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 \u003d 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

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 highest value functions. Thus, the most favorable aspect ratio of the site under the given conditions of the problem is 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 platform have 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.

Similar information.

Instruction

Find the scope of the function. For example, the function sin(x) is defined on the entire interval from -∞ to +∞, and the function 1/x is defined from -∞ to +∞, except for the point x = 0.

Define areas of continuity and break points. Usually a function is continuous in the same domain where it is defined. To detect discontinuities, you need to calculate when the argument approaches isolated points inside the domain of definition. For example, the function 1/x tends to infinity when x→0+ and to minus infinity when x→0-. This means that at the point x = 0 it has a discontinuity of the second kind.
If the limits at the discontinuity point are finite but not equal, then this is a discontinuity of the first kind. If they are equal, then the function is considered continuous, although it is not defined at an isolated point.

Find the vertical asymptotes, if any. The calculations from the previous step will help you here, since the vertical asymptote is almost always at the discontinuity point of the second kind. However, sometimes it is not individual points that are excluded from the domain of definition, but entire intervals of points, and then the vertical asymptotes can be located at the edges of these intervals.

Check if the function has special properties: even, odd, and periodic.
The function will be even if for any x in the domain f(x) = f(-x). For example, cos(x) and x^2 are even functions.

Periodicity is a property that says that there is a certain number T called a period, which for any x f(x) = f(x + T). For example, all major trigonometric functions(sine, cosine, tangent) - periodic.

Find points. To do this, calculate the derivative of given function and find those x values ​​where it vanishes. For example, the function f(x) = x^3 + 9x^2 -15 has a derivative g(x) = 3x^2 + 18x that vanishes at x = 0 and x = -6.

To determine which extremum points are maxima and which are minima, trace the change in the signs of the derivative in the found zeros. g(x) changes sign from plus at x = -6 and back from minus to plus at x = 0. Therefore, the function f(x) has a minimum at the first point and a minimum at the second.

Thus, you have also found areas of monotonicity: f(x) increases monotonically on the interval -∞;-6, decreases monotonically on -6;0 and increases again on 0;+∞.

Find the second derivative. Its roots will show where the graph of a given function will be convex, and where it will be concave. For example, the second derivative of the function f(x) will be h(x) = 6x + 18. It vanishes at x = -3, changing its sign from minus to plus. Therefore, the graph f (x) before this point will be convex, after it - concave, and this point itself will be an inflection point.

A function may have other asymptotes, except for vertical ones, but only if its domain of definition includes . To find them, calculate the limit of f(x) when x→∞ or x→-∞. If it is finite, then you have found the horizontal asymptote.

The oblique asymptote is a straight line of the form kx + b. To find k, calculate the limit of f(x)/x as x→∞. To find b - limit (f(x) – kx) with the same x→∞.

Plot the function on the computed data. Label the asymptotes, if any. Mark the extremum points and the function values ​​in them. For greater accuracy of the graph, calculate the function values ​​at several more intermediate points. Research completed.

For a complete study of the function and plotting its graph, it is recommended to use the following scheme:

1) find the scope of the function;

2) find the discontinuity points of the function and vertical asymptotes (if they exist);

3) investigate the behavior of the function at infinity, find the horizontal and oblique asymptotes;

4) investigate the function for evenness (oddity) and for periodicity (for trigonometric functions);

5) find extrema and intervals of monotonicity of the function;

6) determine the intervals of convexity and inflection points;

7) find points of intersection with the coordinate axes, if possible, and some additional points that refine the graph.

The study of the function is carried out simultaneously with the construction of its graph.

Example 9 Explore the function and build a graph.

1. Domain of definition: ;

2. The function breaks at points
,
;

We investigate the function for the presence of vertical asymptotes.

;
,
─ vertical asymptote.

;
,
─ vertical asymptote.

3. We investigate the function for the presence of oblique and horizontal asymptotes.

Straight
─ oblique asymptote, if
,
.

,
.

Straight
─ horizontal asymptote.

4. The function is even because
. The parity of the function indicates the symmetry of the graph with respect to the y-axis.

5. Find the intervals of monotonicity and extrema of the function.

Let's find the critical points, i.e. points where the derivative is 0 or does not exist:
;
. We have three points
;

. These points divide the entire real axis into four intervals. Let's define the signs on each of them.

On the intervals (-∞; -1) and (-1; 0) the function increases, on the intervals (0; 1) and (1; +∞) it decreases. When passing through a point
the derivative changes sign from plus to minus, therefore, at this point, the function has a maximum
.

6. Let's find convexity intervals, inflection points.

Let's find the points where is 0, or does not exist.

has no real roots.
,
,

points
and
divide the real axis into three intervals. Let's define the sign at every interval.

Thus, the curve on the intervals
and
convex downwards, on the interval (-1;1) convex upwards; there are no inflection points, since the function at the points
and
not determined.

