Conduct a complete research into the features of the online solution. Exploring Functions and Graphing a Function Using Derivatives

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Today we invite you to explore and build a graph of a function with us. After carefully studying this article, you will not have to sweat for long to complete this type of task. It is not easy to study and construct a graph of a function; it is a voluminous work that requires maximum attention and accuracy of calculations. To make the material easier to understand, we will study the same function step by step and explain all our actions and calculations. Welcome to the amazing and fascinating world of mathematics! Go!

Domain

In order to explore and graph a function, you need to know several definitions. Function is one of the main (basic) concepts in mathematics. It reflects the dependence between several variables (two, three or more) during changes. The function also shows the dependence of sets.

Imagine that we have two variables that have a certain range of change. So, y is a function of x, provided that each value of the second variable corresponds to one value of the second. In this case, the variable y is dependent, and it is called a function. It is customary to say that the variables x and y are in For greater clarity of this dependence, a graph of the function is built. What is a graph of a function? This is a set of points on the coordinate plane, where each x value corresponds to one y value. Graphs can be different - straight line, hyperbola, parabola, sine wave, and so on.

It is impossible to graph a function without research. Today we will learn how to conduct research and build a graph of a function. It is very important to take notes during the study. This will make the task much easier to cope with. The most convenient research plan:

1. Domain.
2. Continuity.
3. Even or odd.
4. Periodicity.
5. Asymptotes.
6. Zeros.
7. Sign constancy.
8. Increasing and decreasing.
9. Extremes.
10. Convexity and concavity.

Let's start with the first point. Let's find the domain of definition, that is, on what intervals our function exists: y=1/3(x^3-14x^2+49x-36). In our case, the function exists for any values ​​of x, that is, the domain of definition is equal to R. This can be written as follows xÎR.

Continuity

Now we will examine the discontinuity function. In mathematics, the term “continuity” appeared as a result of the study of the laws of motion. What is infinite? Space, time, some dependencies (an example is the dependence of the variables S and t in movement problems), the temperature of a heated object (water, frying pan, thermometer, etc.), a continuous line (that is, one that can be drawn without lifting it from the sheet pencil).

A graph is considered continuous if it does not break at some point. One of the most obvious examples of such a graph is a sinusoid, which you can see in the picture in this section. The function is continuous at some point x0 if a number of conditions are met:

• a function is defined at a given point;
• the right and left limits at a point are equal;
• the limit is equal to the value of the function at point x0.

If at least one condition is not met, the function is said to fail. And the points at which the function breaks are usually called break points. An example of a function that will “break” when displayed graphically is: y=(x+4)/(x-3). Moreover, y does not exist at the point x = 3 (since it is impossible to divide by zero).

In the function that we are studying (y=1/3(x^3-14x^2+49x-36)) everything turned out to be simple, since the graph will be continuous.

Even, odd

Now examine the function for parity. First, a little theory. An even function is one that satisfies the condition f(-x)=f(x) for any value of the variable x (from the range of values). Examples include:

• module x (the graph looks like a daw, the bisector of the first and second quarters of the graph);
• x squared (parabola);
• cosine x (cosine).

Note that all of these graphs are symmetrical when viewed with respect to the y-axis (that is, the y-axis).

What then is called an odd function? These are those functions that satisfy the condition: f(-x)=-f(x) for any value of the variable x. Examples:

• hyperbola;
• cubic parabola;
• sinusoid;
• tangent and so on.

Please note that these functions are symmetrical about the point (0:0), that is, the origin. Based on what was said in this section of the article, an even and odd function must have the property: x belongs to the definition set and -x too.

Let's examine the function for parity. We can see that she doesn't fit any of the descriptions. Therefore, our function is neither even nor odd.

Asymptotes

Let's start with a definition. An asymptote is a curve that is as close as possible to the graph, that is, the distance from a certain point tends to zero. In total, there are three types of asymptotes:

• vertical, that is, parallel to the y-axis;
• horizontal, that is, parallel to the x axis;
• inclined.

