Reshebnik Kuznetsova.
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Task 7. Conduct a complete study of the function and build its schedule.

& NBSP & NBSP & NBSP & NBSP Before you start downloading your options, try to solve the sample problem below for option 3. Part of the options are archived in format.rar

& NBSP & NBSP & NBSP & NBSP 7.3 Conduct a full study of the function and build its schedule

Decision.

& NBSP & NBSP & NBSP & NBSP 1) Definition Area: & NBSP & NBSP & NBSP & & NBSP & NBSP & NBSP & NBSP, i.e. & NBSP & NBSP & NBSP & NBSP.
.
Thus: & NBSP & NBSP & NBSP & NBSP.

& NBSP & NBSP & NBSP & NBSP 2) Crossing Points with OX axis. Indeed, the & NBSP & NBSP & NBSP & NBSP equation has no solutions.
Points of intersection with Oy axis no, since & NBSP & NBSP & NBSP & NBSP.

& NBSP & NBSP & NBSP & NBSP 3) The function is either something nor intense. There are no symmetries regarding the axis of the ordinate. There are no symmetries regarding the start of the coordinates. As
.
We see that & NBSP & NBSP & NBSP & NBSP & NBSP & NBSP & NBSP & NBSP.

& NBSP & NBSP & NBSP & NBSP 4) The function is continuous in the definition area
.

; .

; .
Consequently, the point & NBSP & NBSP & NBSP & NBSP is a second-order break point (infinite break).

5) Vertical Asymptotes: & NBSP & NBSP & NBSP & NBSP

We will find inclined asymptotes & NBSP & NBSP & NBSP & NBSP. Here

;
.
Consequently, we have horizontal asymptotes: y \u003d 0.. There is no inclined asymptot.

& NBSP & NBSP & NBSP & NBSP 6) will find the first derivative. First derivative:
.
And that's why
.
Find stationary points where the derivative is zero, that is
.

& NBSP & NBSP & NBSP & NBSP 7) We will find the second derivative. Second derivative:
.
And it is easy to make sure because

How to investigate the function and build its schedule?

It seems that I begin to understand the spiritualized-penetrated face of the leader of the world proletariat, the author of the collection of writings in 55 volumes .... Non-income way began with elementary information about functions and charts, And now, work on time-consuming theme ends with a natural result - article on full study of the function. The long-awaited task is formulated as follows:

Explore the function of differential calculus methods and based on the results of the study to build its schedule

Or shorter: explore the function and build a chart.

Why explore? In simple cases, we will not find it difficult to understand the elementary functions, draw the schedule obtained by elementary geometric transformations etc. However, properties and graphic images of more complex functions are far from obvious, which is why the whole study is necessary.

The main stages of the solution are reduced in reference material. Function research schemeThis is your guide to the section. Teapotes require a step-by-step explanation of the topic, some readers do not know where to start and how to organize a study, and advanced students may be interested only at some moments. But whoever you could, dear visitor, the proposed abstract with pointers to various lessons in the shortest term orientates and will direct you in the direction of interest. Robots slandered \u003d) Guide the swaths in the form of a PDF file and took a well-deserved place on the page Mathematical formulas and tables.

The study of the function I used to break up 5-6 points:

6) Additional points and schedule based on the results of the study.

At the expense of the final action, I think everything is clear to everyone - it will be very disappointing if in a matter of seconds it will be crossed and returned to the refinement. The correct and accurate drawing is the main result of the solution! It is greatly likely to "bind" analytical abrasions, while incorrect and / or negligent chart will deliver problems even with an ideally conducted study.

It should be noted that in other sources the number of research items, the procedure for their implementation and style of registration can differ significantly from the scheme proposed by me, but in most cases it is quite enough. The simplest version of the task consists of only 2-3 stages and is formulated as follows: "Explore the function using a derivative and build a chart" or "explore the function using the 1st and 2nd derivative, build a chart."

Naturally - if your method is disassembled in detail another algorithm or your teacher strictly demands to adhere to his lectures, it will have to make some adjustments to the solution. Not more difficult than replacing the fork with a chainsaw spoon.

