Which function is called odd. 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.
Taking any value taken by the argument x in the formula y=2x^(2)-3, you can calculate only one value of the function 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 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, when f(-x)=f(x) for any x from the domain of definition. Such a function will be symmetrical about the Oy axis.
The function is 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) .
The function is 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)
Intervals where the function is negative, that is, 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))
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 limited function: y=\sin x is limited on the entire number axis, since \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 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 an even function increases, then it decreases for x< 0
b) When an even function decreases at x > 0, then it increases at x< 0
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c) When at x > 0 odd function increases, then it also increases as x< 0
d) When an odd function decreases for x > 0, then it will also decrease for x< 0
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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
- When the derivative changes sign 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 stationary point x_(0) .
The largest and smallest value of a function on an interval
Calculation steps:
- The derivative f"(x) is sought;
- Stationary and critical points of the function are found and those belonging to the segment are selected;
- 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.
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 of general form. 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 \(x =-b\) .
\(\blacktriangleright\) A 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.
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 main period is equal to \(2\pi\), for the functions \(f(x)=\mathrm( tg)\,x\) and \(f(x)=\mathrm(ctg)\,x\) the main period is equal to \(\pi\) .
In order to build 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 the left side of the equation (*) is greater than or equal to \(0+ \mathrm(tg)^2\,1\) .
Thus, equality (*) can only be satisfied 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 with respect to 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? Thus, we have determined that both roots of the equation \((*)\) must lie in the interval \((1;4)\) . How to write this condition? 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 \(f(x)=13|x|-2|5x+12a|\) . 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: 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 met. The next step is to substitute the opposite value (-x) for the argument (x). 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 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: 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 this 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. Converting graphs. Verbal description of the function. Graphic method. The graphical method of specifying a function is the most visual and is often used in technology. In mathematical analysis, the graphical method of specifying functions is used as an illustration. Function graph f is the set of all points (x;y) of the coordinate plane, where y=f(x), and x “runs through” the entire domain of definition of this function. A subset of the coordinate plane is a graph of a function if it has no more than one common point with any straight line parallel to the Oy axis. Example. Are the figures shown below graphs of functions? The advantage of a graphic task is its clarity. You can immediately see how the function behaves, where it increases and where it decreases. From the graph you can immediately find out some important characteristics of the function. In general, analytical and graphical methods of defining a function go hand in hand. Working with the formula helps to build a graph. And the graph often suggests solutions that you wouldn’t even notice in the formula. Almost any student knows the three ways to define a function that we just looked at. Let's try to answer the question: "Are there other ways to specify a function?" There is such a way. The function can be quite unambiguously specified in words. For example, the function y=2x can be specified by the following verbal description: each real value of the argument x is associated with its double value. The rule is established, the function is specified. Moreover, you can verbally specify a function that is extremely difficult, if not impossible, to define using a formula. For example: each value of the natural argument x is associated with the sum of the digits that make up the value of x. For example, if x=3, then y=3. If x=257, then y=2+5+7=14. And so on. It is problematic to write this down in a formula. But the sign is easy to make. The method of verbal description is a rather rarely used method. But sometimes it does. If there is a law of one-to-one correspondence between x and y, then there is a function. What law, in what form it is expressed - a formula, a tablet, a graph, words - does not change the essence of the matter. Let us consider functions whose domains of definition are symmetrical with respect to the origin, i.e. for anyone X from the domain of definition number (- X) also belongs to the domain of definition. Among these functions are even and odd. Definition. The function f is called even, if for any X from its domain of definition Example. Consider the function It is even. Let's check it out. For anyone X equalities are satisfied Thus, both conditions are met, which means the function is even. Below is a graph of this function. Definition. The function f is called odd, if for any X from its domain of definition Example. Consider the function It is odd. Let's check it out. The domain of definition is the entire numerical axis, which means it is symmetrical about the point (0;0). For anyone X equalities are satisfied Thus, both conditions are met, which means the function is odd. Below is a graph of this function. The graphs shown in the first and third figures are symmetrical about the ordinate axis, and the graphs shown in the second and fourth figures are symmetrical about the origin. Which of the functions whose graphs are shown in the figures are even and which are odd? 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). 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. 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.
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
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
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
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: \
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.
h(-x)=-(11^x-11^(-x))=- h(x). Therefore, h(x) is odd.
Graph of an even function
Graph of an odd function
