Conduct a complete study of the function and plot an online solution. How to examine a function and graph it

Conduct a complete study and graph the function

y(x)=x2+81−x.y(x)=x2+81−x.

1) The scope of the function. Since the function is a fraction, we 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 domain of definition of the function 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. Let's find one-sided limits:

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

3) Let us determine the intersection points of the function graph 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 OyOy axis 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) Let's examine the function for periodicity. The function is not periodic, since it is a fractional rational function.

6) Let's examine the function for extrema and monotonicity. To do this, we find the first derivative of the function:

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

We got three critical points: x=−2,x=1,x=4x=−2,x=1,x=4. Let us divide the entire domain of definition of the function into intervals with these 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:

Let us 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;+ ∞) is satisfied by y′′<0y″<0, то есть функция выпуклая.

8) Let us examine 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 using 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 more accurately construct the graph.

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 construct a graph, supplement it with asymptotes x=1x=1 (blue), y=−x−1y=−x−1 (green) and mark the characteristic points (purple intersection with the ordinate axis, orange extrema, black additional points) :

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

Example 3.23. a

Solution. x And y y
y = a - 2×a/4 =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) = 6x 2 - 30x +36 = 6(x ​​-2)(x - 3), then the critical points of the function x 1 = 2 and x 2 = 3. Extrema can only be at these points. So as when passing through the point x 1 = 2 the derivative changes its sign from plus to minus, then at this point the function has a maximum. When passing through the point x 2 = 3 the derivative changes its sign from minus to plus, therefore at the point x 2 = 3 the function has a minimum. Having calculated the function values ​​at the 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 on three sides with wire mesh, and the fourth side is adjacent to the wall. For this there is a linear meters of mesh. At what aspect ratio will the site have the largest area?

Solution. Let us denote the sides of the platform by x And y. The area of ​​the site is S = xy. Let y- this 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 pad cannot be negative). S " = a - 4x, a - 4x = 0 at x = a/4, whence
y = a - 2×a/4 =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 manufacture a closed cylindrical tank with a capacity V=16p ≈ 50 m 3 . What should be the dimensions of the tank (radius R and height H) so that the least amount of material is used 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 N Þ N = V/pR 2 =16p/ pR 2 = 16/ R 2 . This means S(R) = 2p(R 2 +16/R). We find the derivative of this function:
S " (R) = 2p(2R- 16/R 2) = 4p (R- 8/R 2). S " (R) = 0 for R 3 = 8, therefore,
R = 2, H = 16/4 = 4.

Related information.

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

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

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 has a discontinuity are called critical.

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 interval and 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 coincide, 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 satisfied, then x 0 is the maximum point; if the derivative changes sign from minus to plus (to the right of x 0 executed f"(x)<0, то х 0 - точка минимума.

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

Theorem 2 (a necessary sign of 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 the 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 at which the function is continuous and the derivative is zero or does not exist.
3) Consider the neighborhood of each point, 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, substitute the values ​​of the critical points into this function. Using sufficient conditions for the extremum, draw the appropriate conclusions.

Example 18. Examine the function y=x 3 -9x 2 +24x for an extremum

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; This means that apart from the two points found, 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 through the point x=4 - from minus to plus.
4) At point x=2 the function has a maximum y max =20, and at point x=4 - a minimum y min =16.

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

Let f"(x 0) and at the point x 0 there exists f""(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 a segment, the function y=f(x) can reach the smallest (y the least) or the greatest (y the highest) value either at the critical points of the function lying in the interval (a;b), or at the ends of the segment.

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=f(x) at the points obtained in step 2), as well as at the ends of the segment and select the largest and smallest from them: they are, respectively, the largest (y the largest) and the smallest (y the least) values ​​of the function on the interval.

Example 19. Find the largest value of the 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 at which y"=0; we get:
3x 2 -6x-45=0
x 2 -2x-15=0
x 1 =-3; x 2 =5
3) Calculate the value of the function at points x=0 y=225, x=5 y=50, x=6 y=63
The segment contains only the point x=5. The largest of the found values ​​of the function is 225, and the smallest is the number 50. So, y max = 225, y min = 50.

Study of a function on convexity

The figure shows graphs of two functions. The first of them is convex upward, the second is convex downward.

The function y=f(x) is continuous on the segment and differentiable in the interval (a;b), is called convex upward (downward) 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 holds on the interval (a;b), then the function is convex downward on the interval ; if the inequality f""(x)0 holds on the interval (a;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 the points at which f""(x) does not exist or vanishes.
2) Examine the sign f""(x) to the left and right of each point found in the first step.
3) Based on Theorem 4, draw a conclusion.