Check the function on readiness / oddness:

After that, a template recording is followed:
Therefore, this function is not even or odd.

Since the function is continuous on, there are no vertical asymptotes.

No inclined asymptot.

Note : I remind you that is higher growth orderthan, so the final limit is equal to " a plus Infinity. "

Find out how the function behaves on infinity:

In other words, if we go to the right, then the schedule goes infinitely far away, if left is endlessly down. Yes, here are also two limits under a single record. If you have any difficulties with decoding signs, please visit the lesson about infinitely small features.

Thus, the function not limited to from above and not limited to below. Considering that we have no break points, it becomes clear and function values \u200b\u200barea: - Also any valid number.

Useful technical technique

Each setup of task brings new information about the graphTherefore, during the solution, it is convenient to use a kind of layout. I will depict the coordinate system on Cartovka Cartov. What is already known? First, the schedule does not have asymptot, therefore, the direct drawback is not needed. Secondly, we know how the function behaves at infinity. According to the analysis, draw the first approximation:

Note that by virtue continuity Functions on and the fact that the schedule should at least once cross the axis. Or maybe there are several intersection points?

3) zeros and intervals of the alignment.

We will first find the point of intersection of the graph with the axis of the ordinate. It's simple. It is necessary to calculate the value of the function when:

One and a half above sea level.

To find the intersection points with the axis (zeros of the function), it is required to solve the equation, and here we will have an unpleasant surprise:

At the end, a free member was attached, which greatly complicates the task.

Such an equation has at least one valid root, and most often this root is irrational. In the worse fairy tale, we will have three pigs. The equation is solvable using the so-called cardano formulasBut the paper damage is comparable almost with all the study. In this regard, it is more intelligible orally either on the draft to try to choose at least one whole root. Check, are not numbers:
- not suitable;
- there is!

It's lucky here. In the event of failure, it is also possible to test, and if these numbers did not come up, then there is very little chances for a profitable solution to the equation. Then the study item is better to completely skip - maybe it will become something clearer at the final step when additional points will be made. And if the same root (roots) is clearly "bad", then the intervals of the alignment is better in generally modestly silex yes, it is more instructed to fulfill the drawing.

However, we have a beautiful root, so we divide the polynomial on no residue:

The algorithm for dividing the polynomial to the polynomial in detail is disassembled in the first example of the lesson Difficult limits.

As a result, the left part of the source equation folded into the work:

And now a little about a healthy lifestyle. I, of course, understand that quadratic equations You need to decide every day, but today we will make an exception: equation It has two valid root.

On a numeric direct postpone the found values and interval method Determine the features of the function:


flagte thus at intervals Schedule located
below the abscissa axis, and at intervals - Above this axis.

The resulting conclusions allow you to detail our layout, and the second approximation of the graph is as follows:

Please note that the function must necessarily have at least one maximum, and at an interval - at least one minimum. But how many times, where and when will "hide" a schedule, we do not know yet. By the way, the function can have both infinitely many extremes.

4) Ascending, decrease and extremum function.

Find critical points:

This equation has two valid roots. I will postpone them on a numeric direct and define the signs of the derivative:


Consequently, the function increases on and decreases on.
At the point, the feature reaches a maximum: .
At the point, the function reaches a minimum: .

Installed facts pound our template in a rather hard frame:

What to say, differential calculus - a powerful thing. Let's finally deal with the shape of the schedule:

5) bulge, concaveness and point of inflection.

We will find critical points of the second derivative:

Determine the signs:


The function graph is convex on and concave on. Calculate the ordinate of the point of inflection :.

Almost everything turned out.

6) It remains to find additional points that will help more precisely build a schedule and perform self-test. In this case, they are not enough, but we will not neglect:

Perform a drawing:

Green color is marked with a point of inflection, crosses - additional points. The graph of the cubic function is symmetrical about his inflection point, which is always located strictly in the middle between the maximum and minimum.

In the course of performing the task, I brought three hypothetical intermediate drawings. In practice, it is enough to draw the coordinate system, mark the points found and after each study item mentally estimate how the function graph may look like. Students with a good level of training will not be difficult to carry out such an analysis exclusively in the mind without attracting draft.

