steam rooms, since for all \(x\) from її region the calculation is correct: \(f(-x)=f(x)\) .
Graph of a pair function symmetrical along the \(y\) axis:
Example: the function \ (f (x) = x ^ 2 + \ cos x \) is paired, because \(f(-x)=(-x)^2+\cos((-x))=x^2+\cos x=f(x)\).
\(\blacktriangleright\) The function \(f(x)\) is called unpaired, since for all \(x\) from її region the calculation is correct: \(f(-x)=-f(x)\) .
The graph of an unpaired function is symmetrical around the coordinates:
Example: function \ (f (x) = x ^ 3 + x \) is unpaired, because \(f(-x)=(-x)^3+(-x)=-x^3-x=-(x^3+x)=-f(x)\).
\(\blacktriangleright\) Functions that are neither paired nor unpaired are called functions of the glacial view. Such a function can always be submitted in a single order in the form of a sum of paired and unpaired functions.
For example, the function \(f(x)=x^2-x\) is the sum of the paired function \(f_1=x^2\) and the unpaired function \(f_2=-x\) .
\(\blacktriangleright\) Actors of power:
1) Public and private are two functions, however, pairing is a pairing function.
2) The public and the private of two functions of different pairings is an unpaired function.
3) The sum and difference of paired functions is a paired function.
4) The sum and difference of unpaired functions is an unpaired function.
5) If \(f(x)\) is a paired function, then the equation \(f(x)=c \ (c\in \mathbb(R)\) ) is the same root then only if \(x =0\).
6) If \(f(x)\) is a paired or unpaired function, and equal \(f(x)=0\) is the root \(x=b\), then the equal root \( x =-b) .
\(\blacktriangleright\) The function \(f(x)\) is called periodic on \(X\) , since for a real number \(T\ne 0\) it is viconno \(f(x)=f(x+T) \) , De \ (x, x + T \ in X \) . The smallest \(T\) , for which the value is equal, is called the leading (main) period of the function.
If a periodic function has a number of the form \(nT\), then \(n\in \mathbb(Z)\) will also be a period.
Example: whether a trigonometric function is periodic;
the function \(f(x)=\sin x\) and \(f(x)=\cos x\) has a leading period greater than \(2\pi\), the function \(f(x)=\mathrm( tg)\,x\) і \(f(x)=\mathrm(ctg)\,x\) the head period is prior to \(\pi\) .
In order to create a graph of a periodic function, you can create a graph for any segment of the day (T) (head period); Then the graph of all functions is obtained by destroying the required part for the whole number of periods right-handed and left-handed:
\(\blacktriangleright\) The area of significance \(D(f)\) of the function \(f(x)\) is impersonal, which is the sum of all the values of the argument \(x\), in which the function has a sense (is significant).
Example: the function \(f(x)=\sqrt x+1\) has a range of values: \(x\in
Zavdannya 1 #6364
Rhubarb: ancient IDE
For any values of the parameter \(a\) equal
Is there only one solution?
Please note that the fragments \(x^2\) and \(\cos x\) are paired functions, since equal to the root \(x_0\) , it is also equal to the root \(-x_0\) .
True, high \(x_0\) – root, that is jealousy \(2x_0^2+a\mathrm(tg)\,(\cos x_0)+a^2=0\) That's right. Substitutable \(-x_0\) : \(2 (-x_0)^2+a\mathrm(tg)\,(\cos(-x_0))+a^2=2x_0^2+a\mathrm(tg)\,(\cos x_0)+a ^2=0\).
In this way, if \(x_0\ne 0\) , then equalization already has at least two roots. Otzhe, \(x_0 = 0\) . Todi:
We took off two values of the parameter \(a\). We respect that we were vikorized by those who (x=0) are exactly the root of the result. Ale mi nowhere did not vikorize those who are united. Therefore, you need to insert the values of the parameter \(a\) into the output equal and check if the \(a\) root \(x=0\) itself will be united.
1) If \(a=0\) , then the equation will look like \(2x^2=0\) . Obviously, the equation has only one root (x = 0). Well, the value (a = 0) is good enough for us.