Example 20. Find the extremum points and inflection points of the graph of the function 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 when x 1 =0, x 2 =1. When passing through the point x=0, the derivative changes sign from minus to plus, but 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 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 upward) and x=1 is also the inflection point (transition from convexity upward to convexity downward). If x=, then y=; if, then x=1, y=13.

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 a 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 a 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 the asymptote for a function

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

Scheme for studying a function and constructing its graph

I. Find the domain of definition of the function.
II. Find the intersection points of the function graph with the coordinate axes.
III. Find asymptotes.
IV. Find possible extremum points.
V. Find critical points.
VI. Using the auxiliary figure, explore the sign of the first and second derivatives. Determine areas of increasing and decreasing function, find the direction of convexity of the graph, points of extrema and inflection points.
VII. Construct a graph, taking into account the research carried out in paragraphs 1-6.

Example 22: Construct a graph of the function according to the above diagram

Solution.
I. The domain of a function is the set of all real numbers except x=1.
II. Since the equation x 2 +1=0 has no real roots, the graph of the function has no 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. Let us study the behavior of the function near the discontinuity point x=1. Since y → ∞ as x → -∞, y → +∞ as x → 1+, then the straight line x=1 is the 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 obtain two possible extremum points:
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. Let us examine 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 domain 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 find that the function increases at (-∞;1-√2), decreases at (1-√2;1+√2), and increases again at (1+√2;+∞). Extremum points: maximum at x=1-√2, and f(1-√2)=2-2√2 minimum at x=1+√2, and f(1+√2)=2+2√2. At (-∞;1) the graph is convex upward, and at (1;+∞) it is convex downward.
VII Let's make a table of the obtained values

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

If the problem requires a complete study of the function f (x) = 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, you should use the properties and graphs of basic elementary functions. The research algorithm includes the following steps:

Yandex.RTB R-A-339285-1

Finding the domain of definition

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

Example 1

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

4 x 2 - 1 = 0 x = ± 1 2 ⇒ x ∈ - ∞ ; - 1 2 ∪ - 1 2 ; 1 2 ∪ 1 2 ; +∞

As a result, you can get roots, logarithms, and so on. Then the ODZ can be searched for 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) > 0.

Studying the boundaries of the ODZ and finding vertical asymptotes

There are vertical asymptotes at the boundaries of the function, when the one-sided limits at such points are infinite.

Example 2

For example, consider the border points equal to x = ± 1 2.

Then it is necessary to study the function to find the one-sided limit. Then we get that: lim x → - 1 2 - 0 f (x) = lim x → - 1 2 - 0 x 2 4 x 2 - 1 = = lim x → - 1 2 - 0 x 2 (2 x - 1 ) (2 x + 1) = 1 4 (- 2) · - 0 = + ∞ lim x → - 1 2 + 0 f (x) = lim x → - 1 2 + 0 x 2 4 x - 1 = = lim x → - 1 2 + 0 x 2 (2 x - 1) (2 x + 1) = 1 4 (- 2) (+ 0) = - ∞ lim x → 1 2 - 0 f (x) = lim x → 1 2 - 0 x 2 4 x 2 - 1 = = lim x → 1 2 - 0 x 2 (2 x - 1) (2 x + 1) = 1 4 (- 0) 2 = - ∞ lim x → 1 2 - 0 f (x) = lim x → 1 2 - 0 x 2 4 x 2 - 1 = = lim x → 1 2 - 0 x 2 (2 x - 1) (2 x + 1) = 1 4 ( + 0) 2 = + ∞

This shows that the one-sided limits are infinite, which means the straight lines x = ± 1 2 are the vertical asymptotes of the graph.

Study of a function and whether it is even or odd

When the condition y (- x) = y (x) is satisfied, the function is considered even. This suggests that the graph is located symmetrically with respect to Oy. When the condition y (- x) = - y (x) is satisfied, the function is considered odd. This means that the symmetry is relative to the origin of coordinates. If at least one inequality is not satisfied, we obtain a function of general form.

The equality y (- x) = y (x) indicates that the function is even. When constructing, it is necessary to take into account that there will be symmetry with respect to Oy.

To solve the inequality, 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 internal points from the domain of definition where the derivative of the function is equal to zero or does not exist.