For self solutions:

Example 2.

Explore the function and build a chart.

There is a faster and more fun, an exemplary sample of finishing design at the end of the lesson.

A lot of secrets reveals the study of fractional rational functions:

Example 3.

Differential calculus methods explore the function and on the basis of the results of the study to build its schedule.

Decision: The first stage of the study does not differ with something remarkable, with the exception of the hole in the field of definition:

1) The function is defined and continuous on the entire numeric direct except the point, domain: .

It means that this function is not even or odd.

Obviously, the function is non-periodic.

The graph of the function is two continuous branches located in the left and right half-plane - this is perhaps the most important conclusion of the 1st point.

2) Asymptotes, behavior of the function at infinity.

a) With the help of one-way limits, we investigate the behavior of a function near a suspicious point where it is clearly a vertical asymptota:

Indeed, functions tolerate infinite break At the point,
and the straight (axis) is vertical Asimptota graphics .

b) check whether oblique asymptotes exist:

Yes, direct is inclined asymptoto Graphics, if.

The limits to analyze makes no sense, since it is so clear that the function in an embrace with its inclined asymptota not limited to from above and not limited to below.

The second research point brought many important information about the function. Perform a draft sketch:

Conclusion number 1 concerns the intervals of the alignment. On the "minus infinity", the graph of the function is uniquely located below the abscissa axis, and on the "plus infinity" - above this axis. In addition, one-sided limits reported to us as the left and right of the function, too, more zero. Please note that in the left half-plane, the schedule at least once is obliged to cross the axis of the abscissa. In the right half-plane zeros, the functions may not be.

The output number 2 is that the function increases on and to the left (there is a "bottom up"). On the right of this point - the function decreases (there is a "top down"). The right branch of the chart certainly should be at least one minimum. Left extremes are not guaranteed.

Conclusion number 3 gives reliable information about the concavity of the graph in the neighborhood of the point. We can not say anything about the bulge / concavity on infinity, because the line can be pressed to their asymptotes both from above and below. Generally speaking, there is an analytical way to figure it out right now, but the shape of the gift "for nothing" will become clearer at the later stages.

Why so many words? To monitor subsequent research points and prevent errors! Further calculations should not be contrary to the conclusions.

3) Points of intersection of the chart with coordinate axes, the intervals of the symbol function.

The graph of the function does not cross the axis.

Interval method Determine the signs:

, if a ;
, if a .

The results of the point fully correspond to the conclusion number 1. After each stage, look at the draft, mentally referred to the study and draw a function schedule.

In the example under consideration, the numerator is divided into a denominator, which is very beneficial for differentiation:

Actually, it has already been done while the asymptotes are found.

- critical point.

Determine the signs:

increases by and decrease by

At the point, the function reaches a minimum: .

Discussions with conclusion number 2 also did not find out, and most likely, we are on the right track.

So, the function graph is concave throughout the field of definition.

Excellent - and do not draw anything.

There are no inflection points.

Conference is consistent with the conclusion number 3, moreover, indicates that in infinity (and there and there) the graph of the function is located above its inclined asymptotes.

6) In conscientiously pink the task with additional points. Here it will be pretty to work hard, because of the study we are known only two points.

And the picture, which, probably, many have long presented:

During the task, you need to carefully ensure that there are no contradictions between the stages of the study, but sometimes the situation is emergency or even a desperate-dead-end. Here "not converge" analyst - and that's it. In this case, I recommend the emergency reception: we find as many points that belong to the graphics (how much patience is enough), and we note them on the coordinate plane. A graphical analysis of the found values \u200b\u200bin most cases will tell you where the truth, and where is a lie. In addition, the schedule can be previously built using any program, for example, in the same exile (understandable, for this you need skills).

Example 4.

Differential calculus methods explore the function and build its schedule.

This is an example for an independent solution. In it, self-control is enhanced by the function - the graph is symmetric about the axis, and if something contrary to this fact in your study, look for an error.

You can also explore a clear or odd function when, and then use the symmetry of the graph. Such a solution is optimal, but it looks like, in my opinion, is very unusual. Personally, I consider the entire numeric axis, but I find additional points yet on the right:

Example 5.