7. Find the points of intersection with the axes.

with axle
the graph of the function intersects at the point (0; -1), and with the axis
the graph does not intersect, because the numerator of this function has no real roots.

The graph of the given function is shown in Figure 1.

Figure 1 ─ Graph of the function

Application of the concept of derivative in economics. Function elasticity

To study economic processes and solve other applied tasks The concept of elasticity of a function is often used.

Definition. Function elasticity
is called the limit of the ratio of the relative increment of the function to the relative increment of the variable at
, . (VII)

The elasticity of a function shows approximately how many percent the function will change
when changing the independent variable by 1%.

The elasticity of a function is used in the analysis of demand and consumption. If the elasticity of demand (in absolute value)
, then demand is considered elastic if
─ neutral if
─ inelastic with respect to price (or income).

Example 10 Calculate the elasticity of a function
and find the value of the elasticity index for = 3.

Solution: according to the formula (VII) the elasticity of the function:

Let x=3 then
This means that if the independent variable increases by 1%, then the value of the dependent variable will increase by 1.42%.

Example 11 Let the demand function regarding the price has the form
, where ─ constant coefficient. Find the value of the elasticity index of the demand function at the price x = 3 den. units

Solution: calculate the elasticity of the demand function using the formula (VII)

Assuming
monetary units, we get
. This means that at the price
monetary unit a price increase of 1% will cause a decrease in demand by 6%, i.e. demand is elastic.

Today we invite you to explore and plot a function graph with us. After a careful study of this article, you will not have to sweat for a long time to complete this kind of task. It is not easy to explore and build a graph of a function, the work is voluminous, requiring maximum attention and accuracy of calculations. To facilitate the perception of the material, we will gradually study the same function, explain all our actions and calculations. Welcome to amazing and fascinating world mathematics! Go!

Domain

In order to explore and plot a function, you need to know a few definitions. A function is one of the basic (basic) concepts in mathematics. It reflects the dependence between several variables (two, three or more) with changes. The function also shows the dependence of sets.

Imagine that we have two variables that have a certain range of change. So, y is a function of x, provided that each value of the second variable corresponds to one value of the second. In this case, the variable y is dependent, and it is called a function. It is customary to say that the variables x and y are in For greater clarity of this dependence, a graph of the function is built. What is a function graph? This is a set of points coordinate plane where each value of x corresponds to one value of y. Graphs can be different - a straight line, hyperbola, parabola, sinusoid and so on.

A function graph cannot be plotted without exploration. Today we will learn how to conduct research and plot a function graph. It is very important to make notes during the study. So it will be much easier to cope with the task. The most convenient study plan:

1. Domain.
2. Continuity.
3. Even or odd.
4. Periodicity.
5. Asymptotes.
6. Zeros.
7. Constancy.
8. Ascending and descending.
9. Extremes.
10. Convexity and concavity.

Let's start with the first point. Let's find the domain of definition, that is, on what intervals our function exists: y \u003d 1/3 (x ^ 3-14x ^ 2 + 49x-36). In our case, the function exists for any values ​​of x, that is, the domain of definition is R. This can be written as xОR.

Continuity

Now we are going to explore the discontinuity function. In mathematics, the term "continuity" appeared as a result of the study of the laws of motion. What is infinite? Space, time, some dependencies (an example is the dependence of the variables S and t in motion problems), the temperature of the heated object (water, frying pan, thermometer, and so on), a continuous line (that is, one that can be drawn without taking it off the sheet pencil).

A graph is considered continuous if it does not break at some point. One of the most obvious examples of such a graph is a sine wave, which you can see in the picture in this section. The function is continuous at some point x0 if a number of conditions are met:

• a function is defined at a given point;
• the right and left limits at a point are equal;
• the limit is equal to the value of the function at the point x0.

If at least one condition is not met, the function is said to break. And the points at which the function breaks are called break points. An example of a function that will “break” when displayed graphically is: y=(x+4)/(x-3). Moreover, y does not exist at the point x = 3 (since it is impossible to divide by zero).

In the function that we are studying (y \u003d 1/3 (x ^ 3-14x ^ 2 + 49x-36)) everything turned out to be simple, since the graph will be continuous.

Even, odd

Now examine the function for parity. Let's start with a little theory. An even function is a function that satisfies the condition f (-x) = f (x) for any value of the variable x (from the range of values). Examples are:

• module x (the graph looks like a jackdaw, the bisector of the first and second quarters of the graph);
• x squared (parabola);
• cosine x (cosine wave).

Note that all of these graphs are symmetrical when viewed with respect to the y-axis.

What then is called an odd function? These are those functions that satisfy the condition: f (-x) \u003d - f (x) for any value of the variable x. Examples:

• hyperbola;
• cubic parabola;
• sinusoid;
• tangent and so on.

Please note that these functions are symmetric about the point (0:0), that is, the origin. Based on what has been said in this section of the article, the even and odd function must have the property: x belongs to the definition set and -x too.

