Whether the function is even or odd. Even and odd functions

The dependence of a variable y on a variable x, in which each value of x corresponds to a single value of y is called a function. For designation use the notation y=f(x). Each function has a number of basic properties, such as monotonicity, parity, periodicity and others.

Take a closer look at the parity property.

A function y=f(x) is called even if it satisfies the following two conditions:

2. The value of the function at point x, belonging to the domain of definition of the function, must be equal to the value of the function at point -x. That is, for any point x, the following equality must be satisfied from the domain of definition of the function: f(x) = f(-x).

Graph of an even function

If you plot a graph of an even function, it will be symmetrical about the Oy 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 it is symmetrical about point O.

Let's take an arbitrary x=3. f(x)=3^2=9.

f(-x)=(-3)^2=9. Therefore f(x) = f(-x). Thus, both conditions are met, which means the function is even. Below is a graph of the function y=x^2.

The figure shows that the graph is symmetrical about the Oy 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 definition of a given function must be symmetrical with respect to point O. That is, if some point a belongs to the domain of definition of the function, then the corresponding point -a must also belong to the domain of definition of the given function.

2. For any point x, the following equality must be satisfied from the domain of definition of the function: f(x) = -f(x).

The graph of an odd function is symmetrical with respect to point O - the origin of coordinates. 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 it is symmetrical about point O.

Let's take an arbitrary x=2. f(x)=2^3=8.

f(-x)=(-2)^3=-8. Therefore f(x) = -f(x). Thus, both conditions are met, which means 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 about the origin.

even, if for all \(x\) from its domain of definition the following 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 of definition the following is true: \(f(-x)=-f(x)\) .

The graph of an odd function is 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. 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 the even function \(f_1=x^2\) and the odd \(f_2=-x\) .

\(\blacktriangleright\) 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 parities is an odd function.

3) Sum and difference of even functions - even function.

4) Sum and difference of odd functions - 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 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 \(x =-b\) .

\(\blacktriangleright\) The function \(f(x)\) is called periodic on \(X\) if for some number \(T\ne 0\) the following holds: \(f(x)=f(x+T) \) , where \(x, x+T\in X\) . The smallest \(T\) for which this equality is satisfied is called the main (main) period of the function.

U periodic function 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 \(f(x)=\cos x\) main period is equal to \(2\pi\), the functions \(f(x)=\mathrm(tg)\,x\) and \(f(x)=\mathrm(ctg)\,x\) have a main period equal to \ (\pi\) .

In order to construct a graph of 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 left:

\(\blacktriangleright\) The domain \(D(f)\) of the function \(f(x)\) is a 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 Exam

At what values ​​of the parameter \(a\) does the equation

has a single 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. Let's 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 received two values ​​for the parameter \(a\) . Note that we 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, you need to substitute the resulting values ​​of the parameter \(a\) into the original equation and check for which specific \(a\) the root \(x=0\) will really 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 will take the form \ Let's rewrite the equation in the form \ Because \(-1\leqslant \cos x\leqslant 1\), That \(-\mathrm(tg)\,1\leqslant \mathrm(tg)\,(\cos x)\leqslant \mathrm(tg)\,1\). Consequently, 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 left side equation (*) is greater than or equal to \(0+ \mathrm(tg)^2\,1\) .

Thus, equality (*) can only be true 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 Exam

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 symmetrical about the origin, then such a function is odd, that is, \(f(-x)=-f(x)\) holds for any \(x\) from the domain of definition of the function. 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 be satisfied for all \(x\) from the domain of \(f(x)\) , therefore, \(\sin(2\pi a)=0 \Rightarrow a=\dfrac n2, n\in\mathbb(Z)\).

Answer:

\(\dfrac n2, n\in\mathbb(Z)\)

Task 3 #3069

Task level: Equal to the Unified State Exam

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 number 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 ordinate axis, therefore, when \(-\dfrac83\leqslant x\leqslant 0\)\(f(x)=ax^2\) . Thus, when \(-\dfrac83\leqslant x\leqslant \dfrac83\), and this is a segment of length \(\dfrac(16)3\) , 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 suitable.

2) Let \(a<0\) . Тогда картинка окажется симметричной относительно начала координат:


It is necessary that the graph \(g(x)\) passes 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 when \(a=0\) is not suitable, since then \(f(x)=0\) for all \(x\) , \(g(x)=2\sqrtx\) and the equation will have only 1 root.

Answer:

\(a\in \left\(-\dfrac(18)(41);\dfrac(18)(23)\right\)\)

Task 4 #3072

Task level: Equal to the Unified State Exam

Find all values ​​of \(a\) , for each of which the equation \

has at least one root.