When making a decision, the following notes must be taken into account:

• for existing intervals of increasing and decreasing inequalities of the form f " (x) > 0, critical points are not included in the solution;
• points at which the function is defined without a finite derivative must be included in the intervals of increasing and decreasing (for example, y = x 3, where the point x = 0 makes the function defined, the derivative has the value of infinity at this point, y " = 1 3 x 2 3, y "(0) = 1 0 = ∞, x = 0 is included in the increasing interval);
• To avoid disagreements, it is recommended to use mathematical literature recommended by the Ministry of Education.

Inclusion of critical points in intervals of increasing and decreasing if they satisfy the domain of definition of the function.

Definition 2

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

• derivative;
• critical points;
• divide the definition domain into intervals using critical points;
• determine the sign of the derivative on each of the intervals, where + is an increase and - is a decrease.

Example 3

Find the derivative on the domain of definition f " (x) = x 2 " (4 x 2 - 1) - x 2 4 x 2 - 1 " (4 x 2 - 1) 2 = - 2 x (4 x 2 - 1) 2 .

Solution

To solve you need:

• find stationary points, this example has x = 0;
• find the zeros of the denominator, the example takes the value zero at x = ± 1 2.

We place points on the number axis to determine the derivative on 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 depict + on the graph, which means the function is increasing, and - means it is decreasing.

For example, f " (- 1) = - 2 · (- 1) 4 - 1 2 - 1 2 = 2 9 > 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 ; + ∞ .

In the diagram, using + and -, the positivity and negativity of the function are depicted, and the arrows indicate decrease and increase.

Extremum points of a function are points where the function is defined and through which the derivative changes sign.

Example 4

If we consider an example where x = 0, then the value of the function in it is equal to f (0) = 0 2 4 · 0 2 - 1 = 0. When the sign of the derivative changes from + to - and passes through the point x = 0, then the point with coordinates (0; 0) is considered the maximum point. When the sign changes from - to +, we obtain a minimum point.

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

Definition 3

For determining the intervals of concavity and convexity necessary:

• find the second derivative;
• find the zeros of the second derivative function;
• divide the definition area into intervals with the appearing points;
• determine the sign of the interval.

Example 5

Find the second derivative from the domain of definition.

Solution

f "" (x) = - 2 x (4 x 2 - 1) 2 " = = (- 2 x) " (4 x 2 - 1) 2 - - 2 x 4 x 2 - 1 2 " (4 x 2 - 1) 4 = 24 x 2 + 2 (4 x 2 - 1) 3

We find the zeros of the numerator and denominator, where in our example we have that the zeros of the denominator x = ± 1 2

Now you need to plot the points on the number line and determine the sign of the second derivative from each interval. We get that

Answer:

• the function is convex from the interval - 1 2 ; 12 ;
• the function is concave from the intervals - ∞ ; - 1 2 and 1 2; + ∞ .

Definition 4

Inflection point– this is a point of the form x 0 ; f (x 0) . When it has a tangent to the graph of the function, then when it passes through x 0 the function changes sign to the opposite.

In other words, this is a point through which the second derivative passes and changes sign, and at the points themselves it is equal to zero or does not exist. All points are considered to be the domain of the function.

In the example, it was clear that there are no inflection points, since the second derivative changes sign while passing through the points x = ± 1 2. They, in turn, are not included in the 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 asymptotes are depicted using straight lines given by the equation y = k x + b, where k = lim x → ∞ f (x) x and b = lim x → ∞ f (x) - k x.

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

In other words, asymptotes are considered to be lines to which the graph of a function approaches at infinity. This facilitates quick construction of a function graph.

If there are no asymptotes, but the function is defined at both infinities, it is necessary to calculate the limit of the function at these infinities in order to understand how the graph of the function will behave.

Example 6

Let's consider as an example that

k = lim x → ∞ f (x) x = lim x → ∞ x 2 4 x 2 - 1 x = 0 b = lim x → ∞ (f (x) - k x) = lim x → ∞ x 2 4 x 2 - 1 = 1 4 ⇒ y = 1 4

is a horizontal asymptote. After examining the function, you can begin to construct it.

Calculating the value of a function at intermediate points

To make the graph more accurate, it is recommended to find several function values ​​at intermediate points.

Example 7

From the example we considered, it is necessary to find the values ​​of the function at the points x = - 2, x = - 1, x = - 3 4, x = - 1 4. Since the function is even, we get that the values ​​coincide with the values ​​at these points, that is, we get x = 2, x = 1, x = 3 4, x = 1 4.

