The study of the function is carried out according to a clear scheme and requires the student to have solid knowledge of basic mathematical concepts such as the domain of definition and values, the continuity of the function, the asymptote, extremum points, parity, periodicity, etc. The student must freely differentiate functions and solve equations, which are sometimes very intricate.

That is, this task tests a significant layer of knowledge, any gap in which will become an obstacle to obtaining the correct solution. Especially often difficulties arise with the construction of graphs of functions. This mistake immediately catches the eye of the teacher and can greatly ruin your grade, even if everything else was done correctly. Here you can find tasks for the study of the function online: study examples, download solutions, order assignments.

Investigate a Function and Plot: Examples and Solutions Online

We have prepared for you a lot of ready-made feature studies, both paid in the solution book, and free in the Feature Research Examples section. On the basis of these solved tasks, you will be able to get acquainted in detail with the methodology for performing such tasks, by analogy, perform your own research.

We offer ready-made examples of a complete study and plotting of functions of the most common types: polynomials, fractional-rational, irrational, exponential, logarithmic, trigonometric functions. Each solved problem is accompanied by a ready-made graph with selected key points, asymptotes, maxima and minima, the solution is carried out according to the algorithm for studying the function.

The solved examples, in any case, will be a good help for you, as they cover the most popular types of functions. We offer you hundreds of already solved problems, but, as you know, there are an infinite number of mathematical functions in the world, and teachers are great experts in inventing more and more intricate tasks for poor students. So, dear students, qualified assistance will not hurt you.

Solving problems for the study of a function to order

In this case, our partners will offer you another service - full function study online to order. The task will be completed for you in compliance with all the requirements for the algorithm for solving such problems, which will greatly please your teacher.

We will do a complete study of the function for you: we will find the domain of definition and the range of values, examine for continuity and discontinuity, set the parity, check your function for periodicity, find the points of intersection with the coordinate axes. And, of course, further with the help of differential calculus: we will find asymptotes, calculate extrema, inflection points, and build the graph itself.

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.

One of the most important tasks of differential calculus is the development of general examples of the study of the behavior of functions.

If the function y \u003d f (x) is continuous on the interval, and its derivative is positive or equal to 0 on the interval (a, b), then y \u003d f (x) increases by (f "(x) 0). If the function y \u003d f (x) is continuous on the segment , and its derivative is negative or equal to 0 on the interval (a,b), then y=f(x) decreases by (f"(x)0)

The intervals in which the function does not decrease or increase are called intervals of monotonicity of the function. The nature of the monotonicity of a function can change only at those points of its domain of definition, at which the sign of the first derivative changes. The points at which the first derivative of a function vanishes or breaks are called critical points.

Theorem 1 (1st sufficient condition for the existence of an extremum).

Let the function y=f(x) be defined at the point x 0 and let there be a neighborhood δ>0 such that the function is continuous on the segment , differentiable on the interval (x 0 -δ, x 0)u(x 0 , x 0 + δ) , and its derivative retains a constant sign on each of these intervals. Then if on x 0 -δ, x 0) and (x 0, x 0 + δ) the signs of the derivative are different, then x 0 is an extremum point, and if they match, then x 0 is not an extremum point. Moreover, if, when passing through the point x0, the derivative changes sign from plus to minus (to the left of x 0, f "(x)> 0 is performed, then x 0 is the maximum point; if the derivative changes sign from minus to plus (to the right of x 0 is executed by f"(x)<0, то х 0 - точка минимума.

The maximum and minimum points are called the extremum points of the function, and the maxima and minima of the function are called its extreme values.

Theorem 2 (necessary criterion for a local extremum).

If the function y=f(x) has an extremum at the current x=x 0, then either f'(x 0)=0 or f'(x 0) does not exist.
At the extremum points of a differentiable function, the tangent to its graph is parallel to the Ox axis.

Algorithm for studying a function for an extremum:

1) Find the derivative of the function.
2) Find critical points, i.e. points where the function is continuous and the derivative is zero or does not exist.
3) Consider the neighborhood of each of the points, and examine the sign of the derivative to the left and right of this point.
4) Determine the coordinates of the extreme points, for this value of the critical points, substitute into this function. Using sufficient extremum conditions, draw appropriate conclusions.