(Task from subscribers)

Let's 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 and 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, when \(x>0\) the second module will open positively (\(|x|=x\) ), therefore, regardless of how the first module will open, \(f(x)\) will be equal to \( kx+A\) , where \(A\) is the expression of \(a\) and \(k\) is equal to either \(-9\) or \(-3\) . When \(x<0\) наоборот: второй модуль раскроется отрицательно и \(f(x)=kx+A\) , где \(k\) равно либо \(3\) , либо \(9\) .
Let's find the value of \(f\) 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 Exam

Find all values ​​of the parameter \(a\) , for each of which the equation \

has six different solutions.

Let's make the replacement \((\sqrt2)^(x^3-3x^2+4)=t\) , \(t>0\) . 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 a maximum of 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 by making 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 in the set will have no more than three solutions. This means that the entire set will have no more than six solutions.
This means that 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 any -by the decision of the second!)
Obviously, if the quadratic equation \((*)\) has one solution, then we will not get six solutions to the original equation.

Thus, the solution plan becomes clear. Let's write down the conditions that must be met point by point.

1) For the equation \((*)\) to have two different solutions, its discriminant must be positive: \

2) It is also necessary that both roots be positive (since \(t>0\) ). 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 different positive roots \(t_1\) and \(t_2\) .

3) Let's look at this equation \ For what \(t\) will it have three different solutions?
Consider the function \(f(x)=x^3-3x^2+4\) .
Can be factorized: \ Therefore, its zeros are: \(x=-1;2\) .
If we find the derivative \(f"(x)=3x^2-6x\) , then we get two extremum points \(x_(max)=0, x_(min)=2\) .
Therefore, the graph looks like this:


We see that any horizontal line \(y=k\) , where \(0 \(x^3-3x^2+4=\log_(\sqrt2) t\) had three different solutions, it is necessary that \(0<\log_ {\sqrt2}t<4\) .
Thus, you need: \[\begin(cases) 0<\log_{\sqrt2}t_1<4\\ 0<\log_{\sqrt2}t_2<4\end{cases}\qquad (**)\] Let's also 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, which means 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 as follows: \[\begin(cases) 1

Thus, we have determined that both roots of the equation \((*)\) must lie in the interval \((1;4)\) . How to write this condition?
We will not write down the roots explicitly.
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 x-axis (we wrote down this condition in paragraph 1)). What should its graph look like so that the points of intersection with the x-axis are in the interval \((1;4)\)? So:


Firstly, the values ​​\(g(1)\) and \(g(4)\) of the function at 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, we can write the system: \[\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\(a\) always has at least one root \(x=0\) . This means that to fulfill the conditions of the problem it is necessary that the equation \

had four different roots, different from zero, 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, which means that 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, according to Vieta’s theorem:

Let's rewrite the equation in the form \ and consider two functions: \(g(x)=20a-a^2-2^(x^2+2)\) and \(f(x)=13|x|-2|5x+12a|\) .
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: \(x=0\) . When \(x<0\) 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, when \(x>0\) the first module will open positively (\(|x|=x\)), therefore, regardless of how the second module will open, \(f(x)\) will be equal to \( kx+A\) , where \(A\) is the expression of \(a\) , and \(k\) is equal to either \(13-10=3\) or \(13+10=23\) . When \(x<0\) наоборот: первый модуль раскроется отрицательно и \(f(x)=kx+A\) , где \(k\) равно либо \(-3\) , либо \(-23\) .
Let's find the value of \(f\) at the minimum 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 \(-2\)\cup\)

Evenness and oddness of a function are one of its main properties, and parity takes up an impressive part of the school mathematics course. It largely determines the behavior of the function and greatly facilitates the construction of the corresponding graph.

Let's determine the parity of the function. Generally speaking, the function under study is considered even if for opposite values ​​of the independent variable (x) located in its domain of definition, the corresponding values ​​of y (function) turn out to be equal.

Let's give a more strict definition. Consider some function f (x), which is defined in the domain D. It will be even if for any point x located in the domain of definition:

• -x (opposite point) also lies in this scope,

• f(-x) = f(x).

From the above definition follows the condition necessary for the domain of definition of such a function, namely, symmetry with respect to the point O, which is the origin of coordinates, since if some point b is contained in the domain of definition of an even function, then the corresponding point b also lies in this domain. From the above, therefore, the conclusion follows: the even function has a form symmetrical with respect to the ordinate axis (Oy).