Let's write and solve:

F (- 2) = f (2) = 2 2 4 2 2 - 1 = 4 15 ≈ 0, 27 f (- 1) - f (1) = 1 2 4 1 2 - 1 = 1 3 ≈ 0 , 33 f - 3 4 = f 3 4 = 3 4 2 4 3 4 2 - 1 = 9 20 = 0 , 45 f - 1 4 = f 1 4 = 1 4 2 4 1 4 2 - 1 = - 1 12 ≈ - 0.08

To determine the maxima and minima of the function, inflection points, and intermediate points, it is necessary to construct asymptotes. For convenient designation, intervals of increasing, decreasing, convexity, and concavity are recorded. Let's look at the picture below.

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

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

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Today we invite you to explore and build a graph of a function with us. After carefully studying this article, you will not have to sweat for long to complete this type of task. It is not easy to study and construct a graph of a function; it is a voluminous work that requires maximum attention and accuracy of calculations. To make the material easier to understand, we will study the same function step by step and explain all our actions and calculations. Welcome to the amazing and fascinating world of mathematics! Go!

Domain

In order to explore and graph a function, you need to know several definitions. Function is one of the main (basic) concepts in mathematics. It reflects the dependence between several variables (two, three or more) during 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 graph of a function? This is a set of points on the coordinate plane, where each x value corresponds to one y value. Graphs can be different - straight line, hyperbola, parabola, sine wave, and so on.

It is impossible to plot a function without research. Today we will learn how to conduct research and build a graph of a function. It is very important to take notes during the study. This will make the task much easier to cope with. The most convenient research plan:

1. Domain.
2. Continuity.
3. Even or odd.
4. Periodicity.
5. Asymptotes.
6. Zeros.
7. Sign constancy.
8. Increasing and decreasing.
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=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 equal to R. This can be written as follows xÎR.

Continuity

Now we will examine 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 movement problems), the temperature of a heated object (water, frying pan, thermometer, etc.), a continuous line (that is, one that can be drawn without lifting it from 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 sinusoid, which you can see in the picture in this section. A 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 point x0.

If at least one condition is not met, the function is said to fail. And the points at which the function breaks are usually 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=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. First, a little theory. An even function is one that satisfies the condition f(-x)=f(x) for any value of the variable x (from the range of values). Examples include:

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

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

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

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

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

Let's 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 a certain point tends to zero. In total, there are three types of asymptotes:

• vertical, that is, parallel to the y-axis;
• horizontal, that is, parallel to the x axis;
• inclined.

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

• gap;
• ends of the domain of definition.

In our case, the function is continuous, and the domain of definition is equal to 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 exists only if two conditions are met:

• lim(f(x))/x=k;
• lim f(x)-kx=b.

Then it can be found using 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 when studying and constructing a graph of a function, 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 graph the function more accurately. In simple terms, the zero of a function is the value of the variable x at which y = 0. If you are looking for the zeros of a function on a graph, then you should pay attention to the points at which 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 carrying out the necessary calculations, we get the following answer:

Sign constancy

The next stage of research and construction of a function (graph) is finding intervals of constant sign. This means that we must determine at which intervals the function takes a positive value and at which intervals it takes a negative value. The zero functions found in the previous section will help us do this. So, we need to build a straight line (separate 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 has a “-”.

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

• from 1 to 4;
• from 9 to infinity.

Negative meaning:

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

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

Increasing and decreasing function

In order to explore and construct a function, we need to know where the graph will increase (go up along the Oy axis) and where it will fall (crawl down along the y-axis).

A function increases only if a larger value of the variable x corresponds to a 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 with a decreasing function (the more x, the less y). To determine the intervals of increase and decrease, you need to find the following:

• domain of definition (we already have);
• 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 function under study y=1/3(x^3-14x^2+49x-36) is continuous and exists for any value of the variable x. The extremum point shows the maximum and minimum of a given function. In our case there are none, which greatly simplifies the construction task. Otherwise, they can also be found using the derivative function. Once found, do not forget to mark them on the chart.

Convexity and concavity

We continue to further explore the function y(x). Now we need to check it for convexity and concavity. The definitions of these concepts are quite difficult to comprehend; it is better to analyze everything using 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 a second order function. We get: y=1/3(6x-28). Now let's equate the right side to zero and solve the equation. Answer: x=14/3. We found the inflection point, that is, the place where the graph changes from convexity to concavity 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 graph should be smooth and soft, there should be no sharp corners.