Conduct a complete study of the function and build its schedule.

Decision: It rushed hard:

1) The function is defined and continuous on the entire numeric line :.

It means that this function is odd, its graph is symmetrical relative to the start of coordinates.

Obviously, the function is non-periodic.

2) Asymptotes, behavior of the function at infinity.

Since the function is continuous on, then the vertical asymptotes are absent

For a function containing the exhibitor typically separate The study "Plus" and "minus infinity", but our lives facilitates the symmetry of the schedule - either to the left and on the right there is an asymptota, or it is not. Therefore, both infinite limits can be issued under a single record. During the solution we use lopital rule:

Direct (axis) is a horizontal asymptota of the graph with.

Please note how I hit the complete algorithm of finding inclined asymptotes: the limit is completely easily and clarifies the behavior of the function at infinity, and the horizontal asymptota has found "as if at the same time."

From continuity on and existence of horizontal asymptotes follows the fact that the function limited from above and limited from below.

3) the intersection points of the graph with the coordinate axes, the intervals of the alignment.

Here, too, reduce the decision:
The schedule passes through the origin of the coordinates.

There are no other points of intersection with coordinate axes. Moreover, the intervals of the alpopurism are obvious, and the axis can not be drawn:, which means that the function of the function depends only on the "ICA":
, if a ;
, if a .

4) increase, decrease, extremum function.


- Critical points.

The points are symmetrical relative to zero, as it should be.

Determine the signs of the derivative:


The function increases on the interval and decreases at intervals

At the point, the feature reaches a maximum: .

By virtue of the property (Foundming Functions) Minimum can not be calculated:

Since the function decreases on the interval, it is obvious to "minus infinity" the schedule is located under With his asymptota. At the interval, the function also decreases, but here everything is the opposite - after switching through the maximum point, the line approaches the axis already on top.

Of the foregoing, it also follows that the function schedule is convex on the "minus infinity" and concave on the "plus infinity."

After this point of study, the field of values \u200b\u200bof the function was also drawn:

If you have no misunderstanding of any moments, once again I urge to draw coordinate axes in the notebook and with a pencil in the hands to re-analyze each conclusion.

5) Conversion, concaveness, inflection of graphics.

- Critical points.

Symmetry points are preserved, and most likely we are not mistaken.

Determine the signs:


The function graph is convex on And concave on .

The bulge / concaveness in the extreme intervals was confirmed.

In all critical points there are bending geographics. We will find the ordinates of the beggar points, while again will reduce the number of calculations using the oddity of the function:

If the task is to complete a full study of the function f (x) \u003d x 2 4 x 2 - 1 with the construction of its schedule, then consider this principle in detail.

To solve the task of this type, use the properties and graphs of the main elementary functions. The study algorithm includes steps:

Finding a field of definition

Since research is carried out on the field definition area, it is necessary to start from this step.

Example 1.

The specified example implies the foundation of the denominator zeros in order to exclude them from OTZ.

4 x 2 - 1 \u003d 0 x \u003d ± 1 2 ⇒ x ∈ - ∞; - 1 2 ∪ - 1 2; 1 2 ∪ 1 2; + ∞.

As a result, you can get roots, logarithms, and so on. Then the OTZ can be sought for an even degree of type G (x) 4 by inequality g (x) ≥ 0, for logarithm Log A G (x) by inequality G (x)\u003e 0.

Study of border borders and finding vertical asymptot

At the boundaries of the function there are vertical asymptotes when one-sided limits at such points are infinite.

Example 2.

For example, consider border points equal to x \u003d ± 1 2.

