Maintaining your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please review our privacy practices and let us know if you have any questions.
Collection and use of personal information
Personal information refers to data that can be used to identify or contact a specific person.
You may be asked to provide your personal information at any time when you contact us.
Below are some examples of the types of personal information we may collect and how we may use such information.
What personal information do we collect:
- When you submit an application on the site, we may collect various information, including your name, phone number, email address, etc.
How we use your personal information:
- The personal information we collect allows us to contact you with unique offers, promotions and other events and upcoming events.
- From time to time, we may use your personal information to send important notices and communications.
- We may also use personal information for internal purposes, such as conducting audits, data analysis and various research in order to improve the services we provide and provide you with recommendations regarding our services.
- If you participate in a prize draw, contest or similar promotion, we may use the information you provide to administer such programs.
Disclosure of information to third parties
We do not disclose the information received from you to third parties.
Exceptions:
- If necessary - in accordance with the law, judicial procedure, in legal proceedings, and/or on the basis of public requests or requests from government bodies in the Russian Federation - to disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public importance purposes.
- In the event of a reorganization, merger, or sale, we may transfer the personal information we collect to the applicable successor third party.
Protection of personal information
We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as unauthorized access, disclosure, alteration and destruction.
Respecting your privacy at the company level
To ensure that your personal information is secure, we communicate privacy and security standards to our employees and strictly enforce privacy practices.
ABSTRACT
“Full study of a function and construction of its graph.”
INTRODUCTION
Studying the properties of a function and plotting its graph is one of the most wonderful applications of derivatives. This method of studying function has been repeatedly subjected to careful analysis. The main reason is that in applications of mathematics it was necessary to deal with more and more complex functions that appeared when studying new phenomena. Exceptions to the rules developed by mathematics appeared, cases appeared when the rules created were not suitable at all, functions appeared that did not have a derivative at any point.
The purpose of studying the course of algebra and elementary analysis in grades 10-11 is the systematic study of functions, the disclosure of the applied value of general methods of mathematics related to the study of functions.
The development of functional concepts in the course of studying algebra and the beginnings of analysis at the senior level of education helps high school students to obtain visual ideas about the continuity and discontinuities of functions, learn about the continuity of any elementary function in the field of its application, learn to construct their graphs and generalize information about the main elementary functions and understand them role in the study of phenomena of reality, in human practice.
Increasing and decreasing functions
Solving various problems from the fields of mathematics, physics and technology leads to the establishment of a functional relationship between the variables involved in this phenomenon.
If such a functional dependence can be expressed analytically, that is, in the form of one or more formulas, then it becomes possible to study it by means of mathematical analysis.
This refers to the possibility of clarifying the behavior of a function when one or another variable changes (where the function increases, where it decreases, where it reaches a maximum, etc.).
The application of differential calculus to the study of a function is based on a very simple connection that exists between the behavior of a function and the properties of its derivative, primarily its first and second derivatives.
Let's consider how we can find intervals of increasing or decreasing function, that is, intervals of its monotonicity. Based on the definition of a monotonically decreasing and increasing function, it is possible to formulate theorems that allow us to relate the value of the first derivative of a given function to the nature of its monotonicity.
Theorem 1.1. If the function y
=
f
(
x
)
, differentiable on the interval(
a
,
b
)
, increases monotonically on this interval, then at any point
(
x
) >0;
if it decreases monotonically, then at any point in the interval (
x
)<0.
Proof. Let the functiony = f ( x ) monotonically increases by( a , b ) , This means that for anyone small enough > 0 the following inequality holds:
f ( x - ) < f ( x ) < f ( x + ) (Fig. 1.1).
Rice. 1.1
Consider the limit
.
If > 0, then 
> 0 if< 0, то
< 0.
In both cases, the expression under the limit sign is positive, which means the limit is positive, that is ( x )>0 , which was what needed to be proven. The second part of the theorem, related to the monotonic decrease of the function, is proved in a similar way.
Theorem 1.2. If the function y = f ( x ) , continuous on the segment[ a , b ] and is differentiable at all its interior points, and, in addition, ( x ) >0 for anyone x ϵ ( a , b ) , then this function increases monotonically by( a , b ) ; If
( x ) <0 for anyone xϵ ( a , b ), then this function decreases monotonically by( a , b ) .
