Let's apply the rules of even function. Even and odd function

- (math.) A function y = f (x) is called even if it does not change when the independent variable only changes sign, that is, if f (x) = f (x). If f (x) = f (x), then the function f (x) is called odd. For example, y = cosx, y = x2... ...

F(x) = x is an example of an odd function. f(x) = x2 is an example of an even function. f(x) = x3 ... Wikipedia

A function satisfying the equality f (x) = f (x). See Even and Odd Functions... Great Soviet Encyclopedia

F(x) = x is an example of an odd function. f(x) = x2 is an example of an even function. f(x) = x3 ... Wikipedia

F(x) = x is an example of an odd function. f(x) = x2 is an example of an even function. f(x) = x3 ... Wikipedia

F(x) = x is an example of an odd function. f(x) = x2 is an example of an even function. f(x) = x3 ... Wikipedia

F(x) = x is an example of an odd function. f(x) = x2 is an example of an even function. f(x) = x3 ... Wikipedia

Special functions introduced by the French mathematician E. Mathieu in 1868 when solving problems on the oscillation of an elliptical membrane. M. f. are also used in studying the distribution electromagnetic waves in an elliptical cylinder... Great Soviet Encyclopedia

The "sin" request is redirected here; see also other meanings. The "sec" request is redirected here; see also other meanings. The "Sine" request is redirected here; see also other meanings... Wikipedia

. 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 coordinate plane, and then connect these points to graph the function.

• Substitute positive ones into the function numeric values x (\displaystyle x) and corresponding negative numeric values. For example, given the function f (x) = 2 x 2 + 1 (\displaystyle f(x)=2x^(2)+1). Substitute the following values ​​into it x (\displaystyle x):

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 axis. 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.

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.

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):
• 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.

Even function.

Even is a function whose sign does not change when the sign changes x.

x equality holds f(–x) = f(x). Sign x does not affect the sign y.

The graph of an even function is symmetrical about the coordinate axis (Fig. 1).

Examples of an even function:

y=cos x

y = x 2

y = –x 2

y = x 4

y = x 6

y = x 2 + x

Explanation:
Let's take the function y = x 2 or y = –x 2 .
For any value x the function is positive. Sign x does not affect the 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 changes x.

In other words, for any value x equality holds f(–x) = –f(x).

The graph of an odd function is symmetrical with respect to the origin (Fig. 2).

Examples of odd function:

y= sin x

y = x 3

y = –x 3

Explanation:

Let's take the function y = – x 3 .
All meanings at it will have a minus sign. That is a sign x influences 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 functions are even or odd. There are functions that do not obey 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 that describe these processes are called periodic functions. That is, these are functions in whose graphs there are elements that repeat at certain numerical intervals.

Which were familiar to you to one degree or another. 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 = f(x), x є X, is called even if for any value x from the set X the equality f (-x) = f (x) holds.

Definition 2.

The function y = f(x), x є X, is called odd if for any value x from the set X the equality f (-x) = -f (x) holds.

Prove that y = x 4 is an even function.

Solution. We have: f(x) = x 4, f(-x) = (-x) 4. But(-x) 4 = x 4. This means that for any x the equality f(-x) = f(x) holds, i.e. the function is even.

Similarly, it can be proven that the functions y - x 2, y = x 6, y - x 8 are even.

Prove that y = x 3 ~ an odd function.

Solution. We have: f(x) = x 3, f(-x) = (-x) 3. But (-x) 3 = -x 3. This means that for any x the equality f (-x) = -f (x) holds, i.e. the function is odd.

Similarly, it can be proven that the functions y = x, y = x 5, y = x 7 are odd.

You and I have already been convinced more than once that new terms in mathematics most often have an “earthly” origin, i.e. they can be explained somehow. This is the case with both even and odd functions. See: y - x 3, y = x 5, y = x 7 are odd functions, while y = x 2, y = x 4, y = x 6 are even functions. And in general, for any function of the form y = x" (below we will specifically study these functions), where n is a natural number, we can conclude: if n is an odd number, then the function y = x" is odd; if n is an even number, then the function y = xn is even.

There are also functions that are neither even nor odd. Such, for example, is the function y = 2x + 3. Indeed, f(1) = 5, and f (-1) = 1. As you can see, here, therefore, neither the identity f(-x) = 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.

Definitions 1 and 2 refer to the values ​​of the function at points x and -x. This assumes that the function is defined at both point x and point -x. This means that point -x belongs to the domain of definition of the function simultaneously with point x. If a numerical set X, together with each of its elements x, also 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 \).

Since \(x^2\geqslant 0\) , then the left side of the 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 about 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 out 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\)