Function is one of the most important mathematical concepts. Function - variable dependency at from a variable x, if each value X matches a single value at. variable X called the independent variable or argument. variable at called the dependent variable. All values ​​of the independent variable (variable x) form the domain of the function. All values ​​that the dependent variable takes (variable y), form the range of the function.

Function Graph they call the set of all points of the coordinate plane, the abscissas of which are equal to the values ​​of the argument, and the ordinates are equal to the corresponding values ​​of the function, that is, the values ​​of the variable are plotted along the abscissa axis x, and the values ​​of the variable are plotted along the y-axis y. To plot a function, you need to know the properties of the function. The main properties of the function will be discussed below!

To plot a function graph, we recommend using our program - Graphing Functions Online. If you have any questions while studying the material on this page, you can always ask them on our forum. Also on the forum you will be helped to solve problems in mathematics, chemistry, geometry, probability theory and many other subjects!

Basic properties of functions.

1) Function scope and function range.

The scope of a function is the set of all valid valid values ​​of the argument x(variable x) for which the function y = f(x) defined.
The range of a function is the set of all real values y that the function accepts.

In elementary mathematics, functions are studied only on the set of real numbers.

2) Function zeros.

Values X, at which y=0, is called function zeros. These are the abscissas of the points of intersection of the graph of the function with the x-axis.

3) Intervals of sign constancy of a function.

The intervals of sign constancy of a function are such intervals of values x, on which the values ​​of the function y either only positive or only negative are called intervals of sign constancy of the function.

4) Monotonicity of the function.

An increasing function (in a certain interval) is a function in which a larger value of the argument from this interval corresponds to a larger value of the function.

Decreasing function (in some interval) - a function in which a larger value of the argument from this interval corresponds to a smaller value of the function.

5) Even (odd) functions.

An even function is a function whose domain of definition is symmetric with respect to the origin and for any X f(-x) = f(x). The graph of an even function is symmetrical about the y-axis.

An odd function is a function whose domain of definition is symmetric with respect to the origin and for any X from the domain of definition the equality f(-x) = - f(x). The graph of an odd function is symmetrical about the origin.

Even function
1) The domain of definition is symmetrical with respect to the point (0; 0), that is, if the point a belongs to the domain of definition, then the point -a also belongs to the domain of definition.
2) For any value x f(-x)=f(x)
3) The graph of an even function is symmetrical about the Oy axis.

odd function has the following properties:
1) The domain of definition is symmetrical with respect to the point (0; 0).
2) for any value x, which belongs to the domain of definition, the equality f(-x)=-f(x)
3) The graph of an odd function is symmetrical with respect to the origin (0; 0).

Not every function is even or odd. Functions general view are neither even nor odd.

6) Limited and unlimited functions.

A function is called bounded if there exists a positive number M such that |f(x)| ≤ M for all values ​​of x . If there is no such number, then the function is unbounded.

7) Periodicity of the function.

A function f(x) is periodic if there exists a non-zero number T such that for any x from the domain of the function, f(x+T) = f(x). This smallest number is called the period of the function. All trigonometric functions are periodic. (Trigonometric formulas).

Function f is called periodic if there exists a number such that for any x from the domain of definition the equality f(x)=f(x-T)=f(x+T). T is the period of the function.

Every periodic function has an infinite number of periods. In practice, the smallest positive period is usually considered.

The values ​​of the periodic function are repeated after an interval equal to the period. This is used when plotting graphs.

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 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 \(-1\leqslant \cos x\leqslant 1\), That \(-\mathrm(tg)\,1\leqslant \mathrm(tg)\,(\cos x)\leqslant \mathrm(tg)\,1\). 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.
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 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

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

Back forward

Attention! The slide preview is for informational purposes only and may not represent the full extent of the presentation. If you are interested in this work, please download the full version.

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;
• 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. 1 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= ; V) at= .

Solution.

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.

V) 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

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 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

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 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 .

A function is called even (odd) if for any and the equality

.

The graph of an even function is symmetrical about the axis
.

The graph of an odd function is symmetrical about the origin.

Example 6.2. Examine for even or odd functions

1)
; 2)
; 3)
.

Solution.

1) The function is defined with
. Let's find
.

Those.
. So this function is even.

2) The function is defined for

Those.
. Thus, this function is odd.

3) the function is defined for , i.e. For

,
. Therefore, the function is neither even nor odd. Let's call it a general function.

3. Investigation of a function for monotonicity.

Function
is called increasing (decreasing) on ​​some interval if in this interval each larger value of the argument corresponds to a larger (smaller) value of the function.

Functions increasing (decreasing) on ​​some interval are called monotonic.

If the function
differentiable on the interval
and has a positive (negative) derivative
, then the function
increases (decreases) in this interval.

Example 6.3. Find intervals of monotonicity of functions

1)
; 3)
.

Solution.

1) This function is defined on the entire number axis. Let's find the derivative.

The derivative is zero if
And
. Domain of definition - numerical axis, divided by points
,
for intervals. Let us determine the sign of the derivative in each interval.

In the interval
the derivative is negative, the function decreases on this interval.

In the interval
the derivative is positive, therefore, the function is increasing on this interval.

2) This function is defined if
or

.

We determine the sign of the square trinomial in each interval.

Thus, the scope of the function

Let's find the derivative
,
, If
, i.e.
, But
. Let us determine the sign of the derivative in the intervals
.

In the interval
the derivative is negative, therefore, the function decreases on the interval
. In the interval
the derivative is positive, the function increases on the interval
.

4. Investigation of a function for an extremum.

Dot
is called the maximum (minimum) point of the function
, if there is such a neighborhood of the point that for everyone
this neighborhood satisfies the inequality

.

The maximum and minimum points of a function are called extremum points.

If the function
at the point has an extremum, then the derivative of the function at this point is equal to zero or does not exist (a necessary condition for the existence of an extremum).

The points at which the derivative is equal to zero or does not exist are called critical.

5. Sufficient conditions for the existence of an extremum.

Rule 1. If during the transition (from left to right) through the critical point derivative
changes sign from "+" to "-", then at the point function
has a maximum; if from "-" to "+", then the minimum; If
does not change sign, then there is no extremum.

Rule 2. Let at the point
first derivative of the function
zero
, and the second derivative exists and is nonzero. If
, That is the maximum point, if
, That is the minimum point of the function.

Example 6.4 . Explore the maximum and minimum functions:

1)
; 2)
; 3)
;

4)
.

Solution.

1) The function is defined and continuous on the interval
.

Let's find the derivative
and solve the equation
, i.e.
.from here
are critical points.

Let us determine the sign of the derivative in the intervals ,
.

When passing through points
And
the derivative changes sign from “–” to “+”, therefore, according to rule 1
are the minimum points.

When passing through a point
derivative changes sign from "+" to "-", so
is the maximum point.

,
.

2) The function is defined and continuous in the interval
. Let's find the derivative
.

By solving the equation
, find
And
are critical points. If the denominator
, i.e.
, then the derivative does not exist. So,
is the third critical point. Let us determine the sign of the derivative in intervals.

Therefore, the function has a minimum at the point
, maximum at points
And
.

3) A function is defined and continuous if
, i.e. at
.

Let's find the derivative

.

Let's find the critical points:

Neighborhoods of points
do not belong to the domain of definition, so they are not extremum t. So let's explore the critical points
And
.

4) The function is defined and continuous on the interval
. We use rule 2. Find the derivative
.

Let's find the critical points:

Let's find the second derivative
and determine its sign at the points

At points
function has a minimum.

At points
function has a maximum.