Tasks from the collection Kuznetsov L.A. Complete study of a function and plotting Y x 2 x 1 investigate a function

Rehebnik Kuznetsov.
III Charts

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

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

& nbsp & nbsp & nbsp & nbsp 7.3 Conduct a complete study of the function and plot its graph

Decision.

& nbsp & nbsp & nbsp & nbsp 1) Scope: & nbsp & nbsp & nbsp & nbsp or & nbsp & nbsp & nbsp & nbsp, i.e. & nbsp & nbsp & nbsp & nbsp.
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Thus: & nbsp & nbsp & nbsp & nbsp.

& nbsp & nbsp & nbsp & nbsp 2) There are no intersections with the Ox axis. Indeed, the equation & nbsp & nbsp & nbsp & nbsp has no solutions.
There are no intersections with the Oy axis since & nbsp & nbsp & nbsp & nbsp.

& nbsp & nbsp & nbsp & nbsp 3) The function is neither even nor odd. There is no symmetry about the ordinate. There is no symmetry about the origin either. As
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We see that & nbsp & nbsp & nbsp & nbsp and & nbsp & nbsp & nbsp & nbsp.

& nbsp & nbsp & nbsp & nbsp 4) The function is continuous in the domain
.

; .

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Therefore, the point & nbsp & nbsp & nbsp & nbsp is a breakpoint of the second kind (infinite break).

5) Vertical asymptotes: & nbsp & nbsp & nbsp & nbsp

Find the oblique asymptote & nbsp & nbsp & nbsp & nbsp. Here

;
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Therefore, we have a horizontal asymptote: y \u003d 0... There are no oblique asymptotes.

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

& nbsp & nbsp & nbsp & nbsp 7) Find the second derivative. Second derivative:
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And this is easy to be convinced of, since

For some time now, the built-in database of certificates for SSL in TheBat (for some reason) ceases to work correctly.

When checking posts, 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 there is a choice of answers - YES / NO. And so every time you take off your mail.

Decision

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

Since I needed to combine all the files into one, I first converted all doc files into a single pdf file (using the Acrobat program), and then converted them to fb2 through an online converter. You can also convert files separately. 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/en/collage/. Use it to your health. And I will use it myself.

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

The answer was simple. You don't need to measure anything, you can calmly determine what size you need.

Smallest burneris 145 millimeters (14.5 centimeters)

Medium hotplate 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 it, I was soaring with 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 task. I think that I am not the only one.

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

Finding the scope

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

Example 1

The given example assumes finding the zeros of the denominator in order to exclude them from the ODZ.

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 ODV can be sought 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)\u003e 0.

Investigation of the boundaries of ODZ and finding the 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 border points equal to x \u003d ± 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) \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 \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 + ∞

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

Examining a function and for even or odd parity

When the condition y (- x) \u003d y (x) is satisfied, the function is considered even. This suggests that the graph is located symmetrically about O y. When the condition y (- x) \u003d - y (x) is satisfied, the function is considered odd. This means that the symmetry is relative to the origin. If at least one inequality is not satisfied, we obtain a general function.

The equality y (- x) \u003d y (x) indicates that the function is even. When constructing, it is necessary to take into account that there will be symmetry about O y.

To solve the inequality, the 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 interior points from the domain, where the derivative of the function is zero or does not exist.

When deciding, it is necessary to take into account the following notes:

  • with the available intervals of increase and decrease of inequalities of the form f "(x)\u003e 0, critical points are not included in the solution;
  • the points at which the function is defined without a finite derivative must be included in the intervals of increasing and decreasing (for example, y \u003d x 3, where the point x \u003d 0 makes the function definite, 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 controversy, it is recommended to use mathematical literature recommended by the Ministry of Education.

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

  • derivative;
  • critical points;
  • split the definition area using 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) \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 zero at x \u003d ± 1 2.

We expose points on the numerical axis to determine the derivative at each interval. To do this, it is enough to take any point from the interval and perform the calculation. If the result is positive, we plot + 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:

  • 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; + ∞.

On the diagram, using + and - depicts the positivity and negativity of the function, and the arrows - decreasing and increasing.

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 equal to 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 a maximum point. When the sign changes from - to +, we get a minimum point.

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

Definition 3

For determining the intervals of concavity and convexity it is necessary:

  • find the second derivative;
  • find the zeros of the second derivative function;
  • split the definition area by the appeared points into intervals;
  • determine the sign of the gap.

Example 5

Find the second derivative from the domain.

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

Now you need to plot points on the numerical axis 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 Is a point of the form x 0; f (x 0). When it has a tangent to the graph of a function, then when it passes through x 0, the function changes its 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 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 \u003d ± 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 asymptotesare depicted using the straight lines defined 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.

For k \u003d 0 and b not equal to infinity, we find that the oblique asymptote becomes horizontal.

In other words, the asymptotes are the lines to which the graph of the function approaches at infinity. This helps to quickly plot 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

For 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

is the horizontal asymptote. After examining 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 \u200b\u200bof the function at intermediate points.

Example 7

From the example we have considered, it is necessary to find the values \u200b\u200bof 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 \u200b\u200bcoincide with the values \u200b\u200bat these points, that is, we get x \u003d 2, x \u003d 1, x \u003d 3 4, x \u003d 1 4.

Let's write down and solve:

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 maxima and minima of a function, inflection points, intermediate points, it is necessary to plot asymptotes. For convenient designation, the intervals of increasing, decreasing, convexity, concavity are fixed. Consider in the figure below.

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

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

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