Scheme of the study of functions using areas. Scheme for constructing a graph of a function Study of functions for an extremum using higher order derivatives Calculation of the roots of equations using the methods of chords and tangents

One of the possible schemes for studying the function and constructing its graph is decomposed into the following stages of solving the problem: 1. Function domain (O.O.F.). 2. Breakpoints of a function, their nature. Vertical asymptotes. 3. Even, odd, periodic function. 4. Points of intersection of the graph with the coordinate axes. 5. Behavior of the function at infinity. Horizontal and oblique asymptotes. 6. Intervals of monotonicity of a function, points of maximum and minimum. 7. Directions of the convexity of the curve. Inflection points. 8. Graph of the function. Example 1. Plot the function y \u003d 1. (vereiora or curl of Maria Anieei). - the entire numerical axis. 2. There are no break points; there are no vertical asymptotes. 3. The function is even: so that its graph is symmetrical about the Oy axis \ non-periodic. It follows from the parity of the function that it suffices to plot its graph on the half-line x ^ 0, and then mirror it in the y-axis. 4. At x = 0, we have Yx, so that the graph of the function lies in the upper half-plane y > 0. Scheme for constructing a graph of a function Investigation of functions for an extremum using higher-order derivatives Calculation of the roots of equations using chord and tangent methods that the graph has a horizontal asymptote y = O, there are no oblique asymptotes. So the function is increasing as and decreasing when. The point x = 0 is critical. When x passes through the point x \u003d 0, the derivative y "(x) changes sign from minus to plus. Therefore, the point x \u003d 0 is the maximum point, y (Q) \u003d I. This result is quite obvious: / (x) \u003d T ^ IV *. The second derivative vanishes at the points x \u003d. We study the point x \u003d 4- (hereinafter the symmetry argument). At we have. the curve is convex downwards; at we obtain (the curve is convex upwards). Therefore, the point x \u003d \u003d - is the inflection point graph of the function. The results of the study are summarized in a table: Inflection point max Inflection point - the entire real axis, excluding the point 2. The point of discontinuity of the function. So we have the straight line x = 0 - the vertical asymptote. 3. The function is neither even nor odd [function in general position), non-periodic. Assuming we get the graph of the function intersects the axis Ox at the point (-1,0), there are no oblique and horizontal asymptotes. where is the critical point. The second derivative of the function is at a point, so x = is the minimum point. The second derivative turns into uul at a point and changes its sign when passing through this point. Therefore, the point is the inflection point of the curve. For) we have e. the convexity of the curve is directed downward; for -I we have. the convexity of the curve is directed upwards. We summarize the results of the study in a table: Does not exist Does not exist Inflection point Does not exist. The vertical asymptote of the torus derivative vanishes at x = e,/2. and when x passes through this point, y "changes sign Therefore, is the abscissa of the inflection point of the curve. We summarize the results of the study in a table: Inflection point. The graph of the function is shown in Fig. 37. Example 4. Graph the function of the entire numerical axis, excluding the point Point point discontinuity of the 2nd kind of function.Since Km , then the direct vertical asymptote of the graph of the function.The function is in general position, non-periodic.Setting y = 0, we have, whence so that the graph of the function intersects the x-axis at the point Therefore, the graph of the function has an oblique asymptote From the condition we obtain - a critical point. The second derivative of the function y" \u003d D\u003e 0 everywhere in the domain of definition, in particular, at the point - the minimum point of the function. 7. Since, everywhere in the domain of definition of the function, the convexity of its graph is directed downwards. We summarize the results of the study in a table: Does not exist Does not exist Does not exist. x \u003d 0 - vertical asymptote The graph of the function is shown in fig. Example 5. Graph the function of the entire number axis. 2. Continuous everywhere. There are no vertical asymptotes. 3. General position, non-periodic. 4. The function vanishes at 5. Thus, the graph of the function has an oblique asymptote. The derivative vanishes at a point and does not exist at. When x passes through the point) the derivative does not change sign, so there is no extremum at the point x = 0. When the point x passes through the point, the derivative) changes sign from “+” to So, the function has a maximum. When x passes through the point x \u003d 3 (x\u003e I), the derivative y "(x) changes sign, i.e., at the point x \u003d 3, the function has a minimum. 