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Task 7. Conduct a complete study of the function and construct its graph.

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        7.3 Conduct a full study of the function and plot it

Solution.

        1) Scope of definition:         or        , that is        .
.
Thus:         .

        2) There are no points of intersection with the Ox axis. Indeed, the equation         has no solutions.
There are no points of intersection with the Oy axis, since        .

        3) The function is neither even nor odd. There is no symmetry about the ordinate axis. There is also no symmetry about the origin. Because
.
We see that         and        .

        4) The function is continuous in the domain of definition
.

; .

; .
Consequently, the point         is a point of discontinuity of the second kind (infinite discontinuity).

5) Vertical asymptotes:       

Let's find the oblique asymptote        . Here

;
.
Consequently, we have a horizontal asymptote: y=0. There are no oblique asymptotes.

        6) Let's find the first derivative. First derivative:
.
And that's why
.
Let's find stationary points where the derivative is equal to zero, that is
.

        7) Let’s find the second derivative. Second derivative:
.
And this is easy to verify, since

If the problem requires a complete study of the function f (x) = x 2 4 x 2 - 1 with the construction of its graph, then we will consider this principle in detail.

To solve the problem of this type properties and graphs of basic elementary functions should be used. The research algorithm includes the following steps:

Finding the domain of definition

Since research is carried out on the domain of definition of the function, it is necessary to start with this step.

Example 1

Behind this example involves finding the zeros of the denominator in order to exclude them from the ODZ.

4 x 2 - 1 = 0 x = ± 1 2 ⇒ x ∈ - ∞ ; - 1 2 ∪ - 1 2 ; 1 2 ∪ 1 2 ; +∞

As a result, you can get roots, logarithms, and so on. Then the ODZ can be searched for a root of an even degree of type g (x) 4 by the inequality g (x) ≥ 0, for the logarithm log a g (x) by the inequality g (x) > 0.

Studying the boundaries of the ODZ and finding vertical asymptotes

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

Example 2

For example, consider the border points equal to x = ± 1 2.

Then it is necessary to study the function to find the one-sided limit. Then we get that: lim x → - 1 2 - 0 f (x) = lim x → - 1 2 - 0 x 2 4 x 2 - 1 = = lim x → - 1 2 - 0 x 2 (2 x - 1 ) (2 x + 1) = 1 4 (- 2) · - 0 = + ∞ lim x → - 1 2 + 0 f (x) = lim x → - 1 2 + 0 x 2 4 x - 1 = = lim x → - 1 2 + 0 x 2 (2 x - 1) (2 x + 1) = 1 4 (- 2) (+ 0) = - ∞ lim x → 1 2 - 0 f (x) = lim x → 1 2 - 0 x 2 4 x 2 - 1 = = lim x → 1 2 - 0 x 2 (2 x - 1) (2 x + 1) = 1 4 (- 0) 2 = - ∞ lim x → 1 2 - 0 f (x) = lim x → 1 2 - 0 x 2 4 x 2 - 1 = = lim x → 1 2 - 0 x 2 (2 x - 1) (2 x + 1) = 1 4 ( + 0) 2 = + ∞

This shows that the one-sided limits are infinite, which means the straight lines x = ± 1 2 are the vertical asymptotes of the graph.

Study of a function and whether it is even or odd

When the condition y (- x) = y (x) is satisfied, the function is considered even. This suggests that the graph is located symmetrically with respect to Oy. When the condition y (- x) = - y (x) is satisfied, the function is considered odd. This means that the symmetry is relative to the origin of coordinates. If at least one inequality is not satisfied, we obtain a function of general form.

The equality y (- x) = y (x) indicates that the function is even. When constructing, it is necessary to take into account that there will be symmetry with respect to Oy.

To solve the inequality, intervals of increasing and decreasing are used with the conditions f " (x) ≥ 0 and f " (x) ≤ 0, respectively.

Definition 1

Stationary points- these are the points that turn the derivative to zero.

Critical points- these are internal points from the domain of definition where the derivative of the function is equal to zero or does not exist.

When making a decision, the following notes must be taken into account:

• for existing intervals of increasing and decreasing inequalities of the form f " (x) > 0, critical points are not included in the solution;
• points at which the function is defined without a finite derivative must be included in the intervals of increasing and decreasing (for example, y = x 3, where the point x = 0 makes the function defined, the derivative has the value of infinity at this point, y " = 1 3 x 2 3, y "(0) = 1 0 = ∞, x = 0 is included in the increasing interval);
• To avoid disagreements, it is recommended to use mathematical literature recommended by the Ministry of Education.