Proof. Let's take ϵ ( a , b ) And ϵ ( a , b ) , and< . According to Lagrange's theorem
( c ) = .
But ( c )>0 and > 0, which means ( > 0, that is
(. The result obtained indicates a monotonic increase in the function, which was what needed to be proven. The second part of the theorem is proved in a similar way.
Extrema of the function
When studying the behavior of a function, a special role is played by the points that separate from each other the intervals of monotonic increase from the intervals of its monotonic decrease.
Definition 2.1. Dot called the maximum point of the function
y
=
f
(
x
)
, if for any, however small ,
(
< 0
, а точка
is called a minimum point if (
> 0.
The minimum and maximum points are collectively called extremum points. The piecewise monotonic function of such points has a finite number on a finite interval (Fig. 2.1).
Rice. 2.1
Theorem 2.1 (necessary condition for the existence of an extremum). If differentiable on the interval( a , b ) function has at point from this interval is the maximum, then its derivative at this point is equal to zero. The same can be said about the minimum point .
The proof of this theorem follows from Rolle’s theorem, in which it was shown that at the points of minimum or maximum = 0, and the tangent drawn to the graph of the function at these points is parallel to the axisOX .
It follows from Theorem 2.1 that if the functiony = f ( x ) has a derivative at all points, then it can reach an extremum at those points where = 0.
However, this condition is not sufficient, since there are functions for which the specified condition is satisfied, but there is no extremum. For example, the functiony= at a point x = 0 the derivative is zero, but there is no extremum at this point. In addition, the extremum may be at those points where the derivative does not exist. For example, the functiony = | x | there is a minimum at the pointx = 0 , although the derivative does not exist at this point.
Definition 2.2. The points at which the derivative of a function vanishes or has a discontinuity are called critical points of this function.
Consequently, Theorem 2.1 is not sufficient for determining extreme points.
Theorem 2.2 (sufficient condition for the existence of an extremum). Let the function y = f ( x ) continuous on the interval( a , b ) , which contains its critical point , and is differentiable at all points of this interval, with the possible exception of the point itself . Then, if, when moving this point from left to right, the sign of the derivative changes from plus to minus, then this is a maximum point, and, conversely, from minus to plus - a minimum point.
Proof. If the derivative of a function changes its sign when passing a point from left to right from plus to minus, then the function moves from increasing to decreasing, that is, it reaches at the point its maximum and vice versa.
From the above, a scheme for studying a function at an extremum follows:
1) find the domain of definition of the function;
2) calculate the derivative;
3) find critical points;
4) by changing the sign of the first derivative, their character is determined.
The task of studying a function for an extremum should not be confused with the task of determining the minimum and maximum values of a function on a segment. In the second case, it is necessary to find not only the extreme points on the segment, but also compare them with the value of the function at its ends.
Intervals of convex and concave functions
Another characteristic of the graph of a function that can be determined using the derivative is its convexity or concavity.
Definition 3.1. Function y = f ( x ) is called convex on the interval( a , b ) , if its graph is located below any tangent drawn to it on a given interval, and vice versa, it is called concave if its graph is above any tangent drawn to it on a given interval.
Let us prove a theorem that allows us to determine the intervals of convexity and concavity of a function.
Theorem 3.1. If at all points of the interval( a , b ) second derivative of the function ( x ) is continuous and negative, then the functiony = f ( x ) is convex and vice versa, if the second derivative is continuous and positive, then the function is concave.
We carry out the proof for the interval of convexity of the function. Let's take an arbitrary pointϵ ( a , b ) and draw a tangent to the graph of the function at this pointy = f ( x ) (Fig. 3.1).
The theorem will be proven if it is shown that all points of the curve on the interval( a , b ) lie under this tangent. In other words, it is necessary to prove that for the same valuesx curve ordinatesy = f ( x ) less than the ordinate of the tangent drawn to it at the point .
Rice. 3.1
For definiteness, we denote the equation of the curve: = f ( x ) , and the equation of the tangent to it at the point :
- f ( ) = ( )( x - )
or
= f ( ) + ( )( x - ) .
