For some time now, in TheBat (it is not clear for what reason), the built-in certificate database for SSL has ceased to work correctly.

When checking the post, an error pops up:

Unknown CA certificate
The server did not present a root certificate in the session and the corresponding root certificate was not found in the address book.
This connection cannot be secret. Please
contact your server administrator.

And it is offered a choice of answers - YES / NO. And so every time you shoot mail.

Solution

In this case, you need to replace the S/MIME and TLS implementation standard with Microsoft CryptoAPI in TheBat!

Since I needed to merge all the files into one, I first converted all doc files into a single pdf file (using the Acrobat program), and then transferred it to fb2 through an online converter. You can also convert files individually. Formats can be absolutely any (source) and doc, and jpg, and even zip archive!

The name of the site corresponds to the essence:) Online Photoshop.

Update May 2015

I found another great site! Even more convenient and functional for creating a completely arbitrary collage! This site is http://www.fotor.com/ru/collage/ . Use on health. And I will use it myself.

Faced in life with the repair of electric stoves. I already did a lot of things, learned a lot, but somehow I had little to do with tiles. It was necessary to replace the contacts on the regulators and burners. The question arose - how to determine the diameter of the burner on the electric stove?

The answer turned out to be simple. No need to measure anything, you can calmly determine by eye what size you need.

The smallest burner is 145 millimeters (14.5 centimeters)

Medium burner is 180 millimeters (18 centimeters).

And finally the most large burner is 225 millimeters (22.5 centimeters).

It is enough to determine the size by eye and understand what diameter you need a burner. When I didn’t know this, I was soaring with these sizes, I didn’t know how to measure, which edge to navigate, etc. Now I'm wise :) I hope it helped you too!

In my life I faced such a problem. I think I'm not the only one.

If in the task it is necessary to carry out a complete study of the function f (x) \u003d x 2 4 x 2 - 1 with the construction of its graph, then we will consider this principle in detail.

To solve a problem of this type, one should use the properties and graphs of the main elementary functions. The research algorithm includes the following steps:

Finding the domain of definition

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

Example 1

The given example involves finding the zeros of the denominator in order to exclude them from the DPV.

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 the 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 .

Investigation of ODZ boundaries and finding vertical asymptotes

There are vertical asymptotes on 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 that the lines x = ± 1 2 are the vertical asymptotes of the graph.

Investigation of the function and for even or odd

When the condition y (- x) = y (x) is met, the function is considered to be even. This suggests that the graph is located symmetrically with respect to O y. When the condition y (- x) = - y (x) is met, the function is considered odd. This means that the symmetry goes with respect to the origin of coordinates. If at least one inequality fails, we obtain a function of general form.

The fulfillment of 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 O y.

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

Definition 1

Stationary points are points that turn the derivative to zero.

Critical points are interior points from the domain where the derivative of the function is equal to zero or does not exist.

When making a decision, the following points should be taken into account:

for the existing intervals of increase and decrease of the inequality of the form f "(x) > 0, the 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 increase and decrease (for example, y \u003d x 3, where the 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) = 1 0 = ∞ , x = 0 is included in the increase 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 in the intervals of increasing and decreasing in the event that they satisfy the domain of the function.

Definition 2

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

derivative;
critical points;
break the domain of definition with the help of critical points into intervals;
determine the sign of the derivative at each of the intervals, where + is an increase and - is a decrease.

Example 3

Find the derivative on the domain 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 expose points on the numerical axis to determine the derivative on each interval. To do this, it is enough to take any point from the interval and make a calculation. If the result is positive, we draw + on the graph, which means an increase in the function, and - means its decrease.

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

Answer:

there is an increase in the function on the interval - ∞ ; - 1 2 and (- 1 2 ; 0 ] ;
there is a decrease on 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 decreasing and increasing.

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

Example 4

If we consider an example where x \u003d 0, then the value of the function in it is f (0) \u003d 0 2 4 0 2 - 1 \u003d 0. When the sign of the derivative changes from + to - and passes through the point x \u003d 0, then the point with coordinates (0; 0) is considered the maximum point. When the sign is changed from - to +, we get the minimum point.

Convexity and concavity are determined by solving inequalities of the form f "" (x) ≥ 0 and f "" (x) ≤ 0 . Less often they use the name bulge down instead of concavity, and bulge up instead of bulge.

