Function is one of the most important mathematical concepts. Function - variable dependency at from variable x, if each value X matches a single value at. Variable X called the independent variable or argument. Variable at called the dependent variable. All values of the independent variable (variable x) form the domain of definition of the function. All values that the dependent variable takes (variable y), form the range of values of the function.
Function graph call the set of all points of the coordinate plane, the abscissas of which are equal to the values of the argument, and the ordinates are equal to the corresponding values of the function, that is, the values of the variable are plotted along the abscissa axis x, and the values of the variable are plotted along the ordinate axis y. To graph a function, you need to know the properties of the function. The main properties of the function will be discussed below!
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Basic properties of functions.
1) Function domain and function range.
The domain of a function is the set of all valid valid argument values x(variable x), for which the function y = f(x) determined.
The range of a function is the set of all real values y, which the function accepts.
In elementary mathematics, functions are studied only on the set of real numbers.
2) Function zeros.
Values X, at which y=0, called function zeros. These are the abscissas of the points of intersection of the function graph with the Ox axis.
3) Intervals of constant sign of a function.
Intervals of constant sign of a function are such intervals of values x, on which the function values y either only positive or only negative are called intervals of constant sign of the function.
4) Monotonicity of the function.
An increasing function (in a certain interval) is a function in which a larger value of the argument from this interval corresponds to a larger value of the function.
A decreasing function (in a certain interval) is a function in which a larger value of the argument from this interval corresponds to a smaller value of the function.
5) Even (odd) function.
An even function is a function whose domain of definition is symmetrical with respect to the origin and for any X f(-x) = f(x). The graph of an even function is symmetrical about the ordinate.
An odd function is a function whose domain of definition is symmetrical with respect to the origin and for any X from the domain of definition the equality is true f(-x) = - f(x). The graph of an odd function is symmetrical about the origin.
Even function
1) The domain of definition is symmetrical with respect to the point (0; 0), that is, if the point a belongs to the domain of definition, then the point -a also belongs to the domain of definition.
2) For any value x f(-x)=f(x)
3) The graph of an even function is symmetrical about the Oy axis.
Odd function has the following properties:
1) The domain of definition is symmetrical about the point (0; 0).
2) for any value x, belonging to the domain of definition, the equality f(-x)=-f(x)
3) The graph of an odd function is symmetrical with respect to the origin (0; 0).
Not every function is even or odd. Functions general view are neither even nor odd.
6) Limited and unlimited functions.
A function is called bounded if there is a positive number M such that |f(x)| ≤ M for all values of x. If such a number does not exist, then the function is unlimited.
7) Periodicity of the function.
A function f(x) is periodic if there is a non-zero number T such that for any x from the domain of definition of the function the following holds: f(x+T) = f(x). This smallest number is called the period of the function. All trigonometric functions are periodic. (Trigonometric formulas).
Function f is called periodic if there is a number such that for any x from the domain of definition the equality f(x)=f(x-T)=f(x+T). T is the period of the function.
Every periodic function has an infinite number of periods. In practice, the smallest positive period is usually considered.
The values of a periodic function are repeated after an interval equal to the period. This is used when constructing graphs.
Function study.
1) D(y) – Definition domain: the set of all those values of the variable x. for which the algebraic expressions f(x) and g(x) make sense.
If a function is given by a formula, then the domain of definition consists of all values of the independent variable for which the formula makes sense.
2) Properties of the function: even/odd, periodicity:
Odd And even functions are called whose graphs are symmetric with respect to changes in the sign of the argument.
Odd function- a function that changes the value to the opposite when the sign of the independent variable changes (symmetrical relative to the center of coordinates).
Even function- a function that does not change its value when the sign of the independent variable changes (symmetrical about the ordinate).
Neither even nor odd function (general function)- a function that does not have symmetry. This category includes functions that do not fall under the previous 2 categories.
Functions that do not belong to any of the categories above are called neither even nor odd(or general functions).