2) If \(a=-\mathrm(tg)\,1\), then I will see jealousy \ Let's rewrite the rivnyannya at the sight \ So yak \(-1\leqslant \cos x\leqslant 1\), That \(-\mathrm(tg)\,1\leqslant \mathrm(tg)\,(\cos x)\leqslant \mathrm(tg)\,1\). Also, the values of the right side of the line (*) are in line \([-\mathrm(tg)^2\,1; \mathrm(tg)^2\,1]\).
The fragments \(x^2\geqslant 0\) , then the left part of the river (*) is larger or more ancient \(0+ \mathrm(tg)^2\,1\) .
In this way, jealousy (*) can only end if the offending parts of the jealousy develop \(\mathrm(tg)^2\,1\) . And this means that \[\begin(cases) 2x^2+\mathrm(tg)^2\,1=\mathrm(tg)^2\,1 \\ \mathrm(tg)\,1\cdot \mathrm(tg)\ ,(\cos x)=\mathrm(tg)^2\,1 \end(cases) \quad\Leftrightarrow\quad \begin(cases) x=0\\ \mathrm(tg)\,(\cos x) =\mathrm(tg)\,1 \end(cases)\quad\Leftrightarrow\quad x=0\] So, the value (a = - mathrm (tg), 1) is suitable for us.
Subject:
\(a\in \(-\mathrm(tg)\,1;0\)\)
Zavdannya 2 #3923
Rhubarb: ancient IDE
Find all the values of the parameter \(a\) for each function graph \
symmetrical to the cob of coordinates.
If the graph of a function is symmetrical to the coordinate system, then such a function is unpaired, so \(f(-x)=-f(x)\) is found for any \(x\) in the domain of the significant function. Thus, it is necessary to know the values of the parameter for which viconanos \(f(-x)=-f(x).\)
\[\begin(aligned) &3\mathrm(tg)\,\left(-\dfrac(ax)5\right)+2\sin \dfrac(8pi a+3x)4= -\left(3\) mathrm (tg)\,\left(\dfrac(ax)5\right)+2\sin \dfrac(8pi-3x)4\right)\quad \Rightarrow\quad -3\mathrm(tg)\ , \dfrac ( ax) 5 + 2 sin dfrac(8pi-3x)4right)quadRightarrowRightarrowquad &sindfrac(8pia+3x)4+sindfrac(8pi- 3x)4=0 \quad \Rightarrow \quad2\sin \dfrac12\left(\dfrac(8pi a +3x)4+dfrac(8pi-3x)4right)\cdot \cos \dfrac12 \left(\dfrac(8pi a+3x)4-dfrac(8pi-3x)4right)=0 \quad \Rightarrow\quad \sin (2\pi a)\cdot \cos \frac34 x=0 \end(aligned)\]
Remaining jealousy may be for everyone \(x\) from the area of appointment \(f(x)\), so, \(\sin(2\pi a)=0 \Rightarrow a=\dfrac n2, n\in\mathbb(Z)\).
Subject:
\(\dfrac n2, n\in\mathbb(Z)\)
Zavdannya 3 #3069
Rhubarb: ancient IDE
Find all the values of the parameter \(a\), for each case there are 4 solutions, where \(f\) is a pair periodic with the period \(T=\dfrac(16)3\) function, calculated on the entire number line, Moreover, \(f(x)=ax^2\) for \(0\leqslant x\leqslant \dfrac83.\)
(Responsibility for prepayments)
Since \(f(x)\) is a pair function, then its graph is symmetrical along the ordinate axis, therefore, with \(-\dfrac83\leqslant x\leqslant 0\)\(f(x)=ax^2\) . In this manner, with \(-\dfrac83\leqslant x\leqslant \dfrac83\), and this is a shortcut \(\dfrac(16)3\) , function \(f(x)=ax^2\).
