Function research.
1) D(y) - Domain of definition: the set of all those values of the variable x. under which the algebraic expressions f(x) and g(x) make sense.
If the 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) Function properties: even/odd, periodicity:
odd and even are called functions whose graphs are symmetric with respect to the change 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 (symmetric about the center of coordinates).
Even function- a function that does not change its value when the sign of the independent variable changes (symmetric about the y-axis).
Neither even nor odd function (general function) is 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 generic functions).
Odd functions
An odd power where is an arbitrary integer.
Even functions
An even power where is an arbitrary integer.
Periodic function is a function that repeats its values at some regular interval of the argument, i.e., does not change its value when some fixed nonzero number is added to the argument ( period functions) over the entire domain of definition.
3) Zeros (roots) of a function are the points where it vanishes.
Finding the point of intersection 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 where the graph intersects the axis are called function zeros. To find the zeros of the function, you need to solve the equation, that is, find those x values, for which the function vanishes.
4) Intervals of constancy of signs, signs in them.
Intervals where the function f(x) retains its sign.
The constancy interval is the interval at every point in which function is positive or negative.
ABOVE the x-axis.
BELOW axis.
5) Continuity (points of discontinuity, character of 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 breakpoints
If the limit of the function exist, but the function is not defined at this point, or the limit does not match the value of the function at this point:
,
then the point is called break point functions (in complex analysis, a removable singular point).
If we "correct" the function at the point of a removable discontinuity and put 
, then we get a function that is continuous at this point. Such an operation on a function is called extending the function to continuous or extension of the function by continuity, which justifies the name of the point, as points disposable gap.
Discontinuity points of the first and second kind
If the 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 related to the existence of numerical functions unilateral limits:
if both one-sided limits exist and are finite, then such a point is called breaking 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 breaking point of the second kind.
Asymptote - straight, which has the property that the distance from a point of the curve to this straight tends to zero as the point moves 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
.
oblique
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 by the horizontal asymptote formula, 
.
6) Finding intervals of monotonicity. Find monotonicity intervals of a function f(x) (that is, intervals of increase and decrease). 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 the intervals where this inequality is satisfied, the function f(x) increases. Where the reverse inequality holds f(x)0, function f(x) decreases.
Finding a local extremum. Having found the intervals of monotonicity, we can immediately determine the points of a local extremum where the increase is replaced by a decrease, there are local maxima, and where the decrease is replaced by an increase, local minima. 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)
|
1. Find the derivative of a function: f(x). 2. Find points where the derivative is zero: f(x)=0x 1, x 2 ,... 3. Determine the ownership of points X 1 ,X 2 , … segment [ a; b]: let be x 1a;b, a x 2a;b . 4. Find function values at selected points and at the ends of the segment: f(x 1), f(x 2),..., f(x a),f(x b), 5. Selection of the largest and smallest values of the function from those found. Comment. If on the interval [ a; b] there are discontinuity points, then it is necessary to calculate one-sided limits in them, and then take their values into account in choosing the largest and smallest values of the function. |
7) Finding intervals of convexity and concavity. This is done by examining the sign of the second derivative f(x). Find the inflection points at the junctions of the convex and concavity intervals. Calculate the value of the function at the inflection points. If the function has other points of continuity (other than inflection points) at which the second derivative is equal to 0 or does not exist, then at these points it is also useful to calculate the value of the function. Finding f(x) , we solve the inequality f(x)0. On each of the solution intervals, the function will be downward convex. Solving the reverse inequality f(x)0, we find the intervals on which the function is convex upwards (that is, concave). We define inflection points as those points at which the function changes the direction of convexity (and is continuous).
Function inflection point- this is the point at which the function is continuous and when passing through which the function changes the direction of convexity.
Conditions of existence
Necessary condition for the existence of an inflection point: if the function is twice differentiable in some punctured neighborhood of the point , then either 
.
The dependence of the variable y on the variable x, in which each value of x corresponds to a single value of y is called a function. The notation is y=f(x). Each function has a number of basic properties, such as monotonicity, parity, periodicity, and others.