Inclusion of critical points in intervals of increasing and decreasing if they satisfy the domain of definition of the function.

Definition 2

For determining the intervals of increase and decrease of a function, it is necessary to find:

• derivative;
• critical points;
• divide the definition domain into intervals using critical points;
• determine the sign of the derivative on each of the intervals, where + is an increase and - is a decrease.

Example 3

Find the derivative on the domain of definition f " (x) = x 2 " (4 x 2 - 1) - x 2 4 x 2 - 1 " (4 x 2 - 1) 2 = - 2 x (4 x 2 - 1) 2 .

Solution

To solve you need:

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

We place points on the number axis to determine the derivative on each interval. To do this, it is enough to take any point from the interval and perform a calculation. If the result is positive, we depict + on the graph, which means the function is increasing, and - means it is decreasing.

For example, f " (- 1) = - 2 · (- 1) 4 - 1 2 - 1 2 = 2 9 > 0, which means that the first interval on the left has a + sign. Consider on the number line.

Answer:

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

In the diagram, using + and -, the positivity and negativity of the function are depicted, and the arrows indicate decrease and increase.

Extremum points of a function are points where the function is defined and through which the derivative changes sign.

Example 4

If we consider an example where x = 0, then the value of the function in it is equal to f (0) = 0 2 4 · 0 2 - 1 = 0. When the sign of the derivative changes from + to - and passes through the point x = 0, then the point with coordinates (0; 0) is considered the maximum point. When the sign changes from - to +, we obtain a minimum point.

Convexity and concavity are determined by solving inequalities of the form f "" (x) ≥ 0 and f "" (x) ≤ 0. Less commonly used is the name convexity down instead of concavity, and convexity upward instead of convexity.

Definition 3

For determining the intervals of concavity and convexity necessary:

• find the second derivative;
• find the zeros of the second derivative function;
• divide the definition area into intervals with the appearing points;
• determine the sign of the interval.

Example 5

Find the second derivative from the domain of definition.

Solution

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

We find the zeros of the numerator and denominator, where in our example we have that the zeros of the denominator x = ± 1 2

Now you need to plot the points on the number line and determine the sign of the second derivative from each interval. We get that

Answer:

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

Definition 4

Inflection point– this is a point of the form x 0 ; f (x 0) . When it has a tangent to the graph of the function, then when it passes through x 0 the function changes sign to the opposite.

In other words, this is a point through which the second derivative passes and changes sign, and at the points themselves it is equal to zero or does not exist. All points are considered to be the domain of the function.

In the example, it was clear that there are no inflection points, since the second derivative changes sign while passing through the points x = ± 1 2. They, in turn, are not included in the scope of definition.

Finding horizontal and oblique asymptotes

When defining a function at infinity, you need to look for horizontal and oblique asymptotes.

Definition 5

Oblique asymptotes are depicted using straight lines given by the equation y = k x + b, where k = lim x → ∞ f (x) x and b = lim x → ∞ f (x) - k x.

For k = 0 and b not equal to infinity, we find that the oblique asymptote becomes horizontal.

In other words, asymptotes are considered to be lines to which the graph of a function approaches at infinity. This facilitates quick construction of a function graph.

If there are no asymptotes, but the function is defined at both infinities, it is necessary to calculate the limit of the function at these infinities in order to understand how the graph of the function will behave.

Example 6

Let's consider as an example that

k = lim x → ∞ f (x) x = lim x → ∞ x 2 4 x 2 - 1 x = 0 b = lim x → ∞ (f (x) - k x) = lim x → ∞ x 2 4 x 2 - 1 = 1 4 ⇒ y = 1 4

is a horizontal asymptote. After examining the function, you can begin to construct it.

Calculating the value of a function at intermediate points

To make the graph more accurate, it is recommended to find several function values ​​at intermediate points.

Example 7

From the example we considered, it is necessary to find the values ​​of the function at the points x = - 2, x = - 1, x = - 3 4, x = - 1 4. Since the function is even, we get that the values ​​coincide with the values ​​at these points, that is, we get x = 2, x = 1, x = 3 4, x = 1 4.