Example 18. Investigate the function y=x 3 -9x 2 +24x

Solution.
1) y"=3x 2 -18x+24=3(x-2)(x-4).
2) Equating the derivative to zero, we find x 1 =2, x 2 =4. In this case, the derivative is defined everywhere; hence, apart from the two found points, there are no other critical points.
3) The sign of the derivative y "=3(x-2)(x-4) changes depending on the interval as shown in Figure 1. When passing through the point x=2, the derivative changes sign from plus to minus, and when passing through the point x=4 - from minus to plus.
4) At the point x=2, the function has a maximum y max =20, and at the point x=4 - a minimum y min =16.

Theorem 3. (2nd sufficient condition for the existence of an extremum).

Let f "(x 0) and f "" (x 0) exist at the point x 0. Then if f "" (x 0)> 0, then x 0 is the minimum point, and if f "" (x 0)<0, то х 0 – точка максимума функции y=f(x).

On the segment, the function y \u003d f (x) can reach the smallest (at least) or largest (at most) value either at the critical points of the function lying in the interval (a; b), or at the ends of the segment.

The algorithm for finding the largest and smallest values ​​of a continuous function y=f(x) on the segment :

1) Find f "(x).
2) Find the points at which f "(x) = 0 or f" (x) - does not exist, and select from them those that lie inside the segment.
3) Calculate the value of the function y \u003d f (x) at the points obtained in paragraph 2), as well as at the ends of the segment and choose the largest and smallest of them: they are, respectively, the largest (for the largest) and the smallest (for the smallest) function values ​​on the segment .

Example 19. Find the largest value of a continuous function y=x 3 -3x 2 -45+225 on the segment .

1) We have y "=3x 2 -6x-45 on the segment
2) The derivative y" exists for all x. Let's find the points where y"=0; we get:
3x2 -6x-45=0
x 2 -2x-15=0
x 1 \u003d -3; x2=5
3) Calculate the value of the function at the points x=0 y=225, x=5 y=50, x=6 y=63
Only the point x=5 belongs to the segment. The largest of the found values ​​of the function is 225, and the smallest is the number 50. So, at max = 225, at max = 50.

Investigation of a function on convexity

The figure shows the graphs of two functions. The first of them is turned with a bulge up, the second - with a bulge down.

The function y=f(x) is continuous on a segment and differentiable in the interval (a;b), is called convex up (down) on this segment if, for axb, its graph lies no higher (not lower) than the tangent drawn at any point M 0 (x 0 ;f(x 0)), where axb.

Theorem 4. Let the function y=f(x) have a second derivative at any interior point x of the segment and be continuous at the ends of this segment. Then if the inequality f""(x)0 is satisfied on the interval (a;b), then the function is downward convex on the segment ; if the inequality f""(x)0 is satisfied on the interval (а;b), then the function is convex upward on .

Theorem 5. If the function y=f(x) has a second derivative on the interval (a;b) and if it changes sign when passing through the point x 0 , then M(x 0 ;f(x 0)) is an inflection point.

Rule for finding inflection points:

1) Find points where f""(x) does not exist or vanishes.
2) Examine the sign f""(x) to the left and right of each point found at the first step.
3) Based on Theorem 4, draw a conclusion.

Example 20. Find extremum points and inflection points of the function graph y=3x 4 -8x 3 +6x 2 +12.

We have f"(x)=12x 3 -24x 2 +12x=12x(x-1) 2. Obviously, f"(x)=0 for x 1 =0, x 2 =1. The derivative, when passing through the point x=0, changes sign from minus to plus, and when passing through the point x=1, it does not change sign. This means that x=0 is the minimum point (y min =12), and there is no extremum at the point x=1. Next, we find . The second derivative vanishes at the points x 1 =1, x 2 =1/3. The signs of the second derivative change as follows: On the ray (-∞;) we have f""(x)>0, on the interval (;1) we have f""(x)<0, на луче (1;+∞) имеем f""(x)>0. Therefore, x= is the inflection point of the function graph (transition from convexity down to convexity up) and x=1 is also an inflection point (transition from convexity up to convexity down). If x=, then y= ; if, then x=1, y=13.

An algorithm for finding the asymptote of a graph

I. If y=f(x) as x → a , then x=a is a vertical asymptote.
II. If y=f(x) as x → ∞ or x → -∞ then y=A is the horizontal asymptote.
III. To find the oblique asymptote, we use the following algorithm:
1) Calculate . If the limit exists and is equal to b, then y=b is the horizontal asymptote; if , then go to the second step.
2) Calculate . If this limit does not exist, then there is no asymptote; if it exists and is equal to k, then go to the third step.
3) Calculate . If this limit does not exist, then there is no asymptote; if it exists and is equal to b, then go to the fourth step.
4) Write down the equation of the oblique asymptote y=kx+b.