7. Find the second derivative of higher order Calculation of the roots of equations by methods of chords and tangents The second derivative y "(x) does not exist at the point x = 0 and when x passes through the point x = 0 y" changes sign from + to so that the point (0,0) of the curve is a point there is no inflection point with a vertical tangent There is no inflection at the point x = 3. Everywhere in the half-plane x > 0 the convexity of the curve is directed upwards. in fig. 39. §7. Investigation of functions to an extremum using higher order derivatives To find the maximum and minimum points of functions, the Taylor formula can be used. Theorem It. Let the function f(x) in some neighborhood of the point xq have an n-th order derivative continuous at the point xo. Let 0. Then if the number n is odd, then the function f(x) at the point x0 has no extremum; when n is even, then at the point x0 the function f(x) has a maximum if f(n)(x0)< 0, и минимум, если /. В силу определения точек максимума и минимума вопрос о том, имеет ли функция f(x) в точке х0 экстремум, сводится к тому, существует ли такое <5 > 0, which is in the interval, the difference - /(x0) retains its sign. According to the Taylor formula as by condition, then from (1) we obtain 1-condition / (n * (r) is continuous at a point and Ф Therefore, due to the stability of a continuous function, there exists such that in the interval () does not change and coincides with the sign / (n) ( Let's consider the possible cases: 1) n is an even number and / Then I, therefore, by virtue of (2) . According to the definition, this means that the point o is the point of minimum of the function f(r). 2) n is even and. Then we will have i together with this and Therefore, the point i will be in this case the point of maximum of the function f(r). 3) n is an odd number, /- Then, for x > x0, the sign > will coincide with the sign of /(n)(ro), and for r, it will be opposite. Therefore, for arbitrarily small 0, the sign of the difference f(r) - f(r0) will not be the same for all x e (r0 - 6, r0 + t). Consequently, in this case the function f(r) does not have a stremum at the point th. Example. Let us consider the functions A. It is easy to see that the point x = 0 is a critical point of both functions. For the function y = x4, the first of the non-zero derivatives at the point x = 0 is the 4th order derivative: Thus, here n = 4 is an even u. Therefore, at the point x = 0, the function y = x4 has a minimum. For the function y = x), the first of the non-zero derivatives at the point x = 0 is the third-order derivative. So in this case n = 3 is odd, and at the point x = 0 the function y = x3 has no extremum. Comment. Using the Taylor formula, we can prove the following theorem, which expresses the sufficient conditions for the inflection point. "Theorem 12. Let the function /(r) in some neighborhood of the point r0 have a derivative of the n-th order, continuous at the point xq. Mo(x0, f(xo)) is the inflection point of the graph of the function y = f(x).The simplest example is provided by the function §8. Calculation of the roots of equations by the methods of chords and tangents The problem is to find the real root of the equation Suppose that the following conditions are satisfied: 1) the function f(x) is continuous on the segment [a, 6]; 2) the numbers /(a) and f(b) are opposite in sign: 3) on the segment [a, 6] there are derivatives f "(x) and f "(x) that preserve a constant sign on this interval. From conditions 1) and 2), by virtue of the Bolzano-Cauchy theorem (p. 220), it follows that the function f(x) vanishes at least at one point £ € ( a, b), i.e., equation (1) has at least one real root £ in the interval (a, b). sign, then f(x) is monotonic on [a, b] and, therefore, in int rvale (a, b) equation (1) has only one real root Let us consider a method for calculating the approximate value of this unique real root £ € (a, 6) of equation (I) with any degree of accuracy. Four cases are possible (Fig. 40): 1) Fig. 40 For definiteness, let us take the case when f \ x) > 0, f "(x) > 0 on the segment [a, 6) (Fig. 41). Let's connect the points A (a, / (a)) and B (b, f(b)) by a chord A B. This is a segment of a straight line passing through points A and B, the equation of which y \u003d 0, we find From Fig. 41 it is easy to see that the point a \ will always be located on the side from which the signs f (x) and f "(x) are opposite. Let us now draw a tangent to the curve y \u003d f (x) in point B(b, f(b)), i.e., at that end of the arc ^AB at which f(x) and /"(x) have the same sign. This is an essential condition: without it, the intersection point tangent to the x-axis may not give an approximation to the required root at all.The point b\, at which the tangent intersects the x-axis, is located between t and b on the same side as 6, and is a better approximation to than b. This tangent is determined by the equation Assuming in (3) y = 0, we find Functions Investigation of extremum functions with the help of higher order derivatives Calculation of the roots of equations by the methods of chords and tangents Thus, we have Let the absolute approximation error C of the root £ be given in advance. For the absolute error of the approximate values ​​of aj and 6, the root £, we can take the value |6i - ai|. If this error is greater than the permissible one, then, taking the segment as the original one, we find the following approximations of the root where. Continuing this process, we obtain two sequences of approximate values. The sequences (an) and (bn) are monotonic and bounded and, therefore, have limits. Let It can be shown that if the conditions formulated above are satisfied 1 is the only root of the equation / Example. Find the root (equations r2 - 1 = 0 on the segment. Thus, all the conditions that ensure the existence of a single root are met (equations x2 - 1 = 0 on the segment . and the method should work. 8 in our case a = 0, b = 2. When n \u003d I from (4) and (5) we find When n \u003d 2 we get what gives an approximation to the exact value of the root (with absolute error) using higher order derivatives: Answers

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 the 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

The process of researching a function consists of several stages. For the most complete idea of ​​the behavior of the function and the nature of its graph, it is necessary to find:

The scope of the function.

