How to check if a function is even or odd. Even and odd functions

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Goals:

• to form the concept of even and odd functions, to teach the ability to determine and use these properties in the study of functions, plotting graphs;
• to develop the creative activity of students, logical thinking, the ability to compare, generalize;
• to cultivate diligence, mathematical culture; develop communication skills .

Equipment: multimedia installation, interactive whiteboard, handouts.

Forms of work: frontal and group with elements of search and research activities.

Information sources:

1. Algebra class 9 A.G. Mordkovich. Textbook.
2. Algebra Grade 9 A.G. Mordkovich. Task book.
3. Algebra grade 9. Tasks for learning and development of students. Belenkova E.Yu. Lebedintseva E.A.

DURING THE CLASSES

1. Organizational moment

Setting goals and objectives of the lesson.

2. Checking homework

No. 10.17 (Problem book 9th grade A.G. Mordkovich).

a) at = f(X), f(X) =

b) f (–2) = –3; f (0) = –1; f(5) = 69;

c) 1. D( f) = [– 2; + ∞)
2. E( f) = [– 3; + ∞)
3. f(X) = 0 for X ~ 0,4
4. f(X) >0 at X > 0,4 ; f(X) < 0 при – 2 < X < 0,4.
5. The function increases with X € [– 2; + ∞)
6. The function is limited from below.
7. at hire = - 3, at naib doesn't exist
8. The function is continuous.

(Did you use the feature exploration algorithm?) Slide.

2. Let's check the table that you were asked on the slide.

Fill the table
	Domain	Function zeros	Constancy intervals		Coordinates of the points of intersection of the graph with Oy
		x = -5, x = 2	х € (–5;3) U U(2;∞)	х € (–∞;–5) U U (–3;2)
	x ∞ -5, x ≠ 2		х € (–5;3) U U(2;∞)	х € (–∞;–5) U U (–3;2)
	x ≠ -5, x ≠ 2		x € (–∞; –5) U U(2;∞)	x € (–5; 2)

3. Knowledge update

– Functions are given.
– Specify the domain of definition for each function.
– Compare the value of each function for each pair of argument values: 1 and – 1; 2 and - 2.
– For which of the given functions in the domain of definition are the equalities f(– X) = f(X), f(– X) = – f(X)? (put the data in the table) Slide

		f(1) and f(– 1)	f(2) and f(– 2)	charts	f(– X) = –f(X)	f(– X) = f(X)
1. f(X) =
2. f(X) = X 3
3. f(X) = | X |
4.f(X) = 2X – 3
5. f(X) =	X ≠ 0
6. f(X)=	X > –1		and not defined.

4. New material

- While doing this work, guys, we have revealed one more property of the function, unfamiliar to you, but no less important than the others - this is the evenness and oddness of the function. Write down the topic of the lesson: “Even and odd functions”, our task is to learn how to determine the even and odd functions, find out the significance of this property in the study of functions and plotting.
So, let's find the definitions in the textbook and read (p. 110) . Slide

Def. one Function at = f (X) defined on the set X is called even, if for any value XЄ X in progress equality f (–x) = f (x). Give examples.

Def. 2 Function y = f(x), defined on the set X is called odd, if for any value XЄ X the equality f(–х)= –f(х) is satisfied. Give examples.

Where did we meet the terms "even" and "odd"?
Which of these functions will be even, do you think? Why? Which are odd? Why?
For any function of the form at= x n, where n is an integer, it can be argued that the function is odd for n is odd and the function is even for n- even.
– View functions at= and at = 2X– 3 is neither even nor odd, because equalities are not met f(– X) = – f(X), f(– X) = f(X)

The study of the question of whether a function is even or odd is called the study of a function for parity. Slide

Definitions 1 and 2 dealt with the values ​​of the function at x and - x, thus it is assumed that the function is also defined at the value X, and at - X.

ODA 3. If a number set together with each of its elements x contains the opposite element x, then the set X is called a symmetric set.

Examples:

(–2;2), [–5;5]; (∞;∞) are symmetric sets, and , [–5;4] are nonsymmetric.

- Do even functions have a domain of definition - a symmetric set? The odd ones?
- If D( f) is an asymmetric set, then what is the function?
– Thus, if the function at = f(X) is even or odd, then its domain of definition is D( f) is a symmetric set. But is the converse true, if the domain of a function is a symmetric set, then it is even or odd?
- So the presence of a symmetric set of the domain of definition is a necessary condition, but not a sufficient one.
– So how can we investigate the function for parity? Let's try to write an algorithm.

Slide

Algorithm for examining a function for parity

1. Determine whether the domain of the function is symmetrical. If not, then the function is neither even nor odd. If yes, then go to step 2 of the algorithm.