Let us examine the function for parity. We can see that she doesn't fit any of the descriptions. Therefore, our function is neither even nor odd.

Asymptotes

Let's start with a definition. An asymptote is a curve that is as close as possible to the graph, that is, the distance from some point tends to zero. There are three types of asymptotes:

• vertical, that is, parallel to the y axis;
• horizontal, i.e. parallel to the x-axis;
• oblique.

As for the first type, these lines should be looked for at some points:

• gap;
• ends of the domain.

In our case, the function is continuous, and the domain of definition is R. Therefore, there are no vertical asymptotes.

The graph of a function has a horizontal asymptote, which meets the following requirement: if x tends to infinity or minus infinity, and the limit is equal to a certain number (for example, a). In this case, y=a is the horizontal asymptote. There are no horizontal asymptotes in the function we are studying.

An oblique asymptote only exists if two conditions are met:

• lim(f(x))/x=k;
• lim f(x)-kx=b.

Then it can be found by the formula: y=kx+b. Again, in our case there are no oblique asymptotes.

Function zeros

The next step is to examine the graph of the function for zeros. It is also very important to note that the task associated with finding the zeros of a function occurs not only in the study and plotting of a function graph, but also as an independent task, and as a way to solve inequalities. You may be required to find the zeros of a function on a graph or use mathematical notation.

Finding these values ​​will help you plot the function more accurately. In simple terms, the zero of the function is the value of the variable x, at which y \u003d 0. If you are looking for zeros of a function on a graph, then you should pay attention to the points where the graph intersects with the x-axis.

To find the zeros of the function, you need to solve the following equation: y=1/3(x^3-14x^2+49x-36)=0. After doing the necessary calculations, we get the following answer:

sign constancy

The next stage in the study and construction of a function (graphics) is finding intervals of sign constancy. This means that we must determine on which intervals the function takes a positive value, and on which intervals it takes a negative value. The zeros of the functions found in the previous section will help us to do this. So, we need to build a straight line (separately from the graph) and distribute the zeros of the function along it in the correct order from smallest to largest. Now you need to determine which of the resulting intervals has a “+” sign, and which one has a “-”.

In our case, the function takes a positive value on the intervals:

• from 1 to 4;
• from 9 to infinity.

Negative meaning:

• from minus infinity to 1;
• from 4 to 9.

This is fairly easy to determine. Substitute any number from the interval into the function and see what sign the answer is (minus or plus).

Function Ascending and Decreasing

In order to explore and build a function, we need to find out where the graph will increase (go up on Oy), and where it will fall (creep down along the y-axis).

The function increases only if the larger value of the variable x corresponds to the larger value of y. That is, x2 is greater than x1, and f(x2) is greater than f(x1). And we observe a completely opposite phenomenon in a decreasing function (the more x, the less y). To determine the intervals of increase and decrease, you need to find the following:

• scope (we already have it);
• derivative (in our case: 1/3(3x^2-28x+49);
• solve the equation 1/3(3x^2-28x+49)=0.

After calculations, we get the result:

We get: the function increases on the intervals from minus infinity to 7/3 and from 7 to infinity, and decreases on the interval from 7/3 to 7.

Extremes

The investigated function y=1/3(x^3-14x^2+49x-36) is continuous and exists for any values ​​of the variable x. The extremum point shows the maximum and minimum of this function. In our case, there are none, which greatly simplifies the construction task. Otherwise, they are also found using the derivative function. After finding, do not forget to mark them on the chart.

Convexity and concavity

We continue to study the function y(x). Now we need to check it for convexity and concavity. The definitions of these concepts are quite difficult to perceive, it is better to analyze everything with examples. For the test: a function is convex if it is a non-decreasing function. Agree, this is incomprehensible!

We need to find the derivative of the second order function. We get: y=1/3(6x-28). Now equate right side to zero and solve the equation. Answer: x=14/3. We have found the inflection point, that is, the place where the graph changes from convex to concave or vice versa. On the interval from minus infinity to 14/3, the function is convex, and from 14/3 to plus infinity, it is concave. It is also very important to note that the inflection point on the chart should be smooth and soft, no sharp corners should not be present.

Definition of additional points

Our task is to explore and plot the function graph. We have completed the study, it will not be difficult to plot the function now. For a more accurate and detailed reproduction of a curve or a straight line on the coordinate plane, you can find several auxiliary points. It's pretty easy to calculate them. For example, we take x=3, solve the resulting equation and find y=4. Or x=5 and y=-5 and so on. You can take as many additional points as you need to build. At least 3-5 of them are found.

Plotting

We needed to investigate the function (x^3-14x^2+49x-36)*1/3=y. All the necessary marks in the course of calculations were made on the coordinate plane. All that remains to be done is to build a graph, that is, connect all the points to each other. Connecting the dots is smooth and accurate, this is a matter of skill - a little practice and your schedule will be perfect.