Definition 3

For determining the gaps of concavity and convexity necessary:

find the second derivative;
find the zeros of the function of the second derivative;
break the domain of definition by the points that appear into intervals;
determine the sign of the gap.

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, using our example, we have that the zeros of the denominator x = ± 1 2

Now you need to put 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 gaps - ∞ ; - 1 2 and 1 2 ; +∞ .

Definition 4

inflection point is a point of the form x 0 ; f(x0) . 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 such a point through which the second derivative passes and changes sign, and at the points themselves is equal to zero or does not exist. All points are considered to be the domain of the function.

In the example, it was seen 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 domain of definition.

Finding horizontal and oblique asymptotes

When defining a function at infinity, one must look for horizontal and oblique asymptotes.

Definition 5

Oblique asymptotes are drawn using 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, the asymptotes are the lines that the graph of the function approaches at infinity. This contributes to the rapid construction of the graph of the function.

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

As an example, consider 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 researching the function, you can start building it.

Calculating the value of a function at intermediate points

To make the plotting the most accurate, it is recommended to find several values of the function at intermediate points.

Example 7

From the example we have considered, it is necessary to find the values of the function at the points x \u003d - 2, x \u003d - 1, x \u003d - 3 4, x \u003d - 1 4. Since the function is even, we get that the values coincide with the values at these points, that is, we get x \u003d 2, x \u003d 1, x \u003d 3 4, x \u003d 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, intermediate points, it is necessary to build asymptotes. For convenient designation, intervals of increase, decrease, convexity, concavity are fixed. Consider the figure below.

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

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

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

How to investigate a function and plot its graph?

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

Investigate the function by methods of differential calculus and, based on the results of the study, build its graph

Or in short: examine the function and plot it.

Why explore? In simple cases, it will not be difficult for us to deal with elementary functions, draw a graph obtained using elementary geometric transformations etc. However, the properties and graphic 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 section guide. Dummies need a step-by-step explanation of the topic, some readers don't know where to start and how to organize the study, and advanced students may be interested in only a few points. But whoever you are, dear visitor, the proposed summary with pointers to various lessons will orient and direct you in the direction of interest in the shortest possible time. The robots shed a tear =) The manual was made up in the form of a pdf file and took its rightful place on the page Mathematical formulas and tables.

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

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

As for the final action, I think everyone understands everything - 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 very likely to "cover up" analytical oversights, while an incorrect and/or sloppy schedule will cause problems even with a perfectly conducted study.

It should be noted that in other sources, the number of research items, the order of their implementation and the design style may differ significantly from the scheme proposed by me, but in most cases it is quite enough. The simplest version of the problem consists of only 2-3 steps and is formulated something like this: “explore the function using the derivative and plot” or “explore the function using the 1st and 2nd derivative, plot”.

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

Let's check the function for even / odd:

This is followed by a template unsubscribe:
, so this function is neither even nor 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 order of growth than , so the final limit is exactly " a 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, 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 do not have break points, it becomes clear and function range: is also any real number.

USEFUL TECHNIQUE

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

Note that in effect 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, find the intersection point of the graph with the y-axis. It's simple. It is necessary to calculate the value of the function when:

Half above sea level.

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

At the end, a free member lurks, which significantly complicates the task.

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

It's lucky here. In case of failure, you can also test and, and if these numbers do not fit, then I'm afraid there are very few chances for a profitable solution to the equation. Then it is better to skip the research point completely - maybe something will become clearer at the final step, when additional points will break through. And if the root (roots) are clearly “bad”, then it is better to remain modestly silent about the intervals of constancy of signs and to more accurately complete the drawing.

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 expands into a product:

And now a little about a healthy lifestyle. Of course I understand that quadratic equations need to be solved every day, but today we will make an exception: the equation has two real roots.

On the number line, we plot the found values and interval method define the signs of the function:

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

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

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

4) Increasing, decreasing and extrema of the function.

Let's find the 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 its minimum: .

The established facts drive our template into a rather rigid framework:

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

5) Convexity, concavity and inflection points.

Find the critical points of the second derivative:

Let's define signs:

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

Almost everything cleared up.

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

Let's execute 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 exactly in the middle between the maximum and minimum.

In the course of the assignment, I gave three hypothetical intermediate drawings. In practice, it is enough to draw a coordinate system, mark the points found, and after each point of the study, mentally figure out 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 minds without involving a draft.

