Investigation of a function with a detailed solution. How to examine a function and graph it

One of the most important tasks of differential calculus is the development common examples studies of function behavior.

If the function y=f(x) is continuous on the interval , and its derivative is positive or equal to 0 on the interval (a,b), then y=f(x) increases by (f"(x)0). If the function y=f (x) is continuous on the segment , and its derivative is negative or equal to 0 on the interval (a,b), then y=f(x) decreases by (f"(x)0)

Intervals in which the function does not decrease or increase are called intervals of monotonicity of the function. The monotonicity of a function can change only at those points of its domain of definition at which the sign of the first derivative changes. The points at which the first derivative of a function vanishes or has a discontinuity are called critical.

Theorem 1 (1st sufficient condition for the existence of an extremum).

Let the function y=f(x) be defined at the point x 0 and let there be a neighborhood δ>0 such that the function is continuous on the interval and differentiable on the interval (x 0 -δ,x 0)u(x 0 , x 0 +δ) , and its derivative retains a constant sign on each of these intervals. Then if on x 0 -δ,x 0) and (x 0 , x 0 +δ) the signs of the derivative are different, then x 0 is an extremum point, and if they coincide, then x 0 is not an extremum point. Moreover, if, when passing through the point x0, the derivative changes sign from plus to minus (to the left of x 0 f"(x)>0 is satisfied, then x 0 is the maximum point; if the derivative changes sign from minus to plus (to the right of x 0 executed f"(x)<0, то х 0 - точка минимума.

The maximum and minimum points are called the extremum points of the function, and the maximum and minimum of the function are called its extreme values.

Theorem 2 (a necessary sign of a local extremum).

If the function y=f(x) has an extremum at the current x=x 0, then either f’(x 0)=0 or f’(x 0) does not exist.
At the extremum points of the differentiable function, the tangent to its graph is parallel to the Ox axis.

Algorithm for studying a function for an extremum:

1) Find the derivative of the function.
2) Find critical points, i.e. points at which the function is continuous and the derivative is zero or does not exist.
3) Consider the neighborhood of each point, and examine the sign of the derivative to the left and right of this point.
4) Determine the coordinates of the extreme points; for this, substitute the values ​​of the critical points into this function. Using sufficient conditions for the extremum, draw the appropriate conclusions.

Example 18. Examine the function y=x 3 -9x 2 +24x for an extremum

Solution.
1) y"=3x 2 -18x+24=3(x-2)(x-4).
2) Equating the derivative to zero, we find x 1 =2, x 2 =4. In this case, the derivative is defined everywhere; This means that apart from the two points found, there are no other critical points.
3) The sign of the derivative y"=3(x-2)(x-4) changes depending on the interval as shown in Figure 1. When passing through the point x=2, the derivative changes sign from plus to minus, and when passing through through the point x=4 - from minus to plus.
4) At point x=2 the function has a maximum y max =20, and at point x=4 - a minimum y min =16.

Theorem 3. (2nd sufficient condition for the existence of an extremum).

Let f"(x 0) and at the point x 0 there exists f""(x 0). Then if f""(x 0)>0, then x 0 is the minimum point, and if f""(x 0)<0, то х 0 – точка максимума функции y=f(x).

On a segment, the function y=f(x) can reach the smallest (y the least) or the greatest (y the highest) value either at the critical points of the function lying in the interval (a;b), or at the ends of the segment.

Algorithm for finding the largest and smallest values ​​of a continuous function y=f(x) on the segment:

1) Find f"(x).
2) Find the points at which f"(x)=0 or f"(x) does not exist, and select from them those that lie inside the segment.
3) Calculate the value of the function y=f(x) at the points obtained in step 2), as well as at the ends of the segment and select the largest and smallest from them: they are, respectively, the largest (y the largest) and the smallest (y the least) values ​​of the function on the interval.

Example 19. Find the largest value of the continuous function y=x 3 -3x 2 -45+225 on the segment.

1) We have y"=3x 2 -6x-45 on the segment
2) The derivative y" exists for all x. Let's find the points at which y"=0; we get:
3x 2 -6x-45=0
x 2 -2x-15=0
x 1 =-3; x 2 =5
3) Calculate the value of the function at points x=0 y=225, x=5 y=50, x=6 y=63
The segment contains only the point x=5. The largest of the found values ​​of the function is 225, and the smallest is the number 50. So, y max = 225, y min = 50.