Let's make up the difference And :
- = f(x) – f( ) - ( )(x- ).
Apply to differencef ( x ) – f ( ) Lagrange's mean value theorem:
- = ( )( x - ) - ( )( x - ) = ( x - )[ ( ) - ( )] ,
Where ϵ ( , x ).
Let us now apply Lagrange's theorem to the expression in square brackets:
- = ( )( - )( x - ) , Where ϵ ( , ).
As can be seen from the figure,x > , Then x - > 0 And - > 0 . Moreover, according to the theorem, ( )<0.
Multiplying these three factors, we get that , which was what needed to be proven.
Definition 3.2. The point separating the convex interval from the concave interval is called the inflection point.
From Definition 3.1 it follows that at a given point the tangent intersects the curve, that is, on one side the curve is located below the tangent, and on the other, above.
Theorem 3.2. If at the point second derivative of the function
y = f ( x ) is equal to zero or does not exist, and when passing through a point the sign of the second derivative changes to the opposite, then this point is an inflection point.
The proof of this theorem follows from the fact that the signs ( x ) on opposite sides of the point are different. This means that on one side of the point the function is convex, and on the other it is concave. In this case, according to Definition 3.2, the point is the inflection point.
The study of a function for convexity and concavity is carried out according to the same scheme as the study for an extremum.
4. Asymptotes of the function
In the previous paragraphs, methods for studying the behavior of a function using the derivative were discussed. However, among the questions related to the complete study of a function, there are also those that are not related to the derivative.
So, for example, it is necessary to know how a function behaves when a point on its graph moves infinitely away from the origin. This problem can arise in two cases: when the argument of a function goes to infinity and when, during a discontinuity of the second kind at the end point, the function itself goes to infinity. In both of these cases, a situation may arise when the function tends to some straight line, called its asymptote.
Definition . Asymptote of the graph of a functiony = f ( x ) is a straight line that has the property that the distance from the graph to this straight line tends to zero as the graph point moves indefinitely from the origin.
There are two types of asymptotes: vertical and oblique.
Vertical asymptotes include straight linesx
=
, which have the property that the graph of the function in their vicinity goes to infinity, that is, the condition is satisfied: 
.
Obviously, the requirement of the specified definition is satisfied here: the distance from the graph of the curve to the straight linex = tends to zero, and the curve itself goes to infinity. So, at points of discontinuity of the second kind, functions have vertical asymptotes, for example,y= at a point x = 0 . Consequently, determining the vertical asymptotes of a function coincides with finding discontinuity points of the second kind.
Oblique asymptotes are described by the general equation of a straight line on a plane, that isy = kx + b . This means that, unlike vertical asymptotes, here it is necessary to determine the numbersk And b .
So let the curve = f ( x ) has an oblique asymptote, that is, atx the points of the curve come as close as desired to the straight line = kx + b (Fig. 4.1). Let M ( x , y ) - a point located on a curve. Its distance from the asymptote will be characterized by the length of the perpendicular| MN | .
To fully study the function and plot its graph, the following scheme is recommended:
A) find the domain of definition, breakpoints; explore the behavior of a function near discontinuity points (find the limits of the function on the left and right at these points). Indicate the vertical asymptotes.
B) determine whether a function is even or odd and conclude that there is symmetry. If , then the function is even and symmetrical about the OY axis; when the function is odd, symmetrical about the origin; and if is a function of general form.
C) find the intersection points of the function with the coordinate axes OY and OX (if possible), determine the intervals of constant sign of the function. The boundaries of intervals of constant sign of a function are determined by the points at which the function is equal to zero (function zeros) or does not exist and the boundaries of the domain of definition of this function. In intervals where the graph of the function is located above the OX axis, and where - below this axis.
D) find the first derivative of the function, determine its zeros and intervals of constant sign. In intervals where the function increases and where it decreases. Make a conclusion about the presence of extrema (points where a function and derivative exist and when passing through which it changes sign. If the sign changes from plus to minus, then at this point the function has a maximum, and if from minus to plus, then a minimum). Find the values of the function at the extrema points.
D) find the second derivative, its zeros and intervals of constant sign. In intervals where< 0 график функции выпуклый, а где – вогнутый. Сделать заключение о наличии точек перегиба и найти значения функции в этих точках.