Odd functions
Odd power where is an arbitrary integer.
Even functions
Even power where is an arbitrary integer.
Periodic function- a function that repeats its values at some regular argument interval, that is, it does not change its value when adding some fixed non-zero number to the argument ( period functions) over the entire domain of definition.
3) Zeros (roots) of a function are the points where it becomes zero.
Finding the intersection point of the graph with the axis Oy. To do this you need to calculate the value f(0). Find also the points of intersection of the graph with the axis Ox, why find the roots of the equation f(x) = 0 (or make sure there are no roots).
The points at which the graph intersects the axis are called function zeros. To find the zeros of a function you need to solve the equation, that is, find those meanings of "x", at which the function becomes zero.
4) Intervals of constancy of signs, signs in them.
Intervals where the function f(x) maintains sign.
The interval of constancy of sign is the interval at every point of which the function is positive or negative.
ABOVE the x-axis.
BELOW the axle.
5) Continuity (points of discontinuity, nature of the discontinuity, asymptotes).
Continuous function- a function without “jumps”, that is, one in which small changes in the argument lead to small changes in the value of the function.
Removable Break Points
If the limit of the function exists, but the function is not defined at this point, or the limit does not coincide with the value of the function at this point:
,
then the point is called removable break point functions (in complex analysis, a removable singular point).
If we “correct” the function at the point of removable discontinuity and put 
, then we get a function that is continuous at a given point. This operation on a function is called extending the function to continuous or redefinition of the function by continuity, which justifies the name of the point as a point removable rupture.
Discontinuity points of the first and second kind
If a function has a discontinuity at a given point (that is, the limit of the function at a given point is absent or does not coincide with the value of the function at a given point), then for numerical functions there are two possible options associated with the existence of numerical functions unilateral limits:
if both one-sided limits exist and are finite, then such a point is called discontinuity point of the first kind. Removable discontinuity points are discontinuity points of the first kind;
if at least one of the one-sided limits does not exist or is not a finite value, then such a point is called point of discontinuity of the second kind.
Asymptote - straight, which has the property that the distance from a point on the curve to this straight tends to zero as the point moves away along the branch to infinity.
Vertical
Vertical asymptote - limit line 
.
As a rule, when determining the vertical asymptote, they look for not one limit, but two one-sided ones (left and right). This is done in order to determine how the function behaves as it approaches the vertical asymptote from different directions. For example:
Horizontal
Horizontal asymptote - straight species, subject to the existence limit
.
Inclined
Oblique asymptote - straight species, subject to the existence limits
Note: a function can have no more than two oblique (horizontal) asymptotes.
Note: if at least one of the two limits mentioned above does not exist (or is equal to ), then the oblique asymptote at (or ) does not exist.
if in item 2.), then , and the limit is found using the horizontal asymptote formula, 
.
6) Finding intervals of monotonicity. Find intervals of monotonicity of a function f(x)(that is, intervals of increasing and decreasing). This is done by examining the sign of the derivative f(x). To do this, find the derivative f(x) and solve the inequality f(x)0. On intervals where this inequality holds, the function f(x)increases. Where the reverse inequality holds f(x)0, function f(x) is decreasing.
Finding a local extremum. Having found the intervals of monotonicity, we can immediately determine the local extremum points where an increase is replaced by a decrease, local maxima are located, and where a decrease is replaced by an increase, local minima are located. Calculate the value of the function at these points. If a function has critical points that are not local extremum points, then it is useful to calculate the value of the function at these points as well.
Finding the largest and smallest values of the function y = f(x) on a segment(continuation)
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1. Find the derivative of the function: f(x). 2. Find the points at which the derivative is zero: f(x)=0x 1, x 2 ,... 3. Determine the affiliation of points X 1 ,X 2 , … segment [ a; b]: let x 1a;b, A x 2a;b . |
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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 general function, 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 .