How to determine the parity of a function in practice?

Let it be specified using the formula h(x)=11^x+11^(-x). Following the algorithm that follows directly from the definition, we first examine its domain of definition. Obviously, it is defined for all values ​​of the argument, that is, the first condition is satisfied.

The next step is to substitute the opposite value (-x) for the argument (x).
We get:
h(-x) = 11^(-x) + 11^x.
Since addition satisfies the commutative (commutative) law, it is obvious that h(-x) = h(x) and the given functional dependence is even.

Let's check the parity of the function h(x)=11^x-11^(-x). Following the same algorithm, we get that h(-x) = 11^(-x) -11^x. Taking out the minus, in the end we have
h(-x)=-(11^x-11^(-x))=- h(x). Therefore, h(x) is odd.

By the way, it should be recalled that there are functions that cannot be classified according to these criteria; they are called neither even nor odd.

Even functions have a number of interesting properties:

• as a result of adding similar functions, they get an even one;
• as a result of subtracting such functions, an even one is obtained;
• even, also even;
• as a result of multiplying two such functions, an even one is obtained;
• as a result of multiplying odd and even functions, an odd one is obtained;
• as a result of dividing odd and even functions, an odd one is obtained;
• the derivative of such a function is odd;
• If you square an odd function, you get an even one.

The parity of a function can be used to solve equations.

To solve an equation like g(x) = 0, where the left side of the equation is an even function, it will be quite enough to find its solutions for non-negative values ​​of the variable. The resulting roots of the equation must be combined with the opposite numbers. One of them is subject to verification.

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

For example, is there any value of the parameter a for which the equation 2x^6-x^4-ax^2=1 will have three roots?

If we take into account that the variable enters the equation in even powers, then it is clear that replacing x with - x will not change the given equation. It follows that if a certain number is its root, then the opposite number is also the root. The conclusion is obvious: the roots of an equation that are different from zero are included in the set of its solutions in “pairs”.

It is clear that the number itself is not 0, that is, the number of roots of such an equation can only be even and, naturally, for any value of the parameter it cannot have three roots.

But the number of roots of the equation 2^x+ 2^(-x)=ax^4+2x^2+2 can be odd, and for any value of the parameter. Indeed, it is easy to check that the set of roots of a given equation contains solutions “in pairs”. Let's check if 0 is a root. When we substitute it into the equation, we get 2=2. Thus, in addition to “paired” ones, 0 is also a root, which proves their odd number.

To do this, use graph paper or a graphing calculator. Select any number of independent variable values x (\displaystyle x) and plug them into the function to calculate the values ​​of the dependent variable y (\displaystyle y). Plot the found coordinates of the points on the coordinate plane, and then connect these points to build a graph of the function.

• Substitute positive numeric values ​​into the function x (\displaystyle x) and corresponding negative numeric values. For example, given the function . Substitute the following values ​​into it x (\displaystyle x):
  • f (1) = 2 (1) 2 + 1 = 2 + 1 = 3 (\displaystyle f(1)=2(1)^(2)+1=2+1=3) (1 , 3) ​​(\displaystyle (1,3)).
  • f (2) = 2 (2) 2 + 1 = 2 (4) + 1 = 8 + 1 = 9 (\displaystyle f(2)=2(2)^(2)+1=2(4)+1 =8+1=9). We got a point with coordinates (2 , 9) (\displaystyle (2,9)).
  • f (− 1) = 2 (− 1) 2 + 1 = 2 + 1 = 3 (\displaystyle f(-1)=2(-1)^(2)+1=2+1=3). We got a point with coordinates (− 1 , 3) ​​(\displaystyle (-1,3)).
  • f (− 2) = 2 (− 2) 2 + 1 = 2 (4) + 1 = 8 + 1 = 9 (\displaystyle f(-2)=2(-2)^(2)+1=2( 4)+1=8+1=9). We got a point with coordinates (− 2, 9) (\displaystyle (-2,9)).

Check whether the graph of the function is symmetrical about the Y axis. Symmetry means a mirror image of the graph relative to the ordinate. If the part of the graph to the right of the Y-axis (positive values ​​of the independent variable) is the same as the part of the graph to the left of the Y-axis (negative values ​​of the independent variable), the graph is symmetrical about the Y-axis. If the function is symmetrical about the y-axis, the function is even.