As for the first type, these lines should be looked for at some points:

• gap;
• ends of the domain of definition.

In our case, the function is continuous, and the domain of definition is equal to R. Therefore, there are no vertical asymptotes.

The graph of a function has a horizontal asymptote, which meets the following requirement: if x tends to infinity or minus infinity, and the limit is equal to a certain number (for example, a). In this case, y=a is the horizontal asymptote. There are no horizontal asymptotes in the function we are studying.

An oblique asymptote exists only if two conditions are met:

• lim(f(x))/x=k;
• lim f(x)-kx=b.

Then it can be found using the formula: y=kx+b. Again, in our case there are no oblique asymptotes.

Function zeros

The next step is to examine the graph of the function for zeros. It is also very important to note that the task associated with finding the zeros of a function occurs not only when studying and constructing a graph of a function, but also as an independent task and as a way to solve inequalities. You may be required to find the zeros of a function on a graph or use mathematical notation.

Finding these values ​​will help you graph the function more accurately. In simple terms, the zero of a function is the value of the variable x at which y = 0. If you are looking for the zeros of a function on a graph, then you should pay attention to the points at which the graph intersects with the x-axis.

To find the zeros of the function, you need to solve the following equation: y=1/3(x^3-14x^2+49x-36)=0. After carrying out the necessary calculations, we get the following answer:

Sign constancy

The next stage of research and construction of a function (graph) is finding intervals of constant sign. This means that we must determine at which intervals the function takes a positive value and at which intervals it takes a negative value. The zero functions found in the last section will help us do this. So, we need to build a straight line (separate from the graph) and distribute the zeros of the function along it in the correct order from smallest to largest. Now you need to determine which of the resulting intervals has a “+” sign and which has a “-”.

In our case, the function takes a positive value on intervals:

• from 1 to 4;
• from 9 to infinity.

Negative meaning:

• from minus infinity to 1;
• from 4 to 9.

This is quite easy to determine. Substitute any number from the interval into the function and see what sign the answer turns out to have (minus or plus).

Increasing and decreasing function

In order to explore and construct a function, we need to know where the graph will increase (go up along the Oy axis) and where it will fall (crawl down along the y-axis).

A function increases only if a larger value of the variable x corresponds to a larger value of y. That is, x2 is greater than x1, and f(x2) is greater than f(x1). And we observe a completely opposite phenomenon with a decreasing function (the more x, the less y). To determine the intervals of increase and decrease, you need to find the following:

• domain of definition (we already have);
• derivative (in our case: 1/3(3x^2-28x+49);
• solve the equation 1/3(3x^2-28x+49)=0.

After calculations we get the result:

We get: the function increases on the intervals from minus infinity to 7/3 and from 7 to infinity, and decreases on the interval from 7/3 to 7.

Extremes

The function under study y=1/3(x^3-14x^2+49x-36) is continuous and exists for any value of the variable x. The extremum point shows the maximum and minimum of a given function. In our case there are none, which greatly simplifies the construction task. Otherwise, they can also be found using the derivative function. Once found, do not forget to mark them on the chart.

Convexity and concavity

We continue to further explore the function y(x). Now we need to check it for convexity and concavity. The definitions of these concepts are quite difficult to comprehend; it is better to analyze everything using examples. For the test: a function is convex if it is a non-decreasing function. Agree, this is incomprehensible!

We need to find the derivative of a second order function. We get: y=1/3(6x-28). Now let's equate the right-hand side to zero and solve the equation. Answer: x=14/3. We found the inflection point, that is, the place where the graph changes from convexity to concavity or vice versa. On the interval from minus infinity to 14/3 the function is convex, and from 14/3 to plus infinity it is concave. It is also very important to note that the inflection point on the graph should be smooth and soft, there should be no sharp corners.