This concept includes both the domain of values ​​and the scope of a function.

Break points. (If they are available).

Intervals of increase and decrease.

High and low points.

The maximum and minimum value of a function on its domain.

Areas of convexity and concavity.

Inflection points. (If any).

Asymptotes. (If any).

Building a graph.

Let's use this scheme with an example.

Example. Investigate the function and plot its graph.

Find the area of ​​existence of the function. It's obvious that domain of definition function is the area (-; -1)  (-1; 1)  (1; ).

In turn, it can be seen that the lines x = 1, x = -1 are vertical asymptotes crooked.

Value area of this function is the interval (-; ).

break points functions are the points x=1, x=-1.

We find critical points.

Let's find the derivative of the function

Critical points: x = 0; x = -;x = ;x = -1; x = 1.

Let's find the second derivative of the function

Let us determine the convexity and concavity of the curve at the intervals.

- < x < -,y < 0, кривая выпуклая

-

1 < x < 0, y >0, curve concave

0 < x < 1, y < 0, кривая выпуклая

1 < x < ,y >0, curve concave

< x < , y >0, curve concave

Finding gaps increasing and descending functions. To do this, we determine the signs of the derivative of the function on the intervals.

- < x < -,y >0, the function is increasing

-

1 < x < 0, y < 0, функция убывает

0 < x < 1, y < 0, функция убывает

1 < x < ,y < 0, функция убывает

< x < , y >0, the function is increasing

It can be seen that the point x = - is a point maximum, and the point x = is the point minimum. The function values ​​at these points are 3/2 and -3/2, respectively.

About vertical asymptotes has already been said above. Now let's find oblique asymptotes.

So, the oblique asymptote equation is y = x.

Let's build schedule features:

Below we consider several examples of research by methods of differential calculus of various types of functions.

Example: Methods of differential calculus

1. The domain of this function is all real numbers (-; ).

3. Points of intersection with coordinate axes: with the Oy axis: x = 0; y=1;

with the Ox axis: y = 0; x = 1;

4. Discontinuity points and asymptotes: There are no vertical asymptotes.

Oblique asymptotes: general equation y = kx + b;

Total: y \u003d -x - oblique asymptote.

5. Increasing and decreasing functions, extremum points.

It can be seen that у 0 for any x  0, therefore, the function decreases over the entire domain of definition and has no extrema. At the point x = 0, the first derivative of the function is equal to zero, however, at this point, the decrease does not change to increase, therefore, at the point x = 0, the function most likely has an inflection. To find the inflection points, we find the second derivative of the function.

y = 0 for x = 0 and y =  for x = 1.

Points (0,1) and (1,0) are inflection points, because y(1-h)< 0; y(1+h) >0; y(-h) > 0; y(h)< 0 для любого h > 0.

6. Let's build a graph of the function.

Example: Investigate the function and plot its graph.

1. The scope of the function is all values ​​of x, except for x = 0.

2. The function is a function of general form in the sense of even and odd.

3. Points of intersection with the coordinate axes: with the Ox axis: y = 0; x=

with the Oy axis: x = 0; y does not exist.

4. The point x \u003d 0 is a discontinuity point, therefore, the line x \u003d 0 is a vertical asymptote.

Oblique asymptotes are looking for in the form: y = kx + b.

Oblique asymptote y = x.

5. Find the extremum points of the function.

; y = 0 at x = 2, y =  at x = 0.

y > 0 at x  (-, 0) - the function increases,

y< 0 при х  (0, 2) – функция убывает,

y > 0 at x  (2, ) – the function increases.

Thus, the point (2, 3) is the minimum point.

To determine the nature of the convexity / concavity of the function, we find the second derivative.

> 0 for any x  0, therefore, the function is concave over the entire domain of definition.

6. Let's build a graph of the function.

Example: Investigate the function and plot its graph.

The domain of this function is the interval x  (-, ).

In the sense of even and odd, the function is a function of general form.

Points of intersection with the coordinate axes: with the Oy axis: x = 0, y = 0;

with the Ox axis: y = 0, x = 0, x = 1.

Curve asymptotes.

