Problems from the collection of L. A. Kuznetsova

For some time now, TheBat's built-in certificate database for SSL has stopped working correctly (it is not clear for what reason).

When checking the post, an error appears:

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 you are offered a choice of answers - YES / NO. And so every time you remove mail.

Solution

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

Since I needed to combine all the files into one, I first converted everything 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. The formats can be absolutely any (source) - doc, jpg, and even a 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 custom collage! This is the site http://www.fotor.com/ru/collage/. Enjoy it for your health. And I will use it myself.

In my life I came across the problem of repairing an electric stove. I’ve already done a lot of things, learned a lot, but somehow 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 an electric stove?

The answer turned out to be simple. You don’t need to measure anything, you can easily determine by eye what size you need.

Smallest burner- this is 145 millimeters (14.5 centimeters)

Middle burner- this is 180 millimeters (18 centimeters).

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

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

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

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 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 definition 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 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, 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

Solver Kuznetsov.
III Charts

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

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

        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

;
.
Therefore, we have horizontal asymptote: y=0. Oblique asymptotes No.

        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