Then it is necessary to study the function to find a unilateral limit. Then we get that: Lim X → - 1 2 - 0 F (x) \u003d Lim X → - 1 2 - 0 x 2 4 x 2 - 1 \u003d \u003d Lim X → - 1 2 - 0 x 2 (2 x - 1 ) (2 x + 1) \u003d 1 4 (- 2) · 0 \u003d + ∞ Lim X → - 1 2 + 0 F (x) \u003d Lim X → - 1 2 + 0 x 2 4 x - 1 \u003d lim X → - 1 2 + 0 x 2 (2 x - 1) (2 x + 1) \u003d 1 4 (- 2) · (+ 0) \u003d - ∞ Lim X → 1 2 - 0 F (x) \u003d Lim x → 1 2 - 0 x 2 4 x 2 - 1 \u003d \u003d Lim x → 1 2 - 0 x 2 (2 x - 1) (2 x + 1) \u003d 1 4 (- 0) · 2 \u003d - ∞ Lim x → 1 2 - 0 F (x) \u003d Lim X → 1 2 - 0 x 2 4 x 2 - 1 \u003d \u003d Lim X → 1 2 - 0 x 2 (2 x - 1) (2 x + 1) \u003d 1 4 ( + 0) · 2 \u003d + ∞

It can be seen that one-sided limits are infinite, which means straight X \u003d ± 1 2 - vertical asymptotes of the graph.

Research function and parity or oddness

When the condition y (- x) \u003d y (x) is satisfied, the function is considered even. This suggests that the schedule is located symmetrically relative to o. When the condition y (- x) \u003d - y (x) is satisfied, the function is considered odd. It means that symmetry comes relative to the start of coordinates. With default, at least one inequality, we obtain a common function.

The implementation of the equality y (- x) \u003d y (x) suggests that the function is even. When constructing it is necessary to take into account that there will be symmetry relative to o.

For solution of increasing and descending gaps with conditions f "(x) ≥ 0 and f" (x) ≤ 0, respectively.

Definition 1.

Stationary points- These are the points that turn the derivative in zero.

Critical points - These are internal points from the definition area, where the derivative of the function is zero or does not exist.

When solving, it is necessary to take into account the following remarks:

• with the extensions of the increasing and descending of the inequality of the form F "(x)\u003e 0, the critical points in the solution are not included;
• the points in which the function is defined without a finite derivative must be included in the gaps of increasing and descending (for example, Y \u003d x 3, where point x \u003d 0 makes the function defined, the derivative has the value of infinity at this point, y "\u003d 1 3 · x 2 3, y "(0) \u003d 1 0 \u003d ∞, x \u003d 0 is included in the increasing interval);
• in order to avoid disagreements, it is recommended to use mathematical literature, which is recommended by the Ministry of Education.

The inclusion of critical points into the gaps of increasing and descending in the event that they satisfy the field definition areas.

Definition 2.

For definitions of gaps of increasing and descending function must be found:

• derivative;
• critical points;
• split the definition area with critical points to the intervals;
• determine the sign of the derivative on each of the gaps, where + is an increase in, and is descending.

Example 3.

Find a derivative on the definition field F "(x) \u003d x 2" (4 x 2 - 1) - x 2 4 x 2 - 1 "(4 x 2 - 1) 2 \u003d - 2 x (4 x 2 - 1) 2 .

Decision

To solve you need:

• find stationary points, this example has x \u003d 0;
• find the zeros of the denominator, the example takes the value of zero at x \u003d ± 1 2.

Test points on the numeric axis to determine the derivative at each interval. To do this, it is enough to take any point from the gap and make a calculation. With a positive result, the graph is depicting +, which means increasing the function, and - means its decrease.

For example, f "(- 1) \u003d - 2 · (- 1) 4 - 1 2 - 1 2 \u003d 2 9\u003e 0, it means that the first interval of the left has a sign +. Consider on a numeric line.

Answer:

• there is an increase in the function in the interval - ∞; - 1 2 and (- 1 2; 0];
• decrease in the interval [0; 1 2) and 1 2; + ∞.

In the diagram with + and - the positiveness and negativity of the function is depicted, and the shooter is decreased and increasing.

Extremum points function - points where the function is defined and through which the derivative changes the sign.

Example 4.

If we consider an example where x \u003d 0, then the function value in it is equal to F (0) \u003d 0 2 4 · 0 2 - 1 \u003d 0. When changing the sign of the derivative with + on - and passing through the point x \u003d 0, then the point with coordinates (0; 0) is considered a maximum point. When changing the sign C - on + we get a minimum point.

Conversion and concavity is determined when solving the inequalities of the form F "" (x) ≥ 0 and f "" (x) ≤ 0. Less often use the name of the bulge down instead of concave, and the bulge up instead of convexity.