Consider the parity property in more detail.
A function y=f(x) is called even if it satisfies the following two conditions:
2. The value of the function at the point x belonging to the scope of the function must be equal to the value of the function at the point -x. That is, for any point x, from the domain of the function, the following equality f (x) \u003d f (-x) must be true.
Graph of an even function
If you build a graph of an even function, it will be symmetrical about the y-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 that it is symmetrical about the point O.
Take an arbitrary x=3. f(x)=3^2=9.
f(-x)=(-3)^2=9. Therefore, f(x) = f(-x). Thus, both conditions are satisfied for us, which means that the function is even. Below is a graph of the function y=x^2.
The figure shows that the graph is symmetrical about the y-axis.
Graph of an odd function
A function y=f(x) is called odd if it satisfies the following two conditions:
1. The domain of the given function must be symmetrical with respect to the point O. That is, if some point a belongs to the domain of the function, then the corresponding point -a must also belong to the domain of the given function.
2. For any point x, from the domain of the function, the following equality f (x) \u003d -f (x) must be satisfied.
The graph of an odd function is symmetrical with respect to the point O - the origin. 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 that it is symmetrical about the point O.
Take an arbitrary x=2. f(x)=2^3=8.
f(-x)=(-2)^3=-8. Therefore f(x) = -f(x). Thus, both conditions are satisfied for us, which means that 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 with respect to the origin.
even, if for all \(x\) from its domain 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 is true: \(f(-x)=-f(x)\) .
The graph of an odd function is symmetrical with respect to 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 generic functions. 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 an even function \(f_1=x^2\) and an odd function \(f_2=-x\) .
\(\blacktriangleright\) 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 parity is an odd function.
3) The sum and difference of even functions is an even function.
4) The sum and difference of odd functions is an 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 if, 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 \(T\ne 0\) we have \(f(x)=f(x+T) \) , where \(x, x+T\in X\) . The smallest \(T\) , for which this equality holds, is called the main (basic) 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 principal period is \(2\pi\) , for the functions \(f(x)=\mathrm( tg)\,x\) and \(f(x)=\mathrm(ctg)\,x\) main period is \(\pi\) .
In order to plot 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 to the left:
\(\blacktriangleright\) The domain \(D(f)\) of the function \(f(x)\) is the 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 Examination
For what values of the parameter \(a\) the equation
has a unique 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. 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 got two parameter values \(a\) . Note that we have 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, it is necessary to substitute the resulting values of the parameter \(a\) into the original equation and check for which \(a\) the root \(x=0\) will indeed 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 takes the form \ We rewrite the equation in the form \ As \(-1\leqslant \cos x\leqslant 1\), then \(-\mathrm(tg)\,1\leqslant \mathrm(tg)\,(\cos x)\leqslant \mathrm(tg)\,1\). Therefore, the values of the right side of the equation (*) belong to the interval \([-\mathrm(tg)^2\,1; \mathrm(tg)^2\,1]\).
Since \(x^2\geqslant 0\) , then the left side of equation (*) is greater than or equal to \(0+ \mathrm(tg)^2\,1\) .
Thus, equality (*) can only hold 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 Examination
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 symmetric with respect to the origin, then such a function is odd, that is, \(f(-x)=-f(x)\) holds for any \(x\) from the function's domain. 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 hold for all \(x\) from the domain \(f(x)\) , hence \(\sin(2\pi a)=0 \Rightarrow a=\dfrac n2, n\in\mathbb(Z)\).
Answer:
\(\dfrac n2, n\in\mathbb(Z)\)
Task 3 #3069
Task level: Equal to the Unified State Examination
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 real 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 y-axis, therefore, when \(-\dfrac83\leqslant x\leqslant 0\)\(f(x)=ax^2\) . Thus, at \(-\dfrac83\leqslant x\leqslant \dfrac83\), and this is a segment of length \(\dfrac(16)3\) , the 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 fine.
