Tg is an even function. Graph of even and odd functions

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Methods for specifying a function

Let the function be given by the formula: y=2x^(2)-3. By assigning any values ​​to the independent variable x, you can calculate, using this formula, the corresponding values ​​of the dependent variable y. For example, if x=-0.5, then, using the formula, we find that the corresponding value of y is y=2 \cdot (-0.5)^(2)-3=-2.5.

y

Using this table, you can see that for the argument value −1 the function value −3 will correspond; and the value x=2 will correspond to y=0, etc. It is also important to know that each argument value in the table corresponds to only one function value.

More functions can be specified using graphs. Using a graph, it is established which value of the function correlates with a certain value x. Most often, this will be an approximate value of the function.

Even and odd function The function is even function

Even and odd function , when f(-x)=f(x) for any x from the domain of definition. Such a function will be symmetrical about the Oy axis. odd function

Even and odd function , when f(-x)=-f(x) for any x from the domain of definition. Such a function will be symmetric about the origin O (0;0) ., not even neither odd and is called function general view

, when it does not have symmetry about the axis or origin.

Let us examine the following function for parity:

f(x)=3x^(3)-7x^(7) D(f)=(-\infty ; +\infty) with a symmetric domain of definition relative to the origin. f(-x)= 3 \cdot (-x)^(3)-7 \cdot (-x)^(7)= -3x^(3)+7x^(7)= -(3x^(3)-7x^(7))=.

-f(x)

This means that the function f(x)=3x^(3)-7x^(7) is odd.

Periodic function The function y=f(x) , in the domain of which the equality f(x+T)=f(x-T)=f(x) holds for any x, is called periodic function

with period T \neq 0 .

Repeating the graph of a function on any segment of the x-axis that has length T.

The intervals where the function is positive, that is, f(x) > 0, are segments of the abscissa axis that correspond to the points of the function graph lying above the abscissa axis. f(x) > 0 on

(x_(1); x_(2)) \cup (x_(3); +\infty)< 0 - отрезки оси абсцисс, которые отвечают точкам графика функции, лежащих ниже оси абсцисс.

Intervals where the function is negative, that is, f(x)< 0 на (-\infty; x_(1)) \cup (x_(2); x_(3))

Limited function

Bounded from below It is customary to call a function y=f(x), x \in X when there is a number A for which the inequality f(x) \geq A holds for any x \in X .

An example of a function bounded from 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 when there is 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 is a number K > 0 for which the inequality \left | f(x)\right | \neq K for any x \in X .

Example of a limited function: y=\sin x is limited over the entire number axis, 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 then when higher value x will correspond to a larger value of the function y=f(x) . It follows that taking two arbitrary values ​​of the argument x_(1) and x_(2) from the interval under consideration, with x_(1) > x_(2) , the result 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 corresponds to a smaller value of the function y(x) . It follows that, taking from the interval under consideration two arbitrary values ​​of the argument x_(1) and x_(2) , and x_(1) > x_(2) , the result will be y(x_(1))< y(x_{2}) .

Function Roots It is customary to call the points at which the function F=y(x) intersects the abscissa axis (they are obtained by solving the equation y(x)=0).

a) If for x > 0 even function increases, then it decreases as x< 0

b) When an even function decreases at x > 0, then it increases at x< 0

c) When an odd function increases at x > 0, then it also increases at x< 0

d) When an odd function decreases for x > 0, then it will also decrease for x< 0

Extrema of the function

Minimum point of the function y=f(x) is usually called a point x=x_(0) whose neighborhood will have other points (except for the point x=x_(0)), and for them the inequality f(x) > f will then be satisfied (x_(0)) . y_(min) - designation of the function at the min point.

Maximum point of the function y=f(x) is usually called a point x=x_(0) whose neighborhood will have other points (except for the point x=x_(0)), and for them the inequality f(x) will then be satisfied< f(x^{0}) . y_{max} - обозначение функции в точке max.

Prerequisite

According to Fermat’s theorem: f"(x)=0 when the function f(x) that is differentiable at the point x_(0) will have an extremum at this point.

Sufficient condition

1. When the derivative changes sign from plus to minus, then x_(0) will be the minimum point;
2. 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 a function on an interval

Calculation steps:

1. The derivative f"(x) is sought;
2. Stationary and critical points of the function are found and those belonging to the segment are selected;
3. The values ​​of the function f(x) are found at stationary and critical points and ends of the segment. The smaller of the results obtained will be lowest value functions, and more - the largest.

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 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)).

Evenness and oddness of a function are one of its main properties, and parity takes up an impressive part school course mathematics. 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- 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 given equation won't change. 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 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.

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.

In definitions 1 and 2 we're talking about about 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 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\)