Study of a function on convexity

The figure shows graphs of two functions. The first of them is convex upward, the second is convex downward.

The function y=f(x) is continuous on an interval and differentiable in the interval (a;b), is called convex upward (downward) on this interval if, for axb, its graph lies no higher (not lower) than the tangent drawn at any point M 0 (x 0 ;f(x 0)), where axb.

Theorem 4. Let the function y=f(x) have a second derivative at any interior point x of the segment and be continuous at the ends of this segment. Then if the inequality f""(x)0 holds on the interval (a;b), then the function is convex downward on the interval ; if the inequality f""(x)0 holds on the interval (a;b), then the function is convex upward on .

Theorem 5. If the function y=f(x) has a second derivative on the interval (a;b) and if it changes sign when passing through the point x 0, then M(x 0 ;f(x 0)) is an inflection point.

Rule for finding inflection points:

1) Find the points at which f""(x) does not exist or vanishes.
2) Examine the sign f""(x) to the left and right of each point found in the first step.
3) Based on Theorem 4, draw a conclusion.

Example 20. Find the extremum points and inflection points of the graph of the function y=3x 4 -8x 3 +6x 2 +12.

We have f"(x)=12x 3 -24x 2 +12x=12x(x-1) 2. Obviously, f"(x)=0 when x 1 =0, x 2 =1. When passing through the point x=0, the derivative changes sign from minus to plus, but when passing through the point x=1 it does not change sign. This means that x=0 is the minimum point (y min =12), and there is no extremum at point x=1. Next, we find . The second derivative vanishes at the points x 1 =1, x 2 =1/3. The signs of the second derivative change as follows: On the ray (-∞;) we have f""(x)>0, on the interval (;1) we have f""(x)<0, на луче (1;+∞) имеем f""(x)>0. Therefore, x= is the inflection point of the function graph (transition from convexity down to convexity upward) and x=1 is also the inflection point (transition from convexity upward to convexity downward). If x=, then y=; if, then x=1, y=13.

Algorithm for finding the asymptote of a graph

I. If y=f(x) as x → a, then x=a is a vertical asymptote.
II. If y=f(x) as x → ∞ or x → -∞, then y=A is a horizontal asymptote.
III. To find the oblique asymptote, we use the following algorithm:
1) Calculate . If the limit exists and is equal to b, then y=b is a horizontal asymptote; if , then go to the second step.
2) Calculate . If this limit does not exist, then there is no asymptote; if it exists and is equal to k, then go to the third step.
3) Calculate . If this limit does not exist, then there is no asymptote; if it exists and is equal to b, then go to the fourth step.
4) Write down the equation of the oblique asymptote y=kx+b.

Example 21: Find the asymptote for a function

1)
2)
3)
4) The equation of the oblique asymptote has the form

Scheme for studying a function and constructing its graph

I. Find the domain of definition of the function.
II. Find the points of intersection of the graph of the function with the coordinate axes.
III. Find asymptotes.
IV. Find possible extremum points.
V. Find critical points.
VI. Using the auxiliary figure, explore the sign of the first and second derivatives. Determine areas of increasing and decreasing function, find the direction of convexity of the graph, points of extrema and inflection points.
VII. Construct a graph, taking into account the research carried out in paragraphs 1-6.

Example 22: Construct a graph of the function according to the above diagram

Solution.
I. The domain of a function is the set of all real numbers except x=1.
II. Since the equation x 2 +1=0 has no real roots, the graph of the function has no points of intersection with the Ox axis, but intersects the Oy axis at the point (0;-1).
III. Let us clarify the question of the existence of asymptotes. Let us study the behavior of the function near the discontinuity point x=1. Since y → ∞ as x → -∞, y → +∞ as x → 1+, then the straight line x=1 is the vertical asymptote of the graph of the function.
If x → +∞(x → -∞), then y → +∞(y → -∞); therefore, the graph does not have a horizontal asymptote. Further, from the existence of limits

Solving the equation x 2 -2x-1=0 we obtain two possible extremum points:
x 1 =1-√2 and x 2 =1+√2

V. To find the critical points, we calculate the second derivative:

Since f""(x) does not vanish, there are no critical points.
VI. Let us examine the sign of the first and second derivatives. Possible extremum points to be considered: x 1 =1-√2 and x 2 =1+√2, divide the domain of existence of the function into intervals (-∞;1-√2),(1-√2;1+√2) and (1+√2;+∞).