An example of a 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 then, when a larger value of x corresponds 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 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
.png)
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 the 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 the smallest value of the function, 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 \(x =-b\) .
\(\blacktriangleright\) 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.
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 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 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? 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\) Definition 1. The function is called even
(odd
), if together with each variable value Thus, a function can be even or odd only if its domain of definition is symmetrical about the origin of coordinates on the number line (number X And - X belong at the same time Function Function Function The graph of an even function is symmetrical about the axis OU, because if the point When proving whether a function is even or odd, the following statements are useful. Theorem 1. a) The sum of two even (odd) functions is an even (odd) function. b) The product of two even (odd) functions is an even function. c) The product of an even and odd function is an odd function. d) If f– even function on the set X, and the function g
defined on the set d) If f– odd function on the set X, and the function g
defined on the set Proof. Let us prove, for example, b) and d). b) Let d) Let f
is an even function. Then. The remaining statements of the theorem can be proved in a similar way. The theorem has been proven. Theorem 2. Any function Proof. Function Function Definition 2. Function Such a number T called period
functions From Definition 1 it follows that if T– period of the function Definition 3. The smallest of the positive periods of a function is called its main
period. Theorem 3. If T– main period of the function f, then the remaining periods are multiples of it. Proof. Let us assume the opposite, that is, that there is a period that is It is well known that trigonometric functions are periodic. Main period (because Meaning T, determined from the first equality, cannot be a period, since it depends on X, i.e. is a function of X, and not a constant number. The period is determined from the second equality: An example of a more complex periodic function is the Dirichlet function Note that if T is a rational number, then for any rational number T. Therefore, any rational number T is the period of the Dirichlet function. It is clear that this function does not have a main period, since there are positive rational numbers that are arbitrarily close to zero (for example, a rational number can be made by choosing n arbitrarily close to zero). Theorem 4. If the function f
defined on the set X and has a period T, and the function g
defined on the set Proof. We have, therefore that is, the statement of the theorem is proven. For example, since cos
x
has a period Definition 4. Functions that are not periodic are called non-periodic
.
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 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
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: \
meaning - X also belongs 
and the equality holds
). For example, the function 
is neither even nor odd, since its domain of definition 
not symmetrical about the origin.
even, because 
symmetrical about the origin and.
odd, because 
And 
.
is not even and odd, since although 
and is symmetrical with respect to the origin, equalities (11.1) are not satisfied. For example,.
also belongs to the schedule. The graph of an odd function is symmetrical about the origin, since if 
belongs to the graph, then the point 
also belongs to the schedule.
, then the function 
– even.
and even (odd), then the function 
– even (odd).
And 
– even functions. Then, therefore. The case of odd functions is treated similarly 
And 
.
, defined on the set X, symmetrical about the origin, can be represented as a sum of even and odd functions.
can be written in the form
.
– even, because 
, and the function 
– odd, because. Thus, 
, Where 
– even, and 
– odd functions. The theorem has been proven.
called periodic
, if there is a number 
, such that for any 
numbers 
And 
also belong to the domain of definition 
and the equalities are satisfied
.
, then the number – T Same
is the period of the function
(since when replacing T on - T equality is maintained). Using the method of mathematical induction it can be shown that if T– period of the function f, then 
, is also a period. It follows that if a function has a period, then it has infinitely many periods.
functions f
(
>0), not multiple T. Then, dividing 
on T with the remainder, we get 
, Where 
. That's why
– period of the function f, and 
, and this contradicts the fact that T– main period of the function f. The statement of the theorem follows from the resulting contradiction. The theorem has been proven.
And 
equals 
,
And 
. Let's find the period of the function 
. Let 
- the period of this function. Then
.
oror 
.
. There are infinitely many periods, with 
the smallest positive period is obtained at 
:
. This is the main period of the function 
.
And 
are rational numbers for rational X and irrational when irrational X. That's why
, then a complex function 
also has a period T.
, then the functions 
have a period 
.
