t_0\ ) must also be in the interval \((1;4)\) . Therefore, the system can be written: \[\begin(cases) 1+a-10+12-a>0\\ 4^2+(a-10)\cdot 4+12-a>0\\ 1<\dfrac{-(a-10)}2<4\end{cases}\quad\Leftrightarrow\quad 4
meaning - X also belongs 
and the equality
). For example, the function 
is neither even nor odd, since its domain of definition 
not symmetrical about the origin.
even, because 
symmetrical with respect to the origin of coordinates and.
odd because 
and 
.
is neither even nor odd, since although 
and is symmetric with respect to the origin, equalities (11.1) are not satisfied. For example,.
also belongs to the graph. The graph of an odd function is symmetrical about the origin, because if 
belongs to the graph, then the point 
also belongs to the graph.
, then the function 
- even.
and even (odd), then the function 
- even (odd).
and 
are even functions. Then, therefore. The case of odd functions is considered similarly 
and 
.
, defined on the set X, which is symmetric with respect to the origin, can be represented as the sum of an even and an odd function.
can be written in the form
.
is even, since 
, and the function 
is odd because. In this way, 
, where 
- even, and 
is an odd function. The theorem has been proven.
called periodical
if there is a number 
, such that for any 
numbers 
and 
also belong to the domain of definition 
and the equalities
.
, then the number T too
is the period of the function
(because when replacing T on the - T equality is maintained). Using the method of mathematical induction, it can be shown that if T– function period f, then and 
, is also a period. It follows that if a function has a period, then it has infinitely many periods.
functions f
(
>0), not multiple T. Then, dividing 
on the T with the remainder, we get 
, where 
. That's why
– function period f, and 
, which contradicts the fact that T is the main period of the function f. The assertion of the theorem follows from the obtained contradiction. The theorem has been proven.
and 
equals 
,
and 
. Find the period of the function 
. Let 
is the period of this function. Then
.
ororor 
.
. There are infinitely many periods 
the smallest positive period is obtained when 
:
. This is the main period of the function 
.
and 
are rational numbers under rational X and irrational when irrational X. That's why
, then the complex function 
also has a period T.
, then the functions 
have a period 
.
Back forward
1. Algebra class 9 A.G. Mordkovich. Textbook.
2. Algebra Grade 9 A.G. Mordkovich. Task book.
3. Algebra grade 9. Tasks for learning and development of students. Belenkova E.Yu. Lebedintseva E.A.
2. E( f) = [– 3; + ∞)
3. f(X) = 0 for X ~ 0,4
4. f(X) >0 at 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 hire = - 3, at naib doesn't exist
8. The function is continuous.Fill the table
Coordinates of the points of intersection of the graph with Oy
x = 2
U(2;∞)
U (–3;2)
x ≠ 2
U(2;∞)
U (–3;2)
x ≠ 2
U(2;∞)
– Specify the domain of definition for each function.
– Compare the value of each function for each pair of argument values: 1 and – 1; 2 and - 2.
– For which of the given functions in the domain of definition are the equalities f(– X)
= f(X), f(– X) = – f(X)? (put the data in 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) =
6. f(X)=
X >
–1
So, let's find the definitions in the textbook and read (p. 110) . Slide
Which of these functions will be even, do you think? Why? Which are odd? Why?
For any function of the form at= x n, where n is an integer, it can be argued that the function is odd for n is odd and the function is even for n- even.
– View functions at= and at = 2X– 3 is neither even nor odd, because equalities are not met f(– X) = – f(X), f(–
X) = f(X)
- If D( f) is an asymmetric set, then what is the function?
– Thus, if the function at = f(X) is even or odd, then its domain of definition is D( f) is a symmetric set. But 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 the domain of definition is a necessary condition, but not a sufficient one.
– So how can we investigate the function for parity? Let's try to write an algorithm.
a); b) y \u003d x (5 - x 2).2. Examine the function for parity:
3. In fig. plotted at = f(X), for all X, satisfying the condition X?
0.
Plot the Function at = f(X), if at = f(X) is an even function. 
3. In fig. plotted at = f(X), for all x satisfying x? 0.
Plot the Function at = f(X), if at = f(X) is an odd function. 
Ways to set a function
x −2
−1
0
1
2
3
y −4
−3
−2
−1
0
1
Even and odd function
Periodic function
Function limitation
Increasing and decreasing function
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Function extremes
Necessary condition
Sufficient condition
The largest and smallest value of the function on the interval
Let's take a function y = x 2 or y = –x 2 .
For any value x the function is positive. Sign x does not affect sign y. The graph is symmetrical about the coordinate axis. This is an even function.
All values at it will have a minus sign. That is the sign x affects the sign y. If the independent variable is a positive number, then the function is positive; if the independent variable is a negative number, then the function is negative: f(–x) = –f(x).
The graph of the function is symmetrical about the origin. This is an odd function.