Let's write and solve:

F (- 2) = f (2) = 2 2 4 2 2 - 1 = 4 15 ≈ 0, 27 f (- 1) - f (1) = 1 2 4 1 2 - 1 = 1 3 ≈ 0 , 33 f - 3 4 = f 3 4 = 3 4 2 4 3 4 2 - 1 = 9 20 = 0 , 45 f - 1 4 = f 1 4 = 1 4 2 4 1 4 2 - 1 = - 1 12 ≈ - 0.08

To determine the maxima and minima of the function, inflection points, and intermediate points, it is necessary to construct asymptotes. For convenient designation, intervals of increasing, decreasing, convexity, and concavity are recorded. Let's look at the picture below.

It is necessary to draw graph lines through the marked points, which will allow you to approach the asymptotes by following the arrows.

This concludes the full exploration of the function. There are cases of constructing some elementary functions for which geometric transformations are used.

If you notice an error in the text, please highlight it and press Ctrl+Enter

How to study a function and build its graph?

It seems that I am beginning to understand the spiritually insightful face of the leader of the world proletariat, the author of collected works in 55 volumes... The long journey began with basic information about functions and graphs, and now work on a labor-intensive topic ends with a logical result - an article about a complete study of the function. The long-awaited task is formulated as follows:

Study a function using differential calculus methods and build its graph based on the results of the study

Or in short: examine the function and build a graph.

Why explore? In simple cases it will not be difficult for us to deal with elementary functions, draw the graph obtained using elementary geometric transformations and so on. However, the properties and graphical representations of more complex functions are far from obvious, which is why a whole study is needed.

The main steps of the solution are summarized in the reference material Function study scheme, this is your guide to the section. Dummies need a step-by-step explanation of a topic, some readers don't know where to start or how to organize their research, and advanced students may only be interested in a few points. But whoever you are, dear visitor, the proposed summary with pointers to various lessons will quickly orient and guide you in the direction of interest. The robots shed tears =) The manual was laid out as a pdf file and took its rightful place on the page Mathematical formulas and tables.

I’m used to breaking down a function’s research into 5-6 points:

6) Additional points and graph based on the research results.

Regarding the final action, I think everything is clear to everyone - it will be very disappointing if in a matter of seconds it is crossed out and the task is returned for revision. A CORRECT AND ACCURATE DRAWING is the main result of the solution! It is likely to “cover up” analytical errors, while an incorrect and/or careless schedule will cause problems even with a perfectly conducted study.

It should be noted that in other sources the number of research points, the order of their implementation and the design style may differ significantly from the scheme I proposed, but in most cases it is quite sufficient. The simplest version of the problem consists of only 2-3 stages and is formulated something like this: “investigate the function using the derivative and build a graph” or “investigate the function using the 1st and 2nd derivatives, build a graph.”

Naturally, if your manual describes another algorithm in detail or your teacher strictly demands that you adhere to his lectures, then you will have to make some adjustments to the solution. No more difficult than replacing a chainsaw fork with a spoon.

Let's check the function for even/odd:

This is followed by a template reply:
, Means, this function is not even or odd.

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

There are no oblique asymptotes either.

Note : I remind you that the higher growth order, than , therefore the final limit is exactly “ plus infinity."

Let's find out how the function behaves at infinity:

In other words, if we go to the right, then the graph goes infinitely far up, if we go to the left, it goes infinitely far down. Yes, there are also two limits under a single entry. If you have difficulty deciphering the signs, please visit the lesson about infinitesimal functions.

So the function not limited from above And not limited from below. Considering that we have no breakpoints, it becomes clear function range: – also any real number.

USEFUL TECHNICAL TECHNIQUE

Each stage of the task brings new information about the graph of the function, therefore, during the solution it is convenient to use a kind of LAYOUT. Let's draw a Cartesian coordinate system on a draft. What is already known for sure? Firstly, the graph has no asymptotes, therefore, there is no need to draw straight lines. Secondly, we know how the function behaves at infinity. According to the analysis, we draw a first approximation:

Please note that due to continuity function on and the fact that the graph must cross the axis at least once. Or maybe there are several points of intersection?

3) Zeros of the function and intervals of constant sign.

First, let's find the point of intersection of the graph with the ordinate axis. It's simple. It is necessary to calculate the value of the function at:

One and a half above sea level.

To find the points of intersection with the axis (zeros of the function), we need to solve the equation, and here an unpleasant surprise awaits us:

There is a free member lurking at the end, which makes the task much more difficult.