Defining additional points

Our task is to investigate and construct a graph of the function. We have completed the study; constructing a graph of the function is now not difficult. For more accurate and detailed reproduction of a curve or straight line on the coordinate plane, you can find several auxiliary points. They are quite easy to calculate. 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 for construction. At least 3-5 of them are found.

Plotting a graph

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

In this article, we will consider a scheme for studying a function, and also give examples of studying extrema, monotonicity, and asymptotes of a given function.

Scheme

1. The domain of existence (DOA) of a function.
2. The intersection of the function (if any) with the coordinate axes, signs of the function, parity, periodicity.
3. Breaking points (their kind). Continuity. The asymptotes are vertical.
4. Monotonicity and extreme points.
5. Inflection points. Convex.
6. Study of a function at infinity, for asymptotes: horizontal and oblique.
7. Building a graph.

Monotonicity test

Theorem. If the function g continuous on , differentiated by (a; b) And g’(x) ≥ 0 (g’(x)≤0), xє(a; b), That g increasing (decreasing) by .

Example:

y = 1: 3x 3 - 6: 2x 2 + 5x.

ODZ: xєR

y’ = x 2 + 6x + 5.

Let's find the intervals of constant signs y'. Because the y' is an elementary function, then it can change signs only at points where it becomes zero or does not exist. Her ODZ: xєR.

Let's find the points at which the derivative is equal to 0 (zero):

y’ = 0;

x = -1; -5.

So, y growing on (-∞; -5] and on [-1; +∞), y descending on .

Research on extremes

T. x 0 called the maximum point (max) on the set A functions g when the function takes the greatest value at this point g(x 0) ≥ g(x), xєA.

T. x 0 called the minimum point (min) of the function g on a set A when the function takes the smallest value at this point g(x 0) ≤ g(x), xєA.

On set A maximum (max) and minimum (min) points are called extremum points g. Such extrema are also called absolute extrema on the set .

If x 0- extremum point of the function g in some of its districts, then x 0 called the point of local or local extremum (max or min) of the function g.

Theorem (necessary condition). If x 0- extremum point (local) of the function g, then the derivative does not exist or is equal to 0 (zero) in this area.

Definition. Points with a non-existent or equal to 0 (zero) derivative are called critical. It is these points that are suspicious for extremes.

Theorem (sufficient condition No. 1). If the function g continuous in some district i.e. x 0 and the sign changes through this point during the transition of the derivative, then this point is the point of extremum g.

Theorem (sufficient condition No. 2). Let the function in some district of the point be differentiable twice and g’ = 0, and g’’ > 0 (g’’< 0) , then this point is the point of maximum (max) or minimum (min) of the function.

Bulge Test

The function is called downward convex (or concave) on the interval (a, b) when the graph of the function is located no higher than the secant on the interval for any x with (a, b), which passes through these points .

The function will be convex strictly downward at (a, b), if - the graph lies below the secant on the interval.

The function is said to be convex upward (convex) on the interval (a, b), if for any t points With (a, b) the graph of a function on the interval lies not lower than the secant line passing through the abscissa at these points .

The function will be strictly convex upward by (a, b), if - the graph on the interval lies above the secant line.

If a function in some point district continuous and through t. x 0 When transitioning, the function changes its convexity, then this point is called the inflection point of the function.

Study on asymptotes

Definition. The straight line is called an asymptote g(x), if at an infinite distance from the origin of coordinates a point in the graph of the function approaches it: d(M,l).

Asymptotes can be vertical, horizontal and oblique.

Vertical line with equation x = x 0 will be the asymptote of the vertical graph of the function g , if there is an infinite gap at point x 0, then there is at least one left or right boundary at this point - infinity.

Studying a function on a segment for the smallest and largest values

If the function is continuous for , then according to the Weierstrass theorem there is a maximum value and a minimum value on this segment, that is, there are t glasses that belong such that g(x 1) ≤ g(x)< g(x 2), x 2 є . From the theorems about monotonicity and extrema, we obtain the following scheme for studying a function on an interval for the smallest and largest values.

Plan

1. Find the derivative g'(x).
2. Search function value g at these points and at the ends of the segment.
3. Compare the found values ​​and select the smallest and largest.

Comment. If you need to study a function on a finite interval (a, b), or on infinite (-∞; b); (-∞; +∞) on the max and min values, then in the plan, instead of the function values ​​at the ends of the interval, we look for the corresponding one-sided boundaries: instead f(a) looking for f(a+) = limf(x), instead of f(b) looking for f(-b). This way you can find the ODZ of a function on an interval, because absolute extrema do not necessarily exist in this case.

