even, if for all \(x\) from its domain is true: \(f(-x)=f(x)\) .
Schedule even function 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: \(f(-x)=-f(x)\) .
Schedule odd function symmetrical about 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. This function can always the only way represent 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 \(\cos x\) are even functions, 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 \(a\) 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 \ As \(-1\leqslant \cos x\leqslant 1\), 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 segment \([-\mathrm(tg)^2\,1; \mathrm(tg)^2\,1]\).
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, \(f(-x)=-f(x)\) holds 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)
Since \(f(x)\) is an even function, its graph is symmetrical with respect to the y-axis, therefore, when \(-\dfrac83\leqslant x\leqslant 0\)\(f(x)=ax^2\) . Thus, at \(-\dfrac83\leqslant x\leqslant \dfrac83\), and this is a segment of length \(\dfrac(16)3\) , the function \(f(x)=ax^2\) .
1) Let \(a>0\) . Then the graph of the function \(f(x)\) will look like this: 
Then, in order for the equation to have 4 solutions, it is necessary that the graph \(g(x)=|a+2|\cdot \sqrtx\) pass through the point \(A\) : 
Hence, \[\dfrac(64)9a=|a+2|\cdot \sqrt8 \quad\Leftrightarrow\quad \left[\begin(gathered)\begin(aligned) &9(a+2)=32a\\ &9(a +2)=-32a \end(aligned) \end(gathered)\right. \quad\Leftrightarrow\quad \left[\begin(gathered)\begin(aligned) &a=\dfrac(18)(23)\\ &a=-\dfrac(18)(41) \end(aligned) \end( gathered)\right.\] Since \(a>0\) , then \(a=\dfrac(18)(23)\) is fine.
2) Let \(a<0\)
. Тогда картинка окажется симметричной относительно начала координат:
We need the graph \(g(x)\) to pass through the point \(B\) : \[\dfrac(64)9a=|a+2|\cdot \sqrt(-8) \quad\Leftrightarrow\quad \left[\begin(gathered)\begin(aligned) &a=\dfrac(18)(23 )\\ &a=-\dfrac(18)(41) \end(aligned) \end(gathered)\right.\] Since \(a<0\)
, то подходит \(a=-\dfrac{18}{41}\)
.
3) The case where \(a=0\) is not suitable, because then \(f(x)=0\) for all \(x\) , \(g(x)=2\sqrtx\) and The equation will only have 1 root.
Answer:
\(a\in \left\(-\dfrac(18)(41);\dfrac(18)(23)\right\)\)
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: \ \\]
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.
Note that the 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 degree, 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? had four distinct non-zero roots, representing together with \(x=0\) an arithmetic progression. Note that the function \(y=25x^4+25(a-1)x^2-4(a-7)\) is even, so if \(x_0\) is the root of the equation \((*)\ ) , then \(-x_0\) will also be its root. Then it is necessary that the roots of this equation be numbers ordered in ascending order: \(-2d, -d, d, 2d\) (then \(d>0\) ). It is then that these five numbers will form an arithmetic progression (with the difference \(d\) ). For these roots to be the numbers \(-2d, -d, d, 2d\) , it is necessary that the numbers \(d^(\,2), 4d^(\,2)\) be the roots of the equation \(25t^2 +25(a-1)t-4(a-7)=0\) . Then by Vieta's theorem: We rewrite the equation in the form \
and consider two functions: \(g(x)=20a-a^2-2^(x^2+2)\) and \(f(x)=13|x|-2|5x+12a|\) . 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 \(-2\)\cup\) Hide Show
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
The function \(g(x)\) has a maximum point \(x=0\) (and \(g_(\text(top))=g(0)=-a^2+20a-4\)):
\(g"(x)=-2^(x^2+2)\cdot \ln 2\cdot 2x\). Zero derivative: \(x=0\) . For \(x<0\)
имеем: \(g">0\) , for \(x>0\) : \(g"<0\)
.
The function \(f(x)\) for \(x>0\) is increasing, and for \(x<0\)
– убывающей, следовательно, \(x=0\)
– точка минимума.
Indeed, for \(x>0\) the first module expands positively (\(|x|=x\) ), therefore, regardless of how the second module expands, \(f(x)\) will be equal to \( kx+A\) , where \(A\) is an expression from \(a\) , and \(k\) is either \(13-10=3\) or \(13+10=23\) . For \(x<0\)
наоборот: первый модуль раскроется отрицательно и \(f(x)=kx+A\)
, где \(k\)
равно либо \(-3\)
, либо \(-23\)
.
Let's find the value \(f\) at the minimum point: \
Ways to set a function
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:
| x | −2 | −1 | 0 | 1 | 2 | 3 |
| y | −4 | −3 | −2 | −1 | 0 | 1 |
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.
Even and odd 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.
Periodic function
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))
Function limitation
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.
Increasing and decreasing function
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 extremes
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 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.