Example 21: Find an asymptote for a function

1)
2)
3)
4) The oblique asymptote equation has the form

The scheme of the study of the function and the construction of its graph

I. Find the domain of the function.
II. Find the points of intersection of the graph of the function with the coordinate axes.
III. Find asymptotes.
IV. Find points of possible extremum.
V. Find critical points.
VI. Using the auxiliary drawing, investigate the sign of the first and second derivatives. Determine the areas of increase and decrease of the function, find the direction of the convexity of the graph, extremum points and inflection points.
VII. Build a graph, taking into account the study conducted in paragraphs 1-6.

Example 22: Plot a function graph according to the above scheme

Solution.
I. The domain of the function is the set of all real numbers, except for x=1.
II. Since the equation x 2 +1=0 does not have real roots, then the graph of the function does not have points of intersection with the Ox axis, but intersects the Oy axis at the point (0; -1).
III. Let us clarify the question of the existence of asymptotes. We investigate the behavior of the function near the discontinuity point x=1. Since y → ∞ for x → -∞, y → +∞ for x → 1+, then the line x=1 is a vertical asymptote of the graph of the function.
If x → +∞(x → -∞), then y → +∞(y → -∞); therefore, the graph does not have a horizontal asymptote. Further, from the existence of limits

Solving the equation x 2 -2x-1=0, we get two points of a possible extremum:
x 1 =1-√2 and x 2 =1+√2

V. To find the critical points, we calculate the second derivative:

Since f""(x) does not vanish, there are no critical points.
VI. We investigate the sign of the first and second derivatives. Possible extremum points to be considered: x 1 =1-√2 and x 2 =1+√2, divide the area of ​​existence of the function into intervals (-∞;1-√2),(1-√2;1+√2) and (1+√2;+∞).

In each of these intervals, the derivative retains its sign: in the first - plus, in the second - minus, in the third - plus. The sequence of signs of the first derivative will be written as follows: +, -, +.
We get that the function on (-∞;1-√2) increases, on (1-√2;1+√2) it decreases, and on (1+√2;+∞) it increases again. Extremum points: maximum at x=1-√2, moreover f(1-√2)=2-2√2 minimum at x=1+√2, moreover f(1+√2)=2+2√2. On (-∞;1) the graph is convex upwards, and on (1;+∞) - downwards.
VII Let's make a table of the obtained values

VIII Based on the data obtained, we build a sketch of the graph of the function

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Conduct a complete study and plot a function graph

y(x)=x2+81−x.y(x)=x2+81−x.

1) Function scope. Since the function is a fraction, you need to find the zeros of the denominator.

1−x=0,⇒x=1.1−x=0,⇒x=1.

We exclude the only point x=1x=1 from the function definition area and get:

D(y)=(−∞;1)∪(1;+∞).D(y)=(−∞;1)∪(1;+∞).

2) Let us study the behavior of the function in the vicinity of the discontinuity point. Find one-sided limits:

Since the limits are equal to infinity, the point x=1x=1 is a discontinuity of the second kind, the line x=1x=1 is a vertical asymptote.

3) Let's determine the intersection points of the graph of the function with the coordinate axes.

Let's find the points of intersection with the ordinate axis OyOy, for which we equate x=0x=0:

Thus, the point of intersection with the axis OyOy has coordinates (0;8)(0;8).

Let's find the points of intersection with the abscissa axis OxOx, for which we set y=0y=0:

The equation has no roots, so there are no points of intersection with the OxOx axis.

Note that x2+8>0x2+8>0 for any xx. Therefore, for x∈(−∞;1)x∈(−∞;1), the function y>0y>0 (takes positive values, the graph is above the x-axis), for x∈(1;+∞)x∈(1; +∞) function y<0y<0 (принимает отрицательные значения, график находится ниже оси абсцисс).

4) The function is neither even nor odd because:

5) We investigate the function for periodicity. The function is not periodic, as it is a fractional rational function.

6) We investigate the function for extremums and monotonicity. To do this, we find the first derivative of the function:

Let us equate the first derivative to zero and find the stationary points (at which y′=0y′=0):

We got three critical points: x=−2,x=1,x=4x=−2,x=1,x=4. We divide the entire domain of the function into intervals by given points and determine the signs of the derivative in each interval:

For x∈(−∞;−2),(4;+∞)x∈(−∞;−2),(4;+∞) the derivative y′<0y′<0, поэтому функция убывает на данных промежутках.

For x∈(−2;1),(1;4)x∈(−2;1),(1;4) the derivative y′>0y′>0, the function increases on these intervals.