2) Let \(a<0\)
. Тогда картинка окажется симметричной относительно начала координат:
We need the graph \(g(x)\) to pass 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 where \(a=0\) is not suitable, because then \(f(x)=0\) for all \(x\) , \(g(x)=2\sqrtx\) and The equation will only have 1 root.
Answer:
\(a\in \left\(-\dfrac(18)(41);\dfrac(18)(23)\right\)\)
Task 4 #3072
Task level: Equal to the Unified State Examination
Find all values \(a\) , for each of which the equation \
has at least one root.
(Task from subscribers)
We 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, 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, for \(x>0\) the second module expands positively (\(|x|=x\) ), therefore, regardless of how the first module expands, \(f(x)\) will be equal to \( kx+A\) , where \(A\) is an expression from \(a\) , and \(k\) is equal to either \(-9\) or \(-3\) . For \(x<0\)
наоборот: второй модуль раскроется отрицательно и \(f(x)=kx+A\)
, где \(k\)
равно либо \(3\)
, либо \(9\)
.
Find the value \(f\) at the maximum point: \ 
In order for the equation to have at least one solution, the graphs of the functions \(f\) and \(g\) must have at least one intersection point. Therefore, you need: \ \\]
Answer:
\(a\in \(-7\)\cup\)
Task 5 #3912
Task level: Equal to the Unified State Examination
Find all values of the parameter \(a\) , for each of which the equation \
has six different solutions.
Let's make the substitution \((\sqrt2)^(x^3-3x^2+4)=t\) , \(t>0\) . 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 at most 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, having made 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 degree, 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 from the set will have no more than three solutions. This means that the whole set will have no more than six solutions.
This means that in order 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 which - or by the decision of the second!)
Obviously, if the quadratic equation \((*)\) has one solution, then we will not get six solutions for the original equation.
Thus, the solution plan becomes clear. Let's write out the conditions that must be met point by point.
1) For the equation \((*)\) to have two different solutions, its discriminant must be positive: \
2) We also need both roots to be positive (because \(t>0\) ). 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 distinct non-zero roots, 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, so 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 by Vieta's theorem: We 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, the graphs of the functions \(f\) and \(g\) must 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 evenness occupies an impressive part of the school course in mathematics. It largely determines the nature of the behavior of the function and greatly facilitates the construction of the corresponding graph. Let us define 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, the corresponding values of y (function) are equal. Let us give a more rigorous 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, the condition necessary for the domain of definition of such a function follows, 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 foregoing, therefore, the conclusion follows: an even function has a form that is symmetrical with respect to the ordinate axis (Oy). How to determine the parity of a function in practice? Let it be given using the formula h(x)=11^x+11^(-x). Following the algorithm that follows directly from the definition, we first of all study 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 argument (x) with its opposite value (-x). Let's check the evenness of the function h(x)=11^x-11^(-x). Following the same algorithm, we get h(-x) = 11^(-x) -11^x. Taking out the minus, as a result, 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 in solving equations. To solve an equation like g(x) = 0, where the left side of the equation is an even function, it will be enough to find its solutions for non-negative values of the variable. The obtained roots of the equation must be combined with opposite numbers. One of them is subject to verification. The same is successfully used to solve non-standard problems with a parameter. For example, is there any value for the parameter a that would make the equation 2x^6-x^4-ax^2=1 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 so is the opposite number. The conclusion is obvious: the roots of the equation, other than zero, are included in the set of its solutions in “pairs”. It is clear that the number 0 itself is not, that is, the number of roots of such an equation can only be even and, naturally, for any value of the parameter it cannot have three roots. But the number of roots of the equation 2^x+ 2^(-x)=ax^4+2x^2+2 can be odd, and for any value of the parameter. Indeed, it is easy to check that the set of roots of a given equation contains solutions in "pairs". Let's check if 0 is a root. When substituting it into the equation, we get 2=2. Thus, in addition to "paired" 0 is also a root, which proves their odd number. Function zeros Zeros are the points of intersection of the graph of the function with the axis Oh.