1) Let go \ (a> 0 \). Then the graph of the function \(f(x)\) looks like this: 
Therefore, in order to make a few decisions, it is necessary to make a graph \(g(x)=|a+2|\cdot \sqrtx\) passing through the point \(A\) : 
Otje, \[\dfrac(64)9a=|a+2|\cdot \sqrt8 \quad\Leftrightarrow\quad \left[\begin(gathered)\begin(aligned) &9(a+2)=32a\&9(a + 2)=-32a\end(aligned)\end(gathered)\right. \quad\Leftrightarrow\quad \left[\begin(gathered)\begin(aligned) &a=\dfrac(18)(23)\\ &a=-\dfrac(18)(41) \end(aligned) \end( gathered) \right.\] So since \ (a> 0\), then go \ (a = \dfrac (18) (23)\).
2) Let it go (a)<0\)
. Тогда картинка окажется симметричной относительно начала координат:
It is necessary for the graph \(g(x)\) to pass through the point \(B\) : \[\dfrac(64)9a=|a+2|\cdot \sqrt(-8) \quad\Leftrightarrow\quad \left[\begin(gathered)\begin(aligned) &a=\dfrac(18)(23 )\\ &a=-\dfrac(18)(41) \end(aligned) \end(gathered)\right.\] Oskolki \(a<0\)
, то подходит \(a=-\dfrac{18}{41}\)
.
3) The fall, if \(a=0\) , does not fit, then \(f(x)=0\) for all \(x\) , \(g(x)=2\sqrtx\) і Rivalry to mother is less than 1 korіn.
Subject:
\(a\in \left\(-\dfrac(18)(41);\dfrac(18)(23)\right\)\)
Zavdannya 4 #3072
Rhubarb: ancient IDE
Find out all the meanings of \(a\), for skin conditions \
I would like one root.
(Responsibility for prepayments)
Let's rewrite the rivnyannya at the sight \
And let's look at 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 paired and points to the minimum point \(x=0\) (and \(g(0)=49\)).
The function \(f(x)\) for \(x>0\) is decaying, and for \(x<0\)
– возрастающей, следовательно, \(x=0\)
– точка максимума.
It is true that when \(x>0\) the other module opens up positively (\(|x|=x\)), and therefore, regardless of how the first module opens up, \(f(x)\) will be more important than \( kx +A\) , where \(A\) is different from \(a\) , 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 \(9\)
.
We know the value of \(f\) at the maximum point: \ 
In order to achieve more than one solution, it is necessary to graph the functions (f) and (g) or one point of the crossbar. Well, it’s necessary: \ \\]
Subject:
\(a\in \(-7\)\cup\)
Zavdannya 5 #3912
Rhubarb: ancient IDE
Find all the values of the parameter \(a\), for skin problems \
There are six different solutions.
It’s important to replace \((\sqrt2)^(x^3-3x^2+4)=t\) , \(t>0\) . Then I see jealousy in the future \
Step by step, write your mind, for which weekend we have six decisions.
Dear, square measure ((*)) can make at most two decisions. If there is a cubic equation (Ax^3+Bx^2+Cx+D=0\) there can be no more than three solutions. Therefore, since the equation \((*)\) has two different decisions (positive!, the fragments \(t\) may be greater than zero) \(t_1\) and \(t_2\), then, having made a reverse substitution, we are rejected : \[\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. So, if a positive number can be represented as \(\sqrt2\) as a world, for example, \(t_1=(\sqrt2)^(\log_(\sqrt2) t_1)\), then first of all the totality will be rewritten as \
As we have already said, even if the cubic equation takes no more than three decisions, then the skin balance from the totality of the matter requires no more than three decisions. This means that the entire total is no more than six solutions.
Therefore, so that the output is a little six decisions, the square equation \((*)\) is responsible for the mother of two different decisions, and the skin is removed cubic equal (from the totality) is guilty of the mother of three different decisions (and moreover It’s not wrong for anyone who is jealous to run away with any decisions of another!)
Obviously, since the square equation \((*)\) is equal to one solution, then we cannot remove six solutions from the output equation.
In this way, the decision plan becomes reasonable. Let's write down point by point what is likely to happen.