E) find inclined (horizontal) asymptotes, the equations of which have the form 
; Where 
.
At 
the graph of the function will have two slanted asymptotes, and each value of x at and can also correspond to two values of b.
G) find additional points to clarify the graph (if necessary) and construct a graph.
Example 1
Explore the function and construct its graph. Solution: A) domain of definition 
; the function is continuous in its domain of definition; – break point, because 
;
. Then – vertical asymptote.
B)
those. y(x) is a function of general form.
C) Find the points of intersection of the graph with the OY axis: set x=0; then y(0)=–1, i.e. the graph of the function intersects the axis at the point (0;-1). Zeros of the function (points of intersection of the graph with the OX axis): set y=0; Then 
.
The discriminant of a quadratic equation is less than zero, which means there are no zeros. Then the boundary of the intervals of constant sign is the point x=1, where the function does not exist.
The sign of the function in each of the intervals is determined by the method of partial values:
It is clear from the diagram that in the interval the graph of the function is located under the OX axis, and in the interval – above the OX axis.
D) We find out the presence of critical points.
.
We find critical points (where or does not exist) from the equalities and .
We get: x1=1, x2=0, x3=2. Let's create an auxiliary table
Table 1 
(The first line contains critical points and the intervals into which these points are divided by the OX axis; the second line indicates the values of the derivative at critical points and the signs on the intervals. The signs are determined by the partial value method. The third line indicates the values of the function y(x) at critical points and shows the behavior of the function - increasing or decreasing at the corresponding intervals of the numerical axis.Additionally, the presence of a minimum or maximum is indicated.
D) Find the intervals of convexity and concavity of the function.
; build a table as in point D); Only in the second line we write down the signs, and in the third we indicate the type of convexity. Because ; then the critical point is one x=1.
table 2 
The point x=1 is the inflection point.
E) Find oblique and horizontal asymptotes 
Then y=x is an oblique asymptote.
G) Based on the data obtained, we build a graph of the function 
1). The scope of the function.
It is obvious that this function is defined on the entire number line, except for the points “” and “”, because at these points the denominator is equal to zero and, therefore, the function does not exist, and straight lines and are vertical asymptotes.
2). The behavior of a function as the argument tends to infinity, the existence of discontinuity points and checking for the presence of oblique asymptotes.
Let's first check how the function behaves as it approaches infinity to the left and to the right. 
Thus, when the function tends to 1, i.e. – horizontal asymptote.
In the vicinity of discontinuity points, the behavior of the function is determined as follows: 
Those. When approaching discontinuity points on the left, the function decreases infinitely, and on the right, it increases infinitely.
We determine the presence of an oblique asymptote by considering the equality: 
There are no oblique asymptotes.
3). Points of intersection with coordinate axes.
Here it is necessary to consider two situations: find the point of intersection with the Ox axis and the Oy axis. The sign of intersection with the Ox axis is the zero value of the function, i.e. it is necessary to solve the equation:
This equation has no roots, therefore, the graph of this function has no points of intersection with the Ox axis.
The sign of intersection with the Oy axis is the value x = 0. In this case
,
those. – the point of intersection of the function graph with the Oy axis.
4).Determination of extremum points and intervals of increase and decrease.
To study this issue, we define the first derivative: 
.
Let us equate the value of the first derivative to zero. 
.
A fraction is equal to zero when its numerator is equal to zero, i.e. .
Let us determine the intervals of increase and decrease of the function.
Thus, the function has one extremum point and does not exist at two points.
Thus, the function increases on the intervals and and decreases on the intervals and .
5). Inflection points and areas of convexity and concavity.
This characteristic of the behavior of a function is determined using the second derivative. Let us first determine the presence of inflection points. The second derivative of the function is equal to 
When and the function is concave;
when and the function is convex.
6). Graphing a function.
Using the found values in points, we will schematically construct a graph of the function:
Example3
Explore function 
and build its graph. Solution
The given function is a non-periodic function of general form. Its graph passes through the origin of coordinates, since .
The domain of definition of a given function is all values of the variable except and for which the denominator of the fraction becomes zero.
Consequently, the points are the discontinuity points of the function.