Function parity Odd function Function Increment Decreasing function Find intervals of monotonicity using the service Intervals of increasing and decreasing functions Local maximum Local minimum Function Periodicity Constancy intervals Function continuity break points Example: Explore the function and build its graph: y = x 3 - 3x
Consider the function \(f(x)=x^3-3x^2+4\) .
Can be multiplied: \
Therefore, its zeros are: \(x=-1;2\) .
If we find the derivative \(f"(x)=3x^2-6x\) , then we get two extreme 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 note right away 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, so 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 like this: \[\begin(cases) 1
We will not explicitly write out the roots.
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 abscissa axis (we wrote this condition in paragraph 1)). How should its graph look like so that the points of intersection with the abscissa axis are in the interval \((1;4)\) ? So: 
Firstly, the values \(g(1)\) and \(g(4)\) of the function at the 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, the system can be written: \[\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: \(x=0\) . For \(x<0\)
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, for \(x>0\) the first module expands positively (\(|x|=x\) ), therefore, regardless of how the second module expands, \(f(x)\) will be equal to \( kx+A\) , where \(A\) is an expression from \(a\) , and \(k\) is either \(13-10=3\) or \(13+10=23\) . For \(x<0\)
наоборот: первый модуль раскроется отрицательно и \(f(x)=kx+A\)
, где \(k\)
равно либо \(-3\)
, либо \(-23\)
.
Let's find the value \(f\) at the minimum point: \
We get:
h(-x) = 11^(-x) + 11^x.
Since addition satisfies the commutative (displacement) law, it is obvious that h(-x) = h(x) and the given functional dependence is even.
h(-x)=-(11^x-11^(-x))=- h(x). Hence h(x) is odd.
The zero of the function is the value X, at which the function becomes 0, that is, f(x)=0.
A function is called even if for any X from the domain of definition, the equality f(-x) = f(x) 
An even function is symmetrical about the axis OU
A function is called odd if for any X from the domain of definition, the equality f(-x) = -f(x) is satisfied. 
An odd function is symmetrical with respect to the origin.
A function that is neither even nor odd is called a general function. 
The function f(x) is called increasing if the larger value of the argument corresponds to the larger value of the function, i.e. x 2 >x 1 → f(x 2)> f(x 1) 
The function f(x) is called decreasing if the larger value of the argument corresponds to the smaller value of the function, i.e. x 2 >x 1 → f(x 2) 
The intervals on which the function either only decreases or only increases are called intervals of monotony. The function f(x) has 3 intervals of monotonicity:
(-∞ x 1), (x 1 , x 2), (x 3 ; +∞) 
Dot x 0 is called a local maximum point if for any X from a neighborhood of a point x 0 the following inequality holds: f(x 0) > f(x) 
Dot x 0 is called a local minimum point if for any X from a neighborhood of a point x 0 the following inequality holds: f(x 0)< f(x).
Local maximum points and local minimum points are called local extremum points. 
x 1 , x 2 - local extremum points.
The function f(x) is called periodic, with period T, if for any X f(x+T) = f(x) . 
Intervals on which the function is either only positive or only negative are called intervals of constant sign. 
f(x)>0 for x∈(x 1 , x 2)∪(x 2 , +∞), f(x)<0 при x∈(-∞,x 1)∪(x 1 , x 2)
The function f(x) is called continuous at the point x 0 if the limit of the function as x → x 0 is equal to the value of the function at this point, i.e. 
.
The points at which the continuity condition is violated are called points of discontinuity of the function. 
x0- breaking point.General scheme for plotting functions
1. Find the domain of the function D(y).
2. Find the intersection points of the graph of functions with the coordinate axes.
3. Investigate the function for even or odd.
4. Investigate the function for periodicity.
5. Find intervals of monotonicity and extremum points of the function.
6. Find intervals of convexity and inflection points of the function.
7. Find the asymptotes of the function.
8. Based on the results of the study, build a graph.
8) Based on the results of the study, we will construct a graph of the function: 
