If in the problem 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 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 the function, it is necessary to start with this step.

Example 1

Per 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 represented by 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, the asymptotes are the lines that the graph of the function approaches at infinity. It contributes fast building 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

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) - kx) = 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

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. You are welcome
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 everything doc files into a single pdf file (using the Acrobat program), and then transferred 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.

Reshebnik Kuznetsov.
III Graphs

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

        Before you start downloading your options, try solving the problem following the example 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) Function 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        .

        4) The function is continuous in the domain
.

; .

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

5) Vertical asymptotes:       

Find the oblique asymptote        . Here

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

        6) 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) Find the second derivative. Second derivative:
.
And this is easy to verify, since