Because 
, 
Because 
,
, then the point is a discontinuity point of the second kind.
The straight lines are the vertical asymptotes of the graph of the function.
Equations of oblique asymptotes, where, 
.
At 
,
.
Thus, for and the graph of the function has one asymptote.
Let's find the intervals of increase and decrease of the function and extrema points.
.
The first derivative of the function at and, therefore, at and the function increases.
When , therefore, when , the function decreases.
does not exist for , . 
, therefore, when 
The graph of the function is concave.
At 
, therefore, when 
The graph of the function is convex. 
When passing through the points , , changes sign. When , the function is not defined, therefore, the graph of the function has one inflection point.
Let's build a graph of the function. 
The study of a function is carried out according to a clear scheme and requires the student to have a solid knowledge of basic mathematical concepts such as the domain of definition and values, continuity of the function, asymptote, extremum points, parity, periodicity, etc. The student must be able to differentiate functions freely and solve equations, which can sometimes be very complex.
That is, this task tests a significant layer of knowledge, any gap in which will become an obstacle to obtaining the correct solution. Particularly often, difficulties arise with constructing graphs of functions. This mistake is immediately noticeable to the teacher and can greatly damage your grade, even if everything else was done correctly. Here you can find online function research problems: study examples, download solutions, order assignments.
Explore a function and plot a graph: examples and solutions online
We have prepared for you a lot of ready-made function studies, both paid in the solution book and free in the section Examples of function studies. Based on these solved tasks, you will be able to familiarize yourself in detail with the methodology for performing similar tasks, and carry out your research by analogy.
We offer ready-made examples of complete research and plotting of functions of the most common types: polynomials, fractional-rational, irrational, exponential, logarithmic, trigonometric functions. Each solved problem is accompanied by a ready-made graph with highlighted key points, asymptotes, maxima and minima; the solution is carried out using an algorithm for studying the function.
In any case, the solved examples will be of great help to you as they cover the most popular types of functions. We offer you hundreds of already solved problems, but, as you know, there are an infinite number of mathematical functions in the world, and teachers are great experts at inventing more and more tricky tasks for poor students. So, dear students, qualified help will not hurt you.
Solving custom function research problems
In this case, our partners will offer you another service - full function research online to order. The task will be completed for you in compliance with all the requirements for an algorithm for solving such problems, which will greatly please your teacher.
We will do a complete study of the function for you: we will find the domain of definition and the domain of values, examine for continuity and discontinuity, establish parity, check your function for periodicity, and find the points of intersection with the coordinate axes. And, of course, further using differential calculus: we will find asymptotes, calculate extrema, inflection points, and construct the graph itself.
Constructing a graph of a function using singular points includes the study of the function itself: determining the range of permissible values of the argument, determining the range of variation of the function, determining whether the function is even or odd, determining the breakpoints of the function, finding intervals of constant sign of the function, finding asymptotes of the graph of the function. Using the first derivative, you can determine the intervals of increase (decrease) of the function and the presence of extremum points. Using the second derivative, you can determine the intervals of convexity (concavity) of the function graph, as well as inflection points. At the same time, we believe that if at some point xo tangent to the graph of the function above the curve, then the graph of the function at this point has convexity; if the tangent is below the curve, then the graph of the function at this point has a concavity.
y(x) = x³/(x²+3)
1. Function study.
a) Range of permissible values of the argument: (-∞,+∞).
b) Area of change of the function: (-∞, +∞).
c) The function is odd, because y(-x) = -y(x), those. the graph of the function is symmetrical about the origin.
d) The function is continuous, there are no discontinuity points, therefore, there are no vertical asymptotes.
e) Finding the equation of oblique asymptote y(x) = k∙x + b, Where
k = /x And b = 
In this example, the asymptote parameters are respectively equal:
k = , because the highest degree of the numerator and denominator are the same, equal to three, and the ratio of the coefficients at these highest degrees is equal to one. When x→ + ∞ the third remarkable limit was used to calculate the limit.
b = = = 0, when calculating the limit at x→ + ∞ used the third remarkable limit. So, the graph of this function has a slanted asymptote y=x.