In each of these intervals, the derivative retains its sign: in the first - plus, in the second - minus, in the third - plus. The sequence of signs of the first derivative will be written as follows: +,-,+.
We find that the function increases at (-∞;1-√2), decreases at (1-√2;1+√2), and increases again at (1+√2;+∞). Extremum points: maximum at x=1-√2, and f(1-√2)=2-2√2 minimum at x=1+√2, and f(1+√2)=2+2√2. At (-∞;1) the graph is convex upward, and at (1;+∞) it is convex downward.
VII Let's make a table of the obtained values

VIII Based on the data obtained, we construct a sketch of the graph of the function

The reference points when studying functions and constructing their graphs are characteristic points - points of discontinuity, extremum, inflection, intersection with coordinate axes. Using differential calculus, it is possible to establish the characteristic features of changes in functions: increase and decrease, maximums and minimums, the direction of convexity and concavity of the graph, the presence of asymptotes.

A sketch of the graph of the function can (and should) be drawn after finding the asymptotes and extremum points, and it is convenient to fill out the summary table of the study of the function as the study progresses.

The following function study scheme is usually used.

1.Find the domain of definition, intervals of continuity and breakpoints of the function.

2.Examine the function for evenness or oddness (axial or central symmetry of the graph.

3.Find asymptotes (vertical, horizontal or oblique).

4.Find and study the intervals of increase and decrease of the function, its extremum points.

5.Find the intervals of convexity and concavity of the curve, its inflection points.

6.Find the intersection points of the curve with the coordinate axes, if they exist.

7.Compile a summary table of the study.

8.A graph is constructed, taking into account the study of the function carried out according to the points described above.

Example. Explore function

and build its graph.

7. Let’s compile a summary table for studying the function, where we will enter all the characteristic points and the intervals between them. Taking into account the parity of the function, we obtain the following table:

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 set of definition 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 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.

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

(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 ; That critical point 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:

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.

Conduct a complete study and graph the function

y(x)=x2+81−x.y(x)=x2+81−x.

1) The scope of the function. Since the function is a fraction, we need to find the zeros of the denominator.

1−x=0,⇒x=1.1−x=0,⇒x=1.

We exclude the only point x=1x=1 from the domain of definition of the function and get:

D(y)=(−∞;1)∪(1;+∞).D(y)=(−∞;1)∪(1;+∞).

2) Let us study the behavior of the function in the vicinity of the discontinuity point. Let's find one-sided limits:

Since the limits are equal to infinity, the point x=1x=1 is a discontinuity of the second kind, the straight line x=1x=1 is a vertical asymptote.

3) Let us determine the intersection points of the function graph with the coordinate axes.

Let's find the points of intersection with the ordinate axis OyOy, for which we equate x=0x=0:

Thus, the point of intersection with the OyOy axis has coordinates (0;8)(0;8).

Let's find the points of intersection with the abscissa axis OxOx, for which we set y=0y=0:

The equation has no roots, so there are no points of intersection with the OxOx axis.

Note that x2+8>0x2+8>0 for any xx. Therefore, for x∈(−∞;1)x∈(−∞;1), the function y>0y>0 (takes positive values, the graph is above the x-axis), for x∈(1;+∞)x∈(1; +∞) function y<0y<0 (принимает отрицательные значения, график находится ниже оси абсцисс).

4) The function is neither even nor odd because:

5) Let's examine the function for periodicity. The function is not periodic, since it is a fractional rational function.

6) Let's examine the function for extrema and monotonicity. To do this, we find the first derivative of the function:

Let's equate the first derivative to zero and find stationary points (at which y′=0y′=0):

We got three critical points: x=−2,x=1,x=4x=−2,x=1,x=4. Let us divide the entire domain of definition of the function into intervals with these points and determine the signs of the derivative in each interval:

For x∈(−∞;−2),(4;+∞)x∈(−∞;−2),(4;+∞) the derivative y′<0y′<0, поэтому функция убывает на данных промежутках.

For x∈(−2;1),(1;4)x∈(−2;1),(1;4) the derivative y′>0y′>0, the function increases on these intervals.