2. Write an expression for f(–X).

3. Compare f(–X).and f(X):

• if f(–X).= f(X), then the function is even;
• if f(–X).= – f(X), then the function is odd;
• if f(–X) ≠ f(X) and f(–X) ≠ –f(X), then the function is neither even nor odd.

Examples:

Investigate the function for parity a) at= x 5 +; b) at= ; in) at= .

Decision.

a) h (x) \u003d x 5 +,

1) D(h) = (–∞; 0) U (0; +∞), symmetric set.

2) h (- x) \u003d (-x) 5 + - x5 - \u003d - (x 5 +),

3) h (- x) \u003d - h (x) \u003d\u003e function h(x)= x 5 + odd.

b) y =,

at = f(X), D(f) = (–∞; –9)? (–9; +∞), asymmetric set, so the function is neither even nor odd.

in) f(X) = , y = f(x),

1) D( f) = (–∞; 3] ≠ ; b) (∞; –2), (–4; 4]?

Option 2

1. Is the given set symmetric: a) [–2;2]; b) (∞; 0], (0; 7) ?

a); b) y \u003d x (5 - x 2). 2. Examine the function for parity:

a) y \u003d x 2 (2x - x 3), b) y \u003d

3. In fig. plotted at = f(X), for all X, satisfying the condition X? 0.
Plot the Function at = f(X), if at = f(X) is an even function.

3. In fig. plotted at = f(X), for all x satisfying x? 0.
Plot the Function at = f(X), if at = f(X) is an odd function.

Mutual check on slide.

6. Homework: №11.11, 11.21,11.22;

Proof of the geometric meaning of the parity property.

*** (Assignment of the USE option).

1. The odd function y \u003d f (x) is defined on the entire real line. For any non-negative value of the variable x, the value of this function coincides with the value of the function g( X) = X(X + 1)(X + 3)(X– 7). Find the value of the function h( X) = at X = 3.

7. Summing up

Which to one degree or another were familiar to you. It was also noted there that the stock of function properties will be gradually replenished. Two new properties will be discussed in this section.

Definition 1.

The function y \u003d f (x), x є X, is called even if for any value x from the set X the equality f (-x) \u003d f (x) is true.

Definition 2.

The function y \u003d f (x), x є X, is called odd if for any value x from the set X the equality f (-x) \u003d -f (x) is true.

Prove that y = x 4 is an even function.

Decision. We have: f (x) \u003d x 4, f (-x) \u003d (-x) 4. But (-x) 4 = x 4 . Hence, for any x, the equality f (-x) = f (x), i.e. the function is even.

Similarly, it can be proved that the functions y - x 2, y \u003d x 6, y - x 8 are even.

Prove that y = x 3 is an odd function.

Decision. We have: f (x) \u003d x 3, f (-x) \u003d (-x) 3. But (-x) 3 = -x 3 . Hence, for any x, the equality f (-x) \u003d -f (x), i.e. the function is odd.

Similarly, it can be proved that the functions y \u003d x, y \u003d x 5, y \u003d x 7 are odd.

You and I have repeatedly convinced ourselves that new terms in mathematics most often have an “earthly” origin, i.e. they can be explained in some way. This is the case for both even and odd functions. See: y - x 3, y \u003d x 5, y \u003d x 7 are odd functions, while y \u003d x 2, y \u003d x 4, y \u003d x 6 are even functions. And in general, for any function of the form y \u003d x "(below we will specifically study these functions), where n is natural number, we can conclude: if n is an odd number, then the function y \u003d x "is odd; if n is an even number, then the function y \u003d xn is even.

There are also functions that are neither even nor odd. Such, for example, is the function y \u003d 2x + 3. Indeed, f (1) \u003d 5, and f (-1) \u003d 1. As you can see, here Hence, neither the identity f (-x) \u003d f ( x), neither identity f(-x) = -f(x).

So, a function can be even, odd, or neither.

The study of the question of whether a given function is even or odd is usually called the study of the function for parity.

Definitions 1 and 2 deal with the values ​​of the function at the points x and -x. This assumes that the function is defined both at the point x and at the point -x. This means that the point -x belongs to the domain of the function at the same time as the point x. If a numerical set X together with each of its elements x contains the opposite element -x, then X is called a symmetric set. Let's say (-2, 2), [-5, 5], (-oo, +oo) are symmetric sets, while \).

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 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?
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 \(x^3-3x^2+4=\log_(\sqrt2) t\) has three different solutions, it is necessary that \(0<\log_ {\sqrt2}t<4\) .
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

Thus, we have determined that both roots of the equation \((*)\) must lie in the interval \((1;4)\) . How to write this condition?
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\(a\) always has at least one root \(x=0\) . So, to fulfill the condition of the problem, it is necessary that the equation \

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 \(f(x)=13|x|-2|5x+12a|\) .
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: \

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\)