For a standalone solution:

Example 2

Explore the function and build a graph.

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

A lot of secrets are revealed by the study of fractional rational functions:

Example 3

Using the methods of differential calculus, investigate the function and, based on the results of the study, construct its graph.

Solution: the first stage of the study does not differ in 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 for the point , domain: .

, so this function is neither even nor odd.

Obviously, the function is non-periodic.

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

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

a) With the help of one-sided limits, we study the behavior of the function near the suspicious point, where the vertical asymptote must clearly be:

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

b) Check if oblique asymptotes exist:

Yes, the line is oblique asymptote graphics if .

It makes no sense to analyze the limits, since it is already clear that the function in an embrace with its oblique asymptote not limited from above and not limited from below.

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

Conclusion No. 1 concerns intervals of sign constancy. At "minus infinity" the graph of the function is uniquely located below the x-axis, and at "plus infinity" it is above this axis. In addition, 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. In the right half-plane, there may be no zeros of the function.

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 gives reliable information about the concavity of the graph in the vicinity of the point. We cannot yet say anything about convexity/concavity at infinity, since the line can be pressed against 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 chart "for nothing" 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 cross the axis.

Using the interval method, we determine the signs:

, if ;
, if .

The results of the paragraph are fully consistent with Conclusion No. 1. After each step, look at the draft, mentally refer to the study, and finish drawing the graph of the function.

In this example, 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 signs:

increases by and decreases to

At the point the function reaches its 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.

Excellent - and you don't need to draw anything.

There are no inflection points.

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

6) We will conscientiously pin the task with additional points. Here we have to work hard, because we know only two points from the study.

And a picture that, probably, many have long submitted:

In the course of the assignment, care must be taken to ensure that there are no contradictions between the stages of the study, but sometimes the situation is urgent or even desperately dead-end. Here the analytics "does not converge" - and that's it. In this case, I recommend an emergency technique: we find as many points belonging to the graph as possible (how much patience is enough), and mark them on the coordinate plane. Graphical analysis of the found values in most cases will tell you where is the truth and where is the lie. In addition, the graph can be pre-built using some program, for example, in the same Excel (it is clear that this requires skills).

Example 4

Using the methods of differential calculus, investigate the function and build its graph.

This is a do-it-yourself example. In it, self-control is enhanced by the evenness of the function - the graph is symmetrical about the axis, and if something in your study contradicts this fact, look for an error.

An even or odd function can only be investigated for , and then the symmetry of the graph can be used. This solution is optimal, but it looks, in my opinion, very unusual. Personally, I consider the entire numerical axis, but I still find additional points only on the right:

Example 5

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

Solution: rushed hard:

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

This means that this function is odd, its graph is symmetrical with respect to the origin.

Obviously, 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, typically separate the study of "plus" and "minus infinity", however, our life is facilitated just by the symmetry of the graph - either there is an asymptote on the left and on the right, or it is not. Therefore, both infinite limits can be arranged under a single entry. In the course of the solution, we use L'Hopital's rule:

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

Pay attention to how I cleverly avoided the full algorithm for finding the oblique asymptote: the limit is quite legal and clarifies the behavior of the function at infinity, and the horizontal asymptote was found "as if at the same time."

It follows from the continuity on and the existence of a horizontal asymptote that the function limited from above and limited from below.

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

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 are obvious, and the axis can not be drawn: , which means that the sign of the function depends only on the "x":
, if ;
, if .

4) Increasing, decreasing, extrema of the function.

are critical points.

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

Let's define the signs of the derivative:

The function increases on the interval and decreases on the intervals

At the point the function reaches its maximum: .

Due to the property (oddity of the function) the minimum can be omitted:

Since the function decreases on the interval , then, obviously, the graph is located at "minus infinity" under with its asymptote. On 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.

It also follows from the above that the function graph is convex at "minus infinity" and concave at "plus infinity".

After this point of the study, the area of \u200b\u200bvalues of the function was also drawn:

If you have a 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, inflections of the graph.

are critical points.

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

Let's define signs:

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

Convexity/concavity at extreme intervals was confirmed.

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

Reshebnik Kuznetsov.
III Graphs

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

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

7.3 Conduct a complete study of the function and plot it

Solution.

1) Scope: or i.e. .
.
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 because .

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