• You can check the symmetry of the graph using individual points. If the value y (\displaystyle y) x (\displaystyle x), matches the value y (\displaystyle y), which corresponds to the value − x (\displaystyle -x), the function is even. In our example with the function f (x) = 2 x 2 + 1 (\displaystyle f(x)=2x^(2)+1) we received the following coordinates of the points:
  • (1.3) and (-1.3)
  • (2.9) and (-2.9)
• Note that for x=1 and x=-1 the dependent variable is y=3, and for x=2 and x=-2 the dependent variable is y=9. Thus the function is even. In fact, to accurately determine the form of the function, you need to consider more than two points, but the described method is a good approximation.

Check whether the graph of the function is symmetrical about the origin. The origin is the point with coordinates (0,0). Symmetry about the origin means that a positive value y (\displaystyle y)(with a positive value x (\displaystyle x)) corresponds to a negative value y (\displaystyle y)(with a negative value x (\displaystyle x)), and vice versa. Odd functions have symmetry about the origin.

• If you substitute several positive and corresponding negative values ​​into the function x (\displaystyle x), values y (\displaystyle y) will differ in sign. For example, given the function f (x) = x 3 + x (\displaystyle f(x)=x^(3)+x). Substitute several values ​​into it x (\displaystyle x):
  • f (1) = 1 3 + 1 = 1 + 1 = 2 (\displaystyle f(1)=1^(3)+1=1+1=2). We got a point with coordinates (1,2).
  • f (− 1) = (− 1) 3 + (− 1) = − 1 − 1 = − 2 (\displaystyle f(-1)=(-1)^(3)+(-1)=-1- 1=-2)
  • f (2) = 2 3 + 2 = 8 + 2 = 10 (\displaystyle f(2)=2^(3)+2=8+2=10)
  • f (− 2) = (− 2) 3 + (− 2) = − 8 − 2 = − 10 (\displaystyle f(-2)=(-2)^(3)+(-2)=-8- 2=-10). We received a point with coordinates (-2,-10).
• Thus, f(x) = -f(-x), that is, the function is odd.

Check if the graph of the function has any symmetry. The last type of function is a function whose graph has no symmetry, that is, there is no mirror image both relative to the ordinate axis and relative to the origin. For example, given the function .

• Substitute several positive and corresponding negative values ​​into the function x (\displaystyle x):
  • f (1) = 1 2 + 2 (1) + 1 = 1 + 2 + 1 = 4 (\displaystyle f(1)=1^(2)+2(1)+1=1+2+1=4 ). We got a point with coordinates (1,4).
  • f (− 1) = (− 1) 2 + 2 (− 1) + (− 1) = 1 − 2 − 1 = − 2 (\displaystyle f(-1)=(-1)^(2)+2 (-1)+(-1)=1-2-1=-2). We got a point with coordinates (-1,-2).
  • f (2) = 2 2 + 2 (2) + 2 = 4 + 4 + 2 = 10 (\displaystyle f(2)=2^(2)+2(2)+2=4+4+2=10 ). We got a point with coordinates (2,10).
  • f (− 2) = (− 2) 2 + 2 (− 2) + (− 2) = 4 − 4 − 2 = − 2 (\displaystyle f(-2)=(-2)^(2)+2 (-2)+(-2)=4-4-2=-2). We got a point with coordinates (2,-2).
• According to the results obtained, there is no symmetry. Values y (\displaystyle y) for opposite values x (\displaystyle x) do not coincide and are not opposite. Thus the function is neither even nor odd.
• Please note that the function f (x) = x 2 + 2 x + 1 (\displaystyle f(x)=x^(2)+2x+1) can be written like this: f (x) = (x + 1) 2 (\displaystyle f(x)=(x+1)^(2)). When written in this form, the function appears even because there is an even exponent. But this example proves that the type of function cannot be quickly determined if the independent variable is enclosed in parentheses. In this case, you need to open the brackets and analyze the obtained exponents.

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Goals:

• formulate the concept of even and odd functions, teach the ability to determine and use these properties when studying functions and constructing graphs;
• develop students’ creative activity, logical thinking, ability to compare and generalize;
• cultivate hard work and mathematical culture; develop communication skills .

Equipment: multimedia installation, interactive whiteboard, handouts.

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

Information sources:

1. Algebra 9th class A.G. Mordkovich. Textbook.
2. Algebra 9th grade A.G. Mordkovich. Problem book.
3. Algebra 9th grade. Tasks for student learning and development. Belenkova E.Yu. Lebedintseva E.A.

DURING THE CLASSES

1. Organizational moment

Setting goals and objectives for the lesson.

2. Checking homework

No. 10.17 (9th grade problem book. A.G. Mordkovich).