Defining additional points

Our task is to investigate and construct a graph of the function. We have completed the study; constructing a graph of the function is now not difficult. For more accurate and detailed reproduction of a curve or straight line on the coordinate plane, you can find several auxiliary points. They are quite easy to calculate. For example, we take x=3, solve the resulting equation and find y=4. Or x=5, and y=-5 and so on. You can take as many additional points as you need for construction. At least 3-5 of them are found.

Plotting a graph

We needed to investigate the function (x^3-14x^2+49x-36)*1/3=y. All necessary marks during the calculations were made on the coordinate plane. All that remains to be done is to build a graph, that is, connect all the dots. Connecting the dots should be smooth and accurate, this is a matter of skill - a little practice and your schedule will be perfect.

Let's study the function \(y= \frac(x^3)(1-x) \) and build its graph.

1. Scope of definition.
The domain of definition of a rational function (fraction) will be: the denominator is not equal to zero, i.e. \(1 -x \ne 0 => x \ne 1\). Domain $$D_f= (-\infty; 1) \cup (1;+\infty)$$

2. Function break points and their classification.
The function has one break point x = 1
Let's examine the point x= 1. Let's find the limit of the function to the right and left of the discontinuity point, to the right $$ \lim_(x \to 1+0) (\frac(x^3)(1-x)) = -\infty $$ and to the left of the point $$ \lim_(x \to 1-0)(\frac(x^3)(1-x)) = +\infty $$ This is a discontinuity point of the second kind because one-sided limits are equal to \(\infty\).

The straight line \(x = 1\) is a vertical asymptote.

3. Function parity.
We check for parity \(f(-x) = \frac((-x)^3)(1+x) \) the function is neither even nor odd.

4. Zeros of the function (points of intersection with the Ox axis). Intervals of constant sign of a function.
Function zeros ( point of intersection with the Ox axis): we equate \(y=0\), we get \(\frac(x^3)(1-x) = 0 => x=0 \). The curve has one intersection point with the Ox axis with coordinates \((0;0)\).

Intervals of constant sign of a function.
On the considered intervals \((-\infty; 1) \cup (1;+\infty)\) the curve has one point of intersection with the Ox axis, so we will consider the domain of definition on three intervals.

Let us determine the sign of the function on intervals of the domain of definition:
interval \((-\infty; 0) \) find the value of the function at any point \(f(-4) = \frac(x^3)(1-x)< 0 \), на этом интервале функция отрицательная \(f(x) < 0 \), т.е. находится ниже оси Ox
interval \((0; 1) \) we find the value of the function at any point \(f(0.5) = \frac(x^3)(1-x) > 0 \), on this interval the function is positive \(f(x ) > 0 \), i.e. is located above the Ox axis.
interval \((1;+\infty) \) find the value of the function at any point \(f(4) = \frac(x^3)(1-x)< 0 \), на этом интервале функция отрицательная \(f(x) < 0 \), т.е. находится ниже оси Ox

5. Intersection points with the Oy axis: we equate \(x=0\), we get \(f(0) = \frac(x^3)(1-x) = 0\). Coordinates of the point of intersection with the Oy axis \((0; 0)\)

6. Intervals of monotony. Extrema of a function.
Let's find the critical (stationary) points, for this we find the first derivative and equate it to zero $$ y" = (\frac(x^3)(1-x))" = \frac(3x^2(1-x) + x^3)( (1-x)^2) = \frac(x^2(3-2x))( (1-x)^2) $$ equal to 0 $$ \frac(x^2(3 -2x))( (1-x)^2) = 0 => x_1 = 0 \quad x_2= \frac(3)(2)$$ Let's find the value of the function at this point \(f(0) = 0\) and \(f(\frac(3)(2)) = -6.75\). We got two critical points with coordinates \((0;0)\) and \((1.5;-6.75)\)

Intervals of monotony.
The function has two critical points (possible extremum points), so we will consider monotonicity on four intervals:
interval \((-\infty; 0) \) find the value of the first derivative at any point in the interval \(f(-4) = \frac(x^2(3-2x))( (1-x)^2) >
interval \((0;1)\) we find the value of the first derivative at any point in the interval \(f(0.5) = \frac(x^2(3-2x))( (1-x)^2) > 0\) , the function increases over this interval.
interval \((1;1.5)\) we find the value of the first derivative at any point in the interval \(f(1.2) = \frac(x^2(3-2x))( (1-x)^2) > 0\) , the function increases over this interval.
interval \((1.5; +\infty)\) find the value of the first derivative at any point in the interval \(f(4) = \frac(x^2(3-2x))( (1-x)^2)< 0\), на этом интервале функция убывает.