Application of the derivative to the solution of applied problems on the extremum of certain quantities

1. Express this quantity in terms of other quantities from the problem statement so that it is a function of only one variable (if possible).
2. Determine the interval of change of this variable.
3. Conduct a study of the function on the interval at max and min values.

Task. It is necessary to build a rectangular platform, using a meter of mesh, against the wall so that on one side it is adjacent to the wall, and on the other three it is fenced with a mesh. At what aspect ratio will the area of ​​such a platform be greatest?

S = xy- function of 2 variables.

S = x(a - 2x)- function of the 1st variable ; x є .

S = ax - 2x 2 ; S" = a - 4x = 0, xєR, x = a: 4.

S(a: 4) = a 2: 8- greatest value;

S(0) =0.

Let's find the other side of the rectangle: at = a: 2.

Aspect Ratio: y: x = 2.

Answer. The largest area will be equal to a 2 /8, if the side that is parallel to the wall is 2 times larger than the other side.

Function study. Examples

Example 1

Available y=x 3: (1-x) 2 . Carry out research.

1. ODZ: xє(-∞; 1) U (1; ∞).
2. A function of general form (neither even nor odd) is not symmetrical with respect to point 0 (zero).
3. Function signs. The function is elementary, therefore it can change sign only at points where it is equal to 0 (zero) or does not exist.
4. The function is elementary, therefore continuous on the ODZ: (-∞; 1) U (1; ∞).

Gap: x = 1;

limx 3: (1- x) 2 = ∞- Discontinuity of the 2nd kind (infinite), therefore there is a vertical asymptote at point 1;

x = 1- equation of vertical asymptote.

5. y’ = x 2 (3 - x) : (1 - x) 3 ;

ODZ (y’): x ≠ 1;

x = 1- critical point.

y’ = 0;

0; 3 - critical points.

6. y’’ = 6x: (1 - x) 4 ;

Critical items: 1, 0;

x = 0 - bend point, y(0) = 0.

7. limx 3: (1 - 2x + x 2) = ∞- there is no horizontal asymptote, but there may be an inclined one.

k = 1- number;

b = 2- number.

Therefore, there is an oblique asymptote y = x + 2 at + ∞ and at - ∞.

Example 2

Given y = (x 2 + 1) : (x - 1). Produce and research. Build a graph.

1. The domain of existence is the entire number line, except the so-called x = 1.

2. y crosses OY (if possible) incl. (0;g(0)). We find y(0) = -1 - t. intersection OY .

Points of intersection of the graph with OX we find by solving the equation y = 0. The equation has no real roots, so this function does not intersect OX.

3. The function is non-periodic. Consider the expression

g(-x) ≠ g(x), and g(-x) ≠ -g(x). This means that it is a general function (neither even nor odd).

4. T. x = 1 the discontinuity is of the second kind. At all other points the function is continuous.

5. Study of a function for an extremum:

(x 2 - 2x - 1) : (x - 1)2 = y"

and solve the equation y" = 0.

So, 1 - √2, 1 + √2, 1 - critical points or points of possible extremum. These points divide the number line into four intervals .

At each interval, the derivative has a certain sign, which can be established by the method of intervals or by calculating the values ​​of the derivative at individual points. At intervals (-∞; 1 - √2 ) U (1 + √2 ; ∞) , positive derivative, which means the function is growing; If xє(1 - √2 ; 1) U(1; 1 + √2 ) , then the function decreases, because on these intervals the derivative is negative. Through t. x 1 during the transition (movement follows from left to right), the derivative sign changes from “+” to “-”, therefore, at this point there is a local maximum, we will find

y max = 2 - 2 √2 .

When passing through x 2 the derivative changes sign from “-” to “+”, therefore, at this point there is a local minimum, and

y mix = 2 + 2√2.

T. x = 1 not the extreme.

6. 4: (x - 1) 3 = y"".

On (-∞; 1 ) 0 > y"" , consequently, on this interval the curve is convex; if xє (1 ; ∞) - the curve is concave. In t point 1 the function is not defined, so this point is not an inflection point.

7. From the results of paragraph 4 it follows that x = 1- vertical asymptote of the curve.

There are no horizontal asymptotes.

x + 1 = y - oblique asymptote of this curve. There are no other asymptotes.

8. Taking into account the research carried out, we build a graph (see figure above).