Such an equation has at least one real root, and most often this root is irrational. In the worst fairy tale, the three little pigs are waiting for us. The equation is solvable using the so-called Cardano formulas, but the damage to paper is comparable to almost the entire study. In this regard, it is wiser to try to select at least one, either verbally or in a draft. whole root. Let's check if these numbers are:
– not suitable;
- There is!

Lucky here. In case of failure, you can also test , and if these numbers do not fit, then I’m afraid there is very little chance of a profitable solution to the equation. Then it is better to skip the research point completely - perhaps something will become clearer at the final step, when additional points will be broken through. And if the root(s) are clearly “bad”, then it is better to remain modestly silent about the intervals of constancy of signs and to draw more carefully.

However, we have a beautiful root, so we divide the polynomial for no remainder:

The algorithm for dividing a polynomial by a polynomial is discussed in detail in the first example of the lesson Complex Limits.

As a result, the left side of the original equation decomposes into the product:

And now a little about healthy way life. I, of course, understand that quadratic equations needs to be solved every day, but today we’ll make an exception: the equation has two real roots.

Let us plot the found values ​​on the number line And interval method Let's define the signs of the function:


og Thus, on the intervals the schedule is located
below the x-axis, and at the intervals – above this axis.

The findings allow us to refine our layout, and the second approximation of the graph looks like this:

Please note that a function must have at least one maximum on an interval, and at least one minimum on an interval. But we don’t yet know how many times, where and when the schedule will loop. By the way, a function can have infinitely many extremes.

4) Increasing, decreasing and extrema of the function.

Let's find critical points:

This equation has two real roots. Let's put them on the number line and determine the signs of the derivative:


Therefore, the function increases by and decreases by .
At the point the function reaches its maximum: .
At the point the function reaches a minimum: .

Established facts drive our template into a fairly rigid framework:

Needless to say, differential calculus is a powerful thing. Let's finally understand the shape of the graph:

5) Convexity, concavity and inflection points.

Let's find the critical points of the second derivative:

Let's define the signs:


The graph of the function is convex on and concave on . Let's calculate the ordinate of the inflection point: .

Almost everything has become clear.

6) It remains to find additional points that will help you more accurately construct a graph and perform self-test. In this case there are few of them, but we will not neglect them:

Let's make the drawing:

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

As the assignment progressed, I provided three hypothetical interim drawings. In practice, it is enough to draw a coordinate system, mark the points found, and after each point of research mentally estimate what the graph of the function might look like. It will not be difficult for students with a good level of preparation to carry out such an analysis solely in their heads without involving a draft.

To solve it yourself:

Example 2

Explore the function and build a graph.

Everything is faster and more fun here, an approximate example of the final design at the end of the lesson.

The study of fractional rational functions reveals many secrets:

Example 3

Use differential calculus methods to study a function and, based on the results of the study, construct its graph.

Solution: the first stage of the study is not distinguished by anything remarkable, with the exception of a hole in the definition area:

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

, which means that this function is not even or odd.

It is obvious that the function is non-periodic.

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

2) Asymptotes, the behavior of a function at infinity.

a) Using one-sided limits, we examine the behavior of the function near a suspicious point, where there should clearly be a vertical asymptote:

Indeed, the functions endure endless gap at the point
and the straight line (axis) is vertical asymptote graphic arts .

b) Let’s check whether oblique asymptotes exist:

Yes, it is straight oblique asymptote graphics , if .

It makes no sense to analyze the limits, since it is already clear that the function embraces its oblique asymptote not limited from above And not limited from below.

The second research point yielded a lot of important information about the function. Let's do a rough sketch:

Conclusion No. 1 concerns intervals of constant sign. At “minus infinity” the graph of the function is clearly located below the x-axis, and at “plus infinity” it is above this axis. In addition, the one-sided limits told us that both to the left and to the right of the point the function is also greater than zero. Please note that in the left half-plane the graph must cross the x-axis at least once. There may not be any zeros of the function in the right half-plane.

Conclusion No. 2 is that the function increases on and to the left of the point (goes “from bottom to top”). To the right of this point, the function decreases (goes “from top to bottom”). The right branch of the graph must certainly have at least one minimum. On the left, extremes are not guaranteed.

Conclusion No. 3 provides reliable information about the concavity of the graph in the vicinity of the point. We cannot yet say anything about convexity/concavity at infinities, since a line can be pressed toward its asymptote both from above and from below. Generally speaking, there is an analytical way to figure this out right now, but the shape of the graph will become clearer at a later stage.