2.
y´= /(x²+3)² - the derivative is calculated using the quotient differentiation formula.
a) Determine the zeros of the derivative and the discontinuity point, equating the numerator and denominator of the derivative to zero, respectively: y´=0, If x=0. The 1st derivative has no discontinuity points.
b) We determine the intervals of constant sign of the derivative, i.e. intervals of monotonicity of the function: at -∞
3. Studying a function using the 2nd derivative.
Using the formula for differentiating quotients and making algebraic transformations, we obtain: y´´ = /(x²+3)³
a) Determine the zeros of the 2nd derivative and intervals of constant sign: y´´ = 0, If x=0 And x= + 3 . The 2nd derivative has no discontinuity points.
b) Let us determine the intervals of constancy of the 2nd derivative, i.e. intervals of convexity or concavity of the graph of a function. At -∞
Example: Explore a function and graph it y(x)=((x-1)²∙(x+1))/x
1.Function study.
a) Range of acceptable values: (-∞,0)U(0,+∞).
b) Area of change of the function: (-∞,+∞).
d) This function has a discontinuity point of the 2nd kind at x=0.
e) Finding asymptotes. Because the function has a discontinuity point of the 2nd kind at x=0, then consequently the function has a vertical asymptote x=0. This function has no oblique or horizontal asymptotes.
2.Studying a function using the 1st derivative.
Let's transform the function by performing all the algebraic operations. As a result, the form of the function will be significantly simplified: y(x)=x²-x-1+(1/x). It’s very easy to take the derivative from the sum of the terms and we get: y´ = 2x – 1 –(1/x²).
a) Determine the zeros and discontinuity points of the 1st derivative. We bring the expressions for the 1st derivative to a common denominator and, equating the numerator and then the denominator to zero, we obtain: y´=0 at x=1, y´ - does not exist when x=0.
b) Let us determine the intervals of monotonicity of the function, i.e. intervals of constant sign of the derivative. At -∞<x<0
And 0
3.
y´´= 2 + 2/x³. Using the 2nd derivative, we determine the intervals of convexity or concavity of the function graph, as well as, if any, inflection points. Let us present the expression for the second derivative to the common denominator, and then, equating the numerator and denominator to zero in turn, we obtain: y´´=0 at x=-1, y´´- does not exist when x=0.
At -∞
rice. 4 fig. 5
Example: Explore a function and graph it y(x) = ln (x²+4x+5)
1.Function study.
a) Range of permissible argument values: the logarithmic function exists only for arguments strictly greater than zero, therefore, x²+4x+5>0 – this condition is satisfied for all values of the argument, i.e. O.D.Z. – (-∞, +∞).
b) Area of change of the function: (0, +∞). Let's transform the expression under the logarithm sign and equate the function to zero: ln((x+2)²+1) =0. Those. the function goes to zero when x=-2. The graph of the function will be symmetrical with respect to the straight line x=-2.
c) The function is continuous and has no breakpoints.
d) The graph of the function has no asymptotes.
2.Studying a function using the 1st derivative.
Using the rule for differentiating a complex function, we get: y´= (2x+4)/(x²+4x+5)
a) Let us determine the zeros and discontinuity points of the derivative: y´=0, at x=-2. The first derivative has no discontinuity points.
b) We determine the intervals of monotonicity of the function, i.e. intervals of constant sign of the first derivative: at -∞<x<-2
derivative y´<0,
therefore, the function decreases; when -2
3.Study of the function in terms of the 2nd derivative.
Let's represent the first derivative in the following form: y´=2∙(x+2)/(1+(x+2)²). y´´=2∙(1-(x+2)²/(1+(x+2)²)².
a) Let us determine the intervals of constant sign of the second derivative. Since the denominator of the 2nd derivative is always non-negative, the sign of the second derivative is determined only by the numerator. y´´=0 at x=-3 And x=-1.
At -∞
Example: Explore a Function and Plot a Graph y(x) = x²/(x+2)²
1.Function study.
a) Range of permissible values of the argument (-∞, -2)U(-2, +∞).
b) Area of change of function².
a) Let us determine the zeros and intervals of constant sign of the second derivative. Because Since the denominator of the fraction is always positive, the sign of the second derivative is completely determined by the numerator. At -∞