Definition 3.

For determining the gaps of concave and bulge Need:

• find the second derivative;
• find zeros of the function of the second derivative;
• split the definition area that appeared on the intervals;
• determine the interval sign.

Example 5.

Find the second derivative from the definition area.

Decision

f "" (x) \u003d - 2 x (4 x 2 - 1) 2 "\u003d \u003d (- 2 x)" (4 x 2 - 1) 2 - - 2 x 4 x 2 - 1 2 "(4 x 2 - 1) 4 \u003d 24 x 2 + 2 (4 x 2 - 1) 3

We find zeros of the numerator and denominator, where on the example of our example we have that zeros of the denominator x \u003d ± 1 2

Now you need to apply points to the numeric axis and define a sign of the second derivative of each gap. We get that

Answer:

• the function is convex from the gap - 1 2; 12 ;
• the function is concave from the gaps - ∞; - 1 2 and 1 2; + ∞.

Definition 4.

Point of inflection - it is a point of type X 0; f (x 0). When it has tangent to the graphics of the function, then when it passes through X 0, the function changes the sign to the opposite.

In other words, this is such a point through which the second derivative passes and changes the sign, and in the points itself equals zero or does not exist. All points are considered a field definition area.

In the example, it was clear that the points of the inflection are absent, since the second derivative changes the sign during passing through the points x \u003d ± 1 2. They, in turn, are not included in the field of definition.

Finding horizontal and inclined asymptotes

When determining the function at infinity, it is necessary to look for horizontal and inclined asymptotes.

Definition 5.

Inclined asymptotespictures are depicted using the direct specified by the equation y \u003d k x + b, where k \u003d lim x → ∞ f (x) x and b \u003d lim x → ∞ f (x) - k x.

At k \u003d 0 and b, not equal to infinity, we obtain that the inclined asymptota becomes horizontal.

In other words, the asymptotes consider the lines to which the schedule of the function is approaching infinity. This contributes to the rapid construction of the function graphics.

If the asymptotes are missing, but the function is determined on both infinitioners, it is necessary to calculate the limit of the function on these infinity, to understand how the function graph itself will be.

Example 6.

On the example, consider that

k \u003d lim x → ∞ f (x) x \u003d lim x → ∞ x 2 4 x 2 - 1 x \u003d 0 b \u003d lim x → ∞ (f (x) - kx) \u003d lim x → ∞ x 2 4 x 2 - 1 \u003d 1 4 ⇒ Y \u003d 1 4

it is horizontal asymptota. After researching, the function can be started to build it.

Calculate function value at intermediate points

To build the schedule is most accurate, it is recommended to find several functions of the function at intermediate points.

Example 7.

From the example we considered, it is necessary to find the values \u200b\u200bof the function at points x \u003d - 2, x \u003d - 1, x \u003d - 3 4, x \u003d - 1 4. Since the function is even, we obtain that the values \u200b\u200bcoincide with the values \u200b\u200bat these points, that is, we obtain x \u003d 2, x \u003d 1, x \u003d 3 4, x \u003d 1 4.

We write and resolve:

F (- 2) \u003d f (2) \u003d 2 2 4 · 2 2 - 1 \u003d 4 15 ≈ 0, 27 f (- 1) - f (1) \u003d 1 2 4 · 1 2 - 1 \u003d 1 3 ≈ 0 , 33 f - 3 4 \u003d F 3 4 \u003d 3 4 2 4 3 4 2 - 1 \u003d 9 20 \u003d 0, 45 F - 1 4 \u003d F 1 4 \u003d 1 4 2 4 · 1 4 2 - 1 \u003d - 1 12 ≈ - 0, 08

To determine the maxims and minima of the function, points of the inflection, intermediate points need to build asymptotes. For convenient designation, the gaps of increasing, decrease, bulge, concaveness are recorded. Consider in the figure shown below.

It is necessary through the marked points to carry out the lines of the graph, which will bring closer to the asymptotam, following the arroges.

This ends the complete study of the function. There are cases of constructing some elementary functions for which geometrical transformations are used.

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