A) at = f(X), f(X) =

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

c) 1. D( f) = [– 2; + ∞)
2. E( f) = [– 3; + ∞)
3. f(X) = 0 at 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 naim = – 3, at naib doesn't exist
8. The function is continuous.

(Have you used a function exploration algorithm?) Slide.

2. Let’s check the table you were asked from the slide.

Fill the table
	Domain	Function zeros	Intervals of sign constancy		Coordinates of the points of intersection of the graph with Oy
		x = –5, x = 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. Updating knowledge

– Functions are given.
– Specify the scope 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 these functions in the domain of definition the equalities hold f(– X) = f(X), f(– X) = – f(X)? (enter the obtained data into the table) Slide

		f(1) and f(– 1)	f(2) and f(– 2)	graphs	f(– X) = –f(X)	f(– X) = f(X)
1. f(X) =
2. f(X) = X 3
3. f(X) = | X |
4.f(X) = 2X – 3
5. f(X) =	X ≠ 0
6. f(X)=	X > –1		and not defined

4. New material

– Carrying out this work, guys, we have identified 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 to determine the evenness and oddness of a function, to find out the significance of this property in the study of functions and plotting graphs.
So, let's find the definitions in the textbook and read (p. 110) . Slide

Def. 1 Function at = f (X), defined on the set X is called even, if for any value XЄ X is executed equality f(–x)= f(x). Give examples.

Def. 2 Function y = f(x), defined on the set X is called odd, if for any value XЄ X the equality f(–х)= –f(х) holds. Give examples.

Where did we meet the terms “even” and “odd”?
Which of these functions will be even, do you think? Why? Which ones are odd? Why?
For any function of the form at= x n, Where n– an integer, it can be argued that the function is odd when n– odd and the function is even when n– even.
– View functions at= and at = 2X– 3 are neither even nor odd, because equalities are not satisfied f(– X) = – f(X), f(– X) = f(X)

The study of whether a function is even or odd is called the study of a function's parity. Slide

In definitions 1 and 2 we were talking about the values ​​of the function at x and – x, thereby it is assumed that the function is also defined at the value X, and at – X.

Def 3. If a numerical set, together with each of its elements x, also contains the opposite element –x, then the set X called a symmetric set.

Examples:

(–2;2), [–5;5]; (∞;∞) are symmetric sets, and , [–5;4] are asymmetric.

– U even functions is the domain of definition a symmetric set? The odd ones?
– If D( f) is an asymmetric set, then what is the function?
– Thus, if the function at = f(X) – even or odd, then its domain of definition is D( f) is a symmetric set. Is the converse statement true: if the domain of definition of a function is a symmetric set, then is it even or odd?
– This means that the presence of a symmetric set of the domain of definition is a necessary condition, but not sufficient.
– So how do you examine a function for parity? Let's try to create an algorithm.

Slide

Algorithm for studying a function for parity

1. Determine whether the domain of definition 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):

• If f(–X).= f(X), then the function is even;
• If f(–X).= – f(X), then the function is odd;
• If f(–X) ≠ f(X) And f(–X) ≠ –f(X), then the function is neither even nor odd.

Examples:

Examine function a) for parity at= x 5 +; b) at= ; V) at= .

Solution.

a) h(x) = x 5 +,

1) D(h) = (–∞; 0) U (0; +∞), symmetric set.

2) h (– x) = (–x) 5 + – x5 –= – (x 5 +),

3) h(– x) = – h (x) => function h(x)= x 5 + odd.

b) y =,

at = f(X), D(f) = (–∞; –9)? (–9; +∞), an asymmetric set, which means the function is neither even nor odd.

V) f(X) = , y = f (x),

1) D( f) = (–∞; 3] ≠ ; b) (∞; –2), (–4; 4]?

Option 2

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

A); b) y = x (5 – x 2). 2. Examine the function for parity:

a) y = x 2 (2x – x 3), b) y =

3. In Fig. a graph has been built at = f(X), for all X, satisfying the condition X? 0.
Graph the Function at = f(X), If at = f(X) is an even function.

3. In Fig. a graph has been built at = f(X), for all x satisfying the condition x? 0.
Graph the Function at = f(X), If at = f(X) is an odd function.

Mutual check on slide.

6. Homework: №11.11, 11.21,11.22;

Proof of the geometric meaning of the parity property.

***(Assignment of the Unified State Examination option).

1. The odd function y = f(x) is defined on the entire number line. For any non-negative value of the variable x, the value of this function coincides with the value of the function g( X) = X(X + 1)(X + 3)(X– 7). Find the value of the function h( X) = at X = 3.

7. Summing up