In this case, x=−2x=−2 is a local minimum point (the function decreases and then increases), x=4x=4 is a local maximum point (the function increases and then decreases).

Let's find the values ​​of the function at these points:

Thus, the minimum point is (−2;4)(−2;4), the maximum point is (4;−8)(4;−8).

7) We examine the function for kinks and convexity. Let's find the second derivative of the function:

Let us equate the second derivative to zero:

The resulting equation has no roots, so there are no inflection points. Moreover, when x∈(−∞;1)x∈(−∞;1) y′′>0y″>0 is satisfied, that is, the function is concave, when x∈(1;+∞)x∈(1;+ ∞) is satisfied by y′′<0y″<0, то есть функция выпуклая.

8) Let us examine the behavior of the function at infinity, that is, at .

Since the limits are infinite, there are no horizontal asymptotes.

Let's try to determine oblique asymptotes of the form y=kx+by=kx+b. We calculate the values ​​of k,bk,b using known formulas:

We found that the function has one oblique asymptote y=−x−1y=−x−1.

9) Additional points. Let's calculate the value of the function at some other points in order to more accurately construct the graph.

y(−5)=5.5;y(2)=−12;y(7)=−9.5.y(−5)=5.5;y(2)=−12;y(7)=−9.5.

10) Based on the data obtained, we will construct a graph, supplement it with asymptotes x=1x=1 (blue), y=−x−1y=−x−1 (green) and mark the characteristic points (purple intersection with the ordinate axis, orange extrema, black additional points) :

Task 4: Geometric, Economic problems (I have no idea what, here is an approximate selection of problems with solutions and formulas)

Example 3.23. a

Solution. x And y y
y = a - 2×a/4 =a/2. Since x = a/4 is the only critical point, let's check whether the sign of the derivative changes when passing through this point. For xa/4 S " > 0, and for x >a/4 S "< 0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед).Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет highest value functions. Thus, the most favorable aspect ratio of the site under the given conditions of the problem is y = 2x.

Example 3.24.

Solution.
R = 2, H = 16/4 = 4.

Example 3.22. Find the extrema of the function f(x) = 2x 3 - 15x 2 + 36x - 14.

Solution. Since f "(x) = 6x 2 - 30x +36 = 6(x ​​-2)(x - 3), then the critical points of the function x 1 = 2 and x 2 = 3. Extrema can only be at these points. So as when passing through the point x 1 = 2 the derivative changes its sign from plus to minus, then at this point the function has a maximum. When passing through the point x 2 = 3 the derivative changes its sign from minus to plus, therefore at the point x 2 = 3 the function has a minimum. Having calculated the function values ​​at the points
x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f(2) = 14 and minimum f(3) = 13.

Example 3.23. We need to build a rectangular platform near stone wall so that it is fenced off on three sides with wire mesh, and the fourth side is adjacent to the wall. For this there is a linear meters grids At what aspect ratio will the site have largest area?

Solution. Let us denote the sides of the platform by x And y. The area of ​​the site is S = xy. Let y- this is the length of the side adjacent to the wall. Then, by condition, the equality 2x + y = a must hold. Therefore y = a - 2x and S = x(a - 2x), where
0 ≤ x ≤ a/2 (the length and width of the pad cannot be negative). S " = a - 4x, a - 4x = 0 at x = a/4, whence
y = a - 2×a/4 =a/2. Since x = a/4 is the only critical point, let's check whether the sign of the derivative changes when passing through this point. For xa/4 S " > 0, and for x >a/4 S "< 0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед).Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет наибольшим значением функции. Таким образом, наиболее выгодным соотношением сторон площадки при данных условиях задачи является y = 2x.

Example 3.24. It is required to produce a closed cylindrical tank with a capacity of V=16p ≈ 50 m 3 . What should be the dimensions of the tank (radius R and height H) so that the least amount of material is used for its manufacture?

Solution. The total surface area of ​​the cylinder is S = 2pR(R+H). We know the volume of the cylinder V = pR 2 N Þ N = V/pR 2 =16p/ pR 2 = 16/ R 2 . This means S(R) = 2p(R 2 +16/R). We find the derivative of this function:
S " (R) = 2p(2R- 16/R 2) = 4p (R- 8/R 2). S " (R) = 0 for R 3 = 8, therefore,
R = 2, H = 16/4 = 4.

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