There are no vertical asymptotes.

Let's try to find oblique asymptotes in the form y = kx + b.

- there are no oblique asymptotes.

Finding extremum points.

To find the critical points, you should solve the equation 4x 3 - 9x 2 + 6x -1 \u003d 0.

To do this, we decompose this polynomial of the third degree into factors.

Selection can determine that one of the roots of this equation is the number

x = 1. Then:

4x 3 – 9x 2 + 6x – 1 x - 1

 4x 3 – 4x 2 4x 2 – 5x + 1

Then we can write (x - 1) (4x 2 - 5x + 1) = 0. Finally, we get two critical points: x = 1 and x = ¼.

Note. The operation of dividing polynomials could be avoided if, when finding the derivative, use the formula for the derivative of the product:

Let's find the second derivative of the function: 12x 2 - 18x + 6. Equating to zero, we find:

We systematize the information received in the table:

Let's build a graph of the function.

Unfortunately, not all students and schoolchildren know and love algebra, but everyone has to prepare homework, solve tests and take exams. It is especially difficult for many to find tasks for plotting function graphs: if somewhere you don’t understand something, don’t finish it, miss it, mistakes are inevitable. But who wants to get bad grades?

Would you like to join the cohort of tailers and losers? To do this, you have 2 ways: sit down for textbooks and fill in the gaps in knowledge, or use a virtual assistant - a service for automatically plotting function graphs according to specified conditions. With or without decision. Today we will introduce you to a few of them.

The best thing about Desmos.com is a highly customizable interface, interactivity, the ability to spread the results into tables and store your work in the resource database for free without time limits. And the disadvantage is that the service is not fully translated into Russian.

Grafikus.ru

Grafikus.ru is another noteworthy Russian-language charting calculator. Moreover, he builds them not only in two-dimensional, but also in three-dimensional space.

Here is an incomplete list of tasks that this service successfully copes with:

• Drawing 2D graphs of simple functions: lines, parabolas, hyperbolas, trigonometric, logarithmic, etc.
• Drawing 2D-graphs of parametric functions: circles, spirals, Lissajous figures and others.
• Drawing 2D graphs in polar coordinates.
• Construction of 3D surfaces of simple functions.
• Construction of 3D surfaces of parametric functions.

The finished result opens in a separate window. The user has options to download, print and copy the link to it. For the latter, you will have to log in to the service through the buttons of social networks.

The Grafikus.ru coordinate plane supports changing the boundaries of the axes, their labels, the grid spacing, as well as the width and height of the plane itself and the font size.

The biggest strength of Grafikus.ru is the ability to create 3D graphs. Otherwise, it works no worse and no better than analogue resources.

Onlinecharts.ru

The Onlinecharts.ru online assistant does not build charts, but charts of almost all existing types. Including:

• Linear.
• Columnar.
• Circular.
• with areas.
• Radial.
• XY charts.
• Bubble.
• Point.
• Polar Bulls.
• Pyramids.
• Speedometers.
• Column-linear.

The resource is very easy to use. The appearance of the chart (background color, grid, lines, pointers, corner shape, fonts, transparency, special effects, etc.) is completely user-defined. Data for building can be entered either manually or imported from a table in a CSV file stored on a computer. The finished result is available for download on a PC as an image, PDF, CSV or SVG file, as well as for saving online on ImageShack.Us photo hosting or in your Onlinecharts.ru personal account. The first option can be used by everyone, the second - only registered ones.

The reference points in the study of functions and the construction of their graphs are characteristic points - points of discontinuity, extremum, inflection, intersection with the coordinate axes. With the help of differential calculus, it is possible to establish the characteristic features of the change in functions: increase and decrease, maxima and minima, the direction of the convexity and concavity of the graph, the presence of asymptotes.

A sketch of the function graph can (and should) be sketched after finding the asymptotes and extremum points, and it is convenient to fill in the summary table of the study of the function in the course of the study.

Usually, the following scheme of function research is used.

1.Find the domain, continuity intervals, and breakpoints of a function.

2.Examine the function for even or odd (axial or central symmetry of the graph.

3.Find asymptotes (vertical, horizontal or oblique).

4.Find and investigate the intervals of increase and decrease of the function, its extremum points.

5.Find the intervals of convexity and concavity of the curve, its inflection points.

6.Find the points of intersection of the curve with the coordinate axes, if they exist.

7.Compile a summary table of the study.

8.Build a graph, taking into account the study of the function, carried out according to the above points.

Example. Explore Function

and plot it.

7. Let's make a summary table of the study of the function, where we will enter all the characteristic points and the intervals between them. Given the parity of the function, we get the following table: