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 general view. 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 - even function.
2) The product and quotient of two functions of different parities is an odd function.
3) The sum and difference of even functions is an 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 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 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\) To do this, use graph paper or a graphing calculator. Select any number of independent variable values x (\displaystyle x) and plug them into the function to calculate the values of the dependent variable y (\displaystyle y). Plot the found coordinates of the points on the coordinate plane, and then connect these points to build a graph of the function. Check whether the graph of the function is symmetrical about the Y axis. Symmetry means a mirror image of the graph relative to the ordinate axis. If the part of the graph to the right of the Y-axis (positive values of the independent variable) is the same as the part of the graph to the left of the Y-axis (negative values of the independent variable), the graph is symmetrical about the Y-axis. If the function is symmetrical about the y-axis, the function is even. Check whether the graph of the function is symmetrical about the origin. The origin is the point with coordinates (0,0). Symmetry about the origin means that a positive value y (\displaystyle y)(with a positive value x (\displaystyle x)) corresponds to a negative value y (\displaystyle y)(with a negative value x (\displaystyle x)), and vice versa. Odd functions have symmetry about the origin. Check if the graph of the function has any symmetry. The last type of function is a function whose graph has no symmetry, that is, there is no mirror image both relative to the ordinate axis and relative to the origin. For example, given the function . Function zeros Zeros are the points of intersection of the function graph with the axis Oh.
Function parity Odd parity function Increasing function Descending function Find intervals of monotonicity using the service Intervals of increasing and decreasing function Local maximum Local minimum Function frequency Intervals of sign constancy Continuity of function Break points General scheme for plotting functions 1. Find the domain of definition of the function D(y). 2. Find the points of intersection of the graph of functions with the coordinate axes. 3. Examine the function for even or odd. 4. Examine the function for periodicity. 5. Find monotonicity intervals and extremum points of the function. 6. Find the convexity intervals and inflection points of the function. 7. Find the asymptotes of the function. 8. Based on the results of the study, construct a graph. Example: Explore the function and plot it: y = x 3 – 3x 1) The function is defined on the entire numerical axis, i.e. its domain of definition is D(y) = (-∞; +∞). 2) Find the points of intersection with the coordinate axes: with the OX axis: solve the equation x 3 – 3x = 0 with OY axis: y(0) = 0 3 – 3*0 = 0 3) Find out whether the function is even or odd: y(-x) = (-x) 3 – 3(-x) = -x 3 + 3x = - (x 3 – 3x) = -y(x) It follows that the function is odd. 4) The function is non-periodic. 5) Let’s find the monotonicity intervals and extremum points of the function: y’ = 3x 2 - 3. Critical points: 3x 2 – 3 = 0, x 2 =1, x= ±1. y(-1) = (-1) 3 – 3(-1) = 2 y(1) = 1 3 – 3*1 = -2 6) Find the convexity intervals and inflection points of the function: y’’ = 6x Critical points: 6x = 0, x = 0. y(0) = 0 3 – 3*0 = 0 7) The function is continuous, it has no asymptotes. 8) Based on the results of the study, we will construct a graph of the function. 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, from the domain of definition of the function the following equality must be satisfied: 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: \
The zero of a function is the value X, at which the function turns to 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) holds 
An even function is symmetrical about the axis Oh
A function is called odd if for any X from the domain of definition the equality f(-x) = -f(x) holds. 
An odd function is symmetric about the origin.
A function that is neither even nor odd is called a general function. 
A function f(x) is said to be increasing if a larger value of the argument corresponds to a larger value of the function, i.e. 
A function f(x) is called decreasing if a larger value of the argument corresponds to a smaller value of the function, i.e. 
Intervals over which the function either only decreases or only increases are called intervals of monotony. The function f(x) has 3 intervals of monotonicity: 
Dot x 0 is called a local maximum point if for any X from the vicinity 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 the vicinity of a point x 0 inequality holds: f(x 0)< f(x).
Local maximum points and local minimum points are called local extremum points. 
local extremum points.
The function f(x) is called periodic, with a period T, if for any X the equality f(x+T) = f(x) holds. 
Intervals on which the function is either only positive or only negative are called intervals of constant sign. 
A function f(x) is called continuous at a 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 function break points. 
x 0- break point.
Graph of an even function
Graph of an odd function