1) If equal \((*)\) there are two different decisions, the discriminant may be positive: \
2) It is also necessary for the offense roots to be positive (shards \(t>0\)). Since the product of two roots is positive and their sum is positive, then the roots themselves will be positive. Well, it’s necessary: \[\begin(cases) 12-a>0\\(a-10)>0\end(cases)\quad\Leftrightarrow\quad a<10\]
In this way, we have already secured two different positive roots \(t_1\) and \(t_2\).
3)
Let's marvel at such jealousy \
For what \(t\) are these three massacres decided? In this way, we determined that the resentment of the roots of jealousy ((*)) is to blame for lying in the intervals ((1; 4)). How can I write down my thoughts? There are few different roots, different from zero, which represent at the same time (x=0) an arithmetic progression. It is important to note that the function \(y=25x^4+25(a-1)x^2-4(a-7)\) is a paired function, since \(x_0\) is the root of \((*)\ ) , then th \(-x_0\) will be its root. It is then necessary that the roots of this number be ordered by increasing numbers: \(-2d, -d, d, 2d\) (so \(d>0\)). The five numbers themselves will create arithmetic progression (with difference (d)). For the roots of the numbers \(-2d, -d, d, 2d\), it is necessary for the numbers \(d^(\,2), 4d^(\,2)\) to be the roots of the numbers \(25t^2 +25(a-1)t-4(a-7)=0\). Following on from Viet's theorem: Let's rewrite the rivnyannya at the sight \
And let's look at two functions: \(g(x)=20a-a^2-2^(x^2+2)\) and \(f(x)=13|x|-2|5x+12a|\) . In order to achieve more than one solution, it is necessary to graph the functions (f) and (g) or one point of the crossbar. Well, it’s necessary: \
Given the totality of systems, we reject the following: \\]
Subject: \(a\in \(-2\)\cup\) Pair function.
Guy is a function whose sign does not change when the sign is changed x. x jealousy ends f(–x) = f(x). Sign x does not fit into the sign y. The graph of a paired function is symmetrical along the coordinate axis (Fig. 1). Applications of paired functions: y=cos x y = x 2 y = –x 2 y = x 4 y = x 6 y = x 2 + x Explanation: Unpaired function.
Unpaired is the name of a function whose sign changes when the sign is changed x. Otherwise, seemingly, for whatever meaning x jealousy ends f(–x) = –f(x). The graph of an unpaired function is symmetrical to the coordinates (Fig. 2). Applications of unpaired functions: y= sin x y = x 3 y = –x 3 Explanation: Let's take the function y = - x 3 . Power of paired and unpaired functions:
NOTE:
Not all functions are paired or unpaired. And functions that are not subject to such gradation. For example, the root function at = √X does not belong to either paired or unpaired functions (Fig. 3). When the authorities re-interpret such functions, give a clear description: neither paired nor unpaired. Periodic functions.
As you know, periodicity is the repetition of previous processes at a regular interval. The functions that describe these processes are called periodic functions. These are the same functions as graphs that have elements that are repeated at regular numerical intervals. Which is what you knew in another world. It was noted there that the supply of power functions would gradually be replenished. This paragraph is about two new powers. Value 1. The function y = f(x), x є X, is called a pair function, since for any value of x from the multiplier X the equality f(-x) = f(x) is determined. Value 2. The function y = f(x), x є X, is called unpaired, since for any value of x from the multiplier X the equality f(-x) = -f(x) is calculated. Bring that y = x 4 is a paired function. Decision. Maєmo: f(x) = x4, f(-x) = (-x)4. Ale(-x) 4 = x4. Well, no matter what x, the jealousy of f(-x) = f(x), then. The function is paired. Similarly, it can be shown that the functions y - x 2, y = x 6, y - x 8 are equal. Show that y = x 3 is an unpaired function. Decision. Maєmo: f(x) = x3, f(-x) = (-x)3. Ale(-x) 3 = -x 3. So, no matter what x, the jealousy of f (-x) = -f (x) is determined. the function is unpaired. Similarly, it can be concluded that the functions y = x, y = x 5, y = x 7 are unpaired. You and I have discussed this many times, new terms in mathematics most often appear on the “earth”, so. They can be explained in some way. This one on the right is with paired and unpaired functions. Watch: y - x 3, y = x 5, y = x 7 - unpaired functions, so y = x 2, y = x 4, y = x 6 - paired functions. And for any function of the form y = x "(below we will specially deal with the development of these functions), where n is a natural number, you can create a formula: if n is an unpaired number, then the function y = x" is unpaired; Since n is a number, then the function y = xn is a number. There are also functions that are neither paired nor unpaired. So, for example, the function y = 2x + 3. In fact, f(1) = 5, and f(-1) = 1. As you see, here it means that we cannot agree on the sameness f(-x) = f (x) , not the same as f(-x) = -f(x). Also, the function can be paired, unpaired, and also similar. Varying nutrition, which has a paired or unpaired function, is called the additional function of pairing. Values 1 and 2 indicate the meaning of the function at points x and -x. Tim himself is conveyed that the function is designated both in point x and in point -x. This means that the point -x lies in the area of the value of the function at the same time as the point x. If the numerical anonymity X is simultaneously removed from its skin element x and the proximal element x, X is called a symmetrical anonymity. Let's say, (-2, 2), [-5, 5], (-oo, +oo) - symmetrical multiplicities, at that time)
Let's look at the function \(f(x)=x^3-3x^2+4\).
Can be divided into multipliers: \
Also, these are zeros: \ (x = -1; 2 \).
Once we have calculated the value \(f"(x)=3x^2-6x\), we subtract two points from the extremum \(x_(max)=0, x_(min)=2\) .
Well, the graph looks like this: 
Mi, if the line \(y=k\) is horizontal, de \(0
In this manner, it is necessary: \[\begin(cases) 0<\log_{\sqrt2}t_1<4\\ 0<\log_{\sqrt2}t_2<4\end{cases}\qquad (**)\]
Let us also be very respectful that since the numbers \(t_1\) and \(t_2\) are different, then the numbers \(\log_(\sqrt2)t_1\) and \(\log_(\sqrt2)t_2\) will be different, that means jealousy \(x^3-3x^2+4=\log_(\sqrt2) t_1\)і \(x^3-3x^2+4=\log_(\sqrt2) t_2\) match the roots that do not coincide with each other.
The system \((**)\) can be rewritten as follows: \[\begin(cases) 1
In obvious cases, we will not write down the roots.
Let's look at the function \(g(t)=t^2+(a-10)t+12-a\) . This graph is a parabola with burnt edges, which has two points across the entire abscissa (we wrote this down in point 1)). How can one look at this graph so that the points intersect with the entire abscissa in the interval \((1;4)\)? So: 
First of all, the values of \(g(1)\) and \(g(4)\) functions at points \(1\) and \(4\) are to blame, in other words, the vertex of the parabola \(t_0\ ) ) is also guilty of being in the interval \((1;4)\). Well, you 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 \(g(x)\) points to the maximum \(x=0\) (and \(g_(\text(top))=g(0)=-a^2+20a-4\)):
\(g"(x)=-2^(x^2+2)\cdot \ln 2\cdot 2x\). Zero value: \(x = 0\). When (x<0\)
имеем: \(g">0\) for \(x>0\) : \(g"<0\)
.
The function \(f(x)\) for \(x>0\) is growing, and for \(x<0\)
– убывающей, следовательно, \(x=0\)
– точка минимума.
It is true that when \(x>0\) the first module opens up positively (\(|x|=x\)), and therefore, regardless of how the other module opens up, \(f(x)\) will be more important than \( kx +A\) , de \(A\) - viraz \(a\) , and \(k\) one 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 \(-23\)
.
We know the value of \(f\) at the minimum point: \
Let's take the function y = x 2 or y = –x 2 .
For whatever it means x the function is positive. Sign x does not fit into the sign y. The graph is symmetrical along the coordinate axis. This is the same function.
All meanings at they will have a minus sign. Tobto sign x flows onto the sign y. If an independent changeable is a positive number, that function is positive, if an independent changeable is a negative number, that function is negative: f(–x) = –f(x).
The graph of the function is symmetrical around the coordinates. This is an unpaired function.