Necessary condition
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.
Sufficient condition
- When the sign of the derivative changes from plus to minus, then x_(0) will be the minimum point;
- x_(0) - will be a maximum point only when the derivative changes sign from minus to plus when passing through the stationary point x_(0) .
The largest and smallest value of the function on the interval
Calculation steps:
- Looking for derivative f"(x) ;
- Stationary and critical points of the function are found and those belonging to the interval are chosen;
- The values of the function f(x) are found at the stationary and critical points and ends of the segment. The smallest of the results will be the smallest value of the function, and more - greatest.
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:
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.
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 .
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.
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.
The dependence of the variable y on the variable x, in which each value of x corresponds to a single value of y is called a function. The notation is y=f(x). Each function has a number of basic properties, such as monotonicity, parity, periodicity, and others.
Consider the parity property in more detail.
A function y=f(x) is called even if it satisfies the following two conditions:
2. The value of the function at the point x belonging to the scope of the function must be equal to the value of the function at the point -x. That is, for any point x, from the domain of the function, the following equality f (x) \u003d f (-x) must be true.
Graph of an even function
If you build a graph of an even function, it will be symmetrical about the y-axis.
For example, the function y=x^2 is even. Let's check it out. The domain of definition is the entire numerical axis, which means that it is symmetrical about the point O.
Take an arbitrary x=3. f(x)=3^2=9.
f(-x)=(-3)^2=9. Therefore, f(x) = f(-x). Thus, both conditions are satisfied for us, which means that the function is even. Below is a graph of the function y=x^2.
The figure shows that the graph is symmetrical about the y-axis.
Graph of an odd function
A function y=f(x) is called odd if it satisfies the following two conditions:
1. The domain of the given function must be symmetrical with respect to the point O. That is, if some point a belongs to the domain of the function, then the corresponding point -a must also belong to the domain of the given function.
2. For any point x, from the domain of the function, the following equality f (x) \u003d -f (x) must be satisfied.
The graph of an odd function is symmetrical with respect to the point O - the origin. For example, the function y=x^3 is odd. Let's check it out. The domain of definition is the entire numerical axis, which means that it is symmetrical about the point O.
Take an arbitrary x=2. f(x)=2^3=8.
f(-x)=(-2)^3=-8. Therefore f(x) = -f(x). Thus, both conditions are satisfied for us, which means that the function is odd. Below is a graph of the function y=x^3.
The figure clearly shows that the odd function y=x^3 is symmetrical with respect to the origin.
Which to one degree or another were familiar to you. It was also noted there that the stock of function properties will be gradually replenished. Two new properties will be discussed in this section.
Definition 1.
The function y \u003d f (x), x є X, is called even if for any value x from the set X the equality f (-x) \u003d f (x) is true.
Definition 2.
The function y \u003d f (x), x є X, is called odd if for any value x from the set X the equality f (-x) \u003d -f (x) is true.
Prove that y = x 4 is an even function.
Decision. We have: f (x) \u003d x 4, f (-x) \u003d (-x) 4. But (-x) 4 = x 4 . Hence, for any x, the equality f (-x) = f (x), i.e. the function is even.
Similarly, it can be proved that the functions y - x 2, y \u003d x 6, y - x 8 are even.
Prove that y = x 3 is an odd function.
Decision. We have: f (x) \u003d x 3, f (-x) \u003d (-x) 3. But (-x) 3 = -x 3 . Hence, for any x, the equality f (-x) \u003d -f (x), i.e. the function is odd.
Similarly, it can be proved that the functions y \u003d x, y \u003d x 5, y \u003d x 7 are odd.
You and I have repeatedly convinced ourselves that new terms in mathematics most often have an “earthly” origin, i.e. they can be explained in some way. This is the case for both even and odd functions. See: y - x 3, y \u003d x 5, y \u003d x 7 are odd functions, while y \u003d x 2, y \u003d x 4, y \u003d x 6 are even functions. And in general, for any function of the form y \u003d x "(below we will specifically study these functions), where n is a natural number, we can conclude: if n is not even number, then the function y \u003d x "is odd; if n is an even number, then the function y \u003d xn is even.
There are also functions that are neither even nor odd. Such, for example, is the function y \u003d 2x + 3. Indeed, f (1) \u003d 5, and f (-1) \u003d 1. As you can see, here Hence, neither the identity f (-x) \u003d f ( x), nor the identity f(-x) = -f(x).
So, a function can be even, odd, or neither.
Studying the question of whether given function even or odd, is usually called the study of a function for parity.
In definitions 1 and 2 we are talking about the values of the function at the points x and -x. This assumes that the function is defined both at the point x and at the point -x. This means that the point -x belongs to the domain of the function at the same time as the point x. If a numerical set X together with each of its elements x contains the opposite element -x, then X is called a symmetric set. Let's say (-2, 2), [-5, 5], (-oo, +oo) are symmetric sets, while )