Extrema of a function.

When studying the function, we obtained two critical (stationary) points on the interval of the domain of definition. Let's determine whether they are extremes. Let us consider the change in the sign of the derivative when passing through critical points:

point \(x = 0\) the derivative changes sign with \(\quad +\quad 0 \quad + \quad\) - the point is not an extremum.
point \(x = 1.5\) the derivative changes sign with \(\quad +\quad 0 \quad - \quad\) - the point is a maximum point.

7. Intervals of convexity and concavity. Inflection points.

To find the intervals of convexity and concavity, we find the second derivative of the function and equate it to zero $$y"" = (\frac(x^2(3-2x))( (1-x)^2))"= \frac(2x (x^2-3x+3))((1-x)^3) $$Equate to zero $$ \frac(2x(x^2-3x+3))((1-x)^3)= 0 => 2x(x^2-3x+3) =0 => x=0$$ The function has one critical point of the second kind with coordinates \((0;0)\).
Let us define convexity on intervals of the domain of definition, taking into account a critical point of the second kind (a point of possible inflection).

interval \((-\infty; 0)\) find the value of the second derivative at any point \(f""(-4) = \frac(2x(x^2-3x+3))((1-x)^ 3)< 0 \), на этом интервале вторая производная функции отрицательная \(f""(x) < 0 \) - функция выпуклая вверх (вогнутая).
interval \((0; 1)\) we find the value of the second derivative at any point \(f""(0.5) = \frac(2x(x^2-3x+3))((1-x)^3) > 0 \), on this interval the second derivative of the function is positive \(f""(x) > 0 \) the function is convex downward (convex).
interval \((1; \infty)\) find the value of the second derivative at any point \(f""(4) = \frac(2x(x^2-3x+3))((1-x)^3)< 0 \), на этом интервале вторая производная функции отрицательная \(f""(x) < 0 \) - функция выпуклая вверх (вогнутая).

Inflection points.

Let us consider the change in the sign of the second derivative when passing through a critical point of the second kind:
At the point \(x =0\), the second derivative changes sign with \(\quad - \quad 0 \quad + \quad\), the graph of the function changes convexity, i.e. this is the inflection point with coordinates \((0;0)\).

8. Asymptotes.

Vertical asymptote. The graph of the function has one vertical asymptote \(x =1\) (see paragraph 2).
Oblique asymptote.
In order for the graph of the function \(y= \frac(x^3)(1-x) \) at \(x \to \infty\) to have a slanted asymptote \(y = kx+b\), it is necessary and sufficient , so that there are two limits $$\lim_(x \to +\infty)=\frac(f(x))(x) =k $$we find it $$ \lim_(x \to \infty) (\frac( x^3)(x(1-x))) = \infty => k= \infty $$ and the second limit $$ \lim_(x \to +\infty)(f(x) - kx) = b$ $, because \(k = \infty\) - there is no oblique asymptote.

Horizontal asymptote: in order for a horizontal asymptote to exist, it is necessary that there be a limit $$\lim_(x \to \infty)f(x) = b$$ let's find it $$ \lim_(x \to +\infty)(\frac( x^3)(1-x))= -\infty$$$$ \lim_(x \to -\infty)(\frac(x^3)(1-x))= -\infty$$
There is no horizontal asymptote.

9. Function graph.

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.

1). Conduct a complete study of the function and construct its graph. Solution.
The scope of the function.

2). 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.
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.


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:

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.