In this case, x=−2x=−2 is a local minimum point (the function decreases and then increases), x=4x=4 is a local maximum point (the function increases and then decreases).

Let's find the values ​​of the function at these points:

Thus, the minimum point is (−2;4)(−2;4), the maximum point is (4;−8)(4;−8).

7) We examine the function for kinks and convexity. Let's find the second derivative of the function:

Equate the second derivative to zero:

The resulting equation has no roots, so there are no inflection points. Moreover, when x∈(−∞;1)x∈(−∞;1) y′′>0y″>0 is satisfied, that is, the function is concave when x∈(1;+∞)x∈(1;+ ∞) y′′<0y″<0, то есть функция выпуклая.

8) We investigate the behavior of the function at infinity, that is, at .

Since the limits are infinite, there are no horizontal asymptotes.

Let's try to determine oblique asymptotes of the form y=kx+by=kx+b. We calculate the values ​​of k,bk,b according to the known formulas:

We found that the function has one oblique asymptote y=−x−1y=−x−1.

9) Additional points. Let's calculate the value of the function at some other points in order to build a graph more accurately.

y(−5)=5.5;y(2)=−12;y(7)=−9.5.y(−5)=5.5;y(2)=−12;y(7)=−9.5.

10) Based on the data obtained, we will build a graph, supplement it with asymptotes x=1x=1 (blue), y=−x−1y=−x−1 (green) and mark the characteristic points (the intersection with the y-axis is purple, extrema are orange, additional points are black) :

Task 4: Geometric, Economic problems (I have no idea what, here is an approximate selection of problems with a solution and formulas)

Example 3.23. a

Solution. x and y y
y \u003d a - 2 × a / 4 \u003d a / 2. Since x = a/4 is the only critical point, let's check whether the sign of the derivative changes when passing through this point. For xa/4 S "> 0, and for x >a/4 S "< 0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед).Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет наибольшим значением функции. Таким образом, наиболее выгодным соотношением сторон площадки при данных условиях задачи является y = 2x.

Example 3.24.

Solution.
R = 2, H = 16/4 = 4.

Example 3.22. Find the extrema of the function f(x) = 2x 3 - 15x 2 + 36x - 14.

Solution. Since f "(x) \u003d 6x 2 - 30x +36 \u003d 6 (x - 2) (x - 3), then the critical points of the function x 1 \u003d 2 and x 2 \u003d 3. Extreme points can only be at these points. So as when passing through the point x 1 \u003d 2, the derivative changes sign from plus to minus, then at this point the function has a maximum.When passing through the point x 2 \u003d 3, the derivative changes sign from minus to plus, therefore, at the point x 2 \u003d 3, the function has a minimum. Calculating the values ​​of the function in points
x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f(2) = 14 and minimum f(3) = 13.

Example 3.23. It is necessary to build a rectangular area near the stone wall so that it is fenced off with wire mesh on three sides, and adjoins the wall on the fourth side. For this there is a linear meters of the grid. At what aspect ratio will the site have the largest area?

Solution. Denote the sides of the site through x and y. The area of ​​the site is S = xy. Let y is the length of the side adjacent to the wall. Then, by condition, the equality 2x + y = a must hold. Therefore y = a - 2x and S = x(a - 2x), where
0 ≤ x ≤ a/2 (the length and width of the area cannot be negative). S "= a - 4x, a - 4x = 0 for x = a/4, whence
y \u003d a - 2 × a / 4 \u003d a / 2. Since x = a/4 is the only critical point, let's check whether the sign of the derivative changes when passing through this point. For xa/4 S "> 0, and for x >a/4 S "< 0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед).Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет наибольшим значением функции. Таким образом, наиболее выгодным соотношением сторон площадки при данных условиях задачи является y = 2x.

Example 3.24. It is required to make a closed cylindrical tank with a capacity of V=16p ≈ 50 m 3 . What should be the dimensions of the tank (radius R and height H) in order to use the least amount of material for its manufacture?

Solution. The total surface area of ​​the cylinder is S = 2pR(R+H). We know the volume of the cylinder V = pR 2 H Þ H = V/pR 2 =16p/ pR 2 = 16/ R 2 . Hence, S(R) = 2p(R 2 +16/R). We find the derivative of this function:
S "(R) \u003d 2p (2R- 16 / R 2) \u003d 4p (R- 8 / R 2). S " (R) \u003d 0 for R 3 \u003d 8, therefore,
R = 2, H = 16/4 = 4.

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