Why so many words? To control subsequent research points and avoid mistakes! Further calculations should not contradict the conclusions drawn.

3) Points of intersection of the graph with the coordinate axes, intervals of constant sign of the function.

The graph of the function does not intersect the axis.

Using the interval method we determine the signs:

, If ;
, If .

The results of this point are fully consistent with Conclusion No. 1. After each stage, look at the draft, mentally check the research and complete the graph of the function.

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

Actually, this has already been done when finding asymptotes.

- critical point.

Let's define the signs:

increases by and decreases by

At the point the function reaches a minimum: .

There were also no discrepancies with Conclusion No. 2, and, most likely, we are on the right track.

This means that the graph of the function is concave over the entire domain of definition.

Great - and you don’t need to draw anything.

There are no inflection points.

Concavity is consistent with Conclusion No. 3, moreover, it indicates that at infinity (both there and there) the graph of the function is located higher its oblique asymptote.

6) We will conscientiously pin the task with additional points. This is where we will have to work hard, since we only know two points from the research.

And a picture that many people probably imagined a long time ago:

During the execution of the task, you need to carefully ensure that there are no contradictions between the stages of the research, but sometimes the situation is urgent or even desperately dead-end. The analytics “doesn’t add up” - that’s all. In this case, I recommend an emergency technique: we find as many points as possible that belong to the graph (as much patience as we have), and mark them on the coordinate plane. A graphical analysis of the values ​​found will in most cases tell you where the truth is and where it is false. In addition, the graph can be pre-built using some program, for example, in Excel (of course, this requires skills).

Example 4

Use differential calculus methods to study a function and construct its graph.

This is an example for you to solve on your own. In it, self-control is enhanced by the parity of the function - the graph is symmetrical about the axis, and if there is something in your research that contradicts this fact, look for an error.

An even or odd function can be studied only at , and then use the symmetry of the graph. This solution is optimal, but, in my opinion, it looks very unusual. Personally, I look at the entire number line, but I still find additional points only on the right:

Example 5

Conduct a complete study of the function and construct its graph.

Solution: things got tough:

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

This means that this function is odd, its graph is symmetrical about the origin.

It is obvious that the function is non-periodic.

2) Asymptotes, the behavior of a function at infinity.

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

For a function containing an exponent, it is typical separate study of “plus” and “minus of infinity”, however, our life is made easier by the symmetry of the graph - either there is an asymptote on both the left and right, or there is none. Therefore, both infinite limits can be written under a single entry. During the solution we use L'Hopital's rule:

The straight line (axis) is the horizontal asymptote of the graph at .

Please note how I cunningly avoided the full algorithm for finding the oblique asymptote: the limit is completely legal and clarifies the behavior of the function at infinity, and the horizontal asymptote was discovered “as if at the same time.”

From the continuity on and the existence of a horizontal asymptote it follows that the function bounded above And bounded below.

3) Points of intersection of the graph with the coordinate axes, intervals of constant sign.

Here we also shorten the solution:
The graph passes through the origin.

There are no other points of intersection with the coordinate axes. Moreover, the intervals of constancy of sign are obvious, and the axis need not be drawn: , which means that the sign of the function depends only on “x”:
, If ;
, If .

4) Increasing, decreasing, extrema of the function.


– critical points.

The points are symmetrical about zero, as it should be.

Let us determine the signs of the derivative:


The function increases on an interval and decreases on intervals

At the point the function reaches its maximum: .

Due to the property (the oddity of the function) the minimum need not be calculated:

Since the function decreases over the interval, then, obviously, the graph is located at “minus infinity” under its asymptote. Over the interval, the function also decreases, but here the opposite is true - after passing through the maximum point, the line approaches the axis from above.

From the above it also follows that the graph of the function is convex at “minus infinity” and concave at “plus infinity”.

After this point of study, the range of function values ​​was drawn:

If you have any misunderstanding of any points, I once again urge you to draw coordinate axes in your notebook and, with a pencil in your hands, re-analyze each conclusion of the task.

5) Convexity, concavity, kinks of the graph.

– critical points.

The symmetry of the points is preserved, and, most likely, we are not mistaken.

Let's define the signs:


The graph of the function is convex on and concave on .

The convexity/concavity at the extreme intervals was confirmed.

At all critical points there are kinks in the graph. Let's find the ordinates of the inflection points, and again reduce the number of calculations using the oddness of the function: