Even to even function. How to determine even and odd functions
In July 2020, NASA launches an expedition to Mars. spacecraft will deliver to Mars an electronic carrier with the names of all registered members of the expedition.
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Another New Year's Eve... frosty weather and snowflakes on the window glass... All this prompted me to write again about... fractals, and what Wolfram Alpha knows about it. On this occasion, there is an interesting article in which there are examples of two-dimensional fractal structures. Here we will look at more complex examples three-dimensional fractals.
A fractal can be visually represented (described) as a geometric figure or body (meaning that both are a set, in this case, a set of points), the details of which have the same shape as the original figure itself. That is, it is a self-similar structure, considering the details of which, when magnified, we will see the same shape as without magnification. Whereas in the case of the usual geometric figure(not a fractal), when zoomed in, we will see details that have a simpler shape than the original figure itself. For example, at a sufficiently high magnification, part of an ellipse looks like a straight line segment. This does not happen with fractals: with any increase in them, we will again see the same complex shape, which with each increase will be repeated again and again.
Benoit Mandelbrot, the founder of the science of fractals, in his article Fractals and Art for Science wrote: "Fractals are geometric shapes that are as complex in their details as they are in their overall form. That is, if part of the fractal will be enlarged to the size of the whole, it will look like the whole, or exactly, or perhaps with a slight deformation.
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 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 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 - 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.
At periodic function 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 \(\cos x\) - even functions, then 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 exactly \(a\) 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 \ Because \(-1\leqslant \cos x\leqslant 1\), 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, \(f(-x)=-f(x)\) is satisfied 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)
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, 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 \(-7\)\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.
notice, that 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 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 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 distinct 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? Thus, we need to intersect the \(a\) parameter values found in the 1st, 2nd and 3rd paragraphs, and we will get the answer: \[\begin(cases) a\in (-\infty;8-2\sqrt3)\cup(8+2\sqrt3;+\infty)\\ a<10\\
4
Definition 1. The function is called even
(odd
) if together with each value of the variable Thus, a function can be even or odd only when its domain of definition is symmetrical with respect to the origin of coordinates on the real line (numbers X and - X simultaneously belong Function Function Function The graph of an even function is symmetrical about the axis OU, since 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 an odd function is an odd function. d) If f is an even function on the set X, and the function g
defined on the set e) If f is an 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 other assertions of the theorem are proved similarly. The theorem has been proven. Theorem 2. Any function Proof. Function . Function Definition 2. Function Such a number T called period
functions Definition 1 implies that if T– function period Definition 3. The smallest of the positive periods of a function is called its main
period. Theorem 3. If T is the main period of the function f, then the remaining periods are multiples of it. Proof. Assume the opposite, that is, that there is a period functions f
(>0), not multiple T. Then, dividing on the T with the remainder, we get that is – function period f, and It is well known that trigonometric functions are periodic. Main period (because ororor Meaning T, determined from the first equality, cannot be a period, since it depends on X, i.e. is a function of X, 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 has no main period, since there are positive rational numbers arbitrarily close to zero (for example, a rational number can be made by choosing n arbitrarily close to zero). Theorem 4. If function f
set on the set X and has a period T, and the function g
set on the set Proof. We have therefore that is, the assertion of the theorem is proved. For example, since cos
x
has a period Definition 4. Functions that are not periodic are called non-periodic
. 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. If the limit of the function exists, 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 asymptote - straight species, subject to the existence limit . 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 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 segment [ 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. Necessary condition for the existence of an inflection point: if the function is twice differentiable in some punctured neighborhood of the point , then either . Check if the graph of the function is symmetrical about the y-axis. Symmetry refers to the mirror image of the graph about the y-axis. If the part of the graph to the right of the y-axis (positive values of the independent variable) matches 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 if 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 with respect to the origin. Check if the graph of the function has any symmetry. The last type of function is a function whose graph does not have symmetry, that is, there is no mirror image both relative to the y-axis and relative to the origin. For example, given a function.
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
meaning - X also belongs
and the equality
). For example, the function
is neither even nor odd, since its domain of definition
not symmetrical about the origin.
even, because
symmetrical with respect to the origin of coordinates and.
odd because
and
.
is neither even nor odd, since although
and is symmetric with respect to the origin, equalities (11.1) are not satisfied. For example,.
also belongs to the graph. The graph of an odd function is symmetrical about the origin, because if
belongs to the graph, then the point
also belongs to the graph.
, then the function
- even.
and even (odd), then the function
- even (odd).
and
are even functions. Then, therefore. The case of odd functions is considered similarly
and
.
, defined on the set X, which is symmetric with respect to the origin, can be represented as the sum of an even and an odd function.
can be written in the form
is even, since
, and the function
is odd because. In this way,
, where
- even, and
is an odd function. The theorem has been proven.
called periodical
if there is a number
, such that for any
numbers
and
also belong to the domain of definition
and the equalities
.
, then the number T too
is the period of the function
(because when replacing T on the - T equality is maintained). Using the method of mathematical induction, it can be shown that if T– function period f, then and
, is also a period. It follows that if a function has a period, then it has infinitely many periods.
, where
. That's why
, which contradicts the fact that T is the main period of the function f. The assertion of the theorem follows from the obtained contradiction. The theorem has been proven.
and
equals
,
and
. Find the period of the function
. Let
is the period of this function. Then
.
.
. There are infinitely many periods
the smallest positive period is obtained when
:
. This is the main period of the function
.
and
are rational numbers under rational X and irrational when irrational X. That's why
, then the complex function
also has a period T.
, then the functions
have a period
.Removable breakpoints
Horizontal
oblique
Conditions of existence