What is called an inflection point. Convexity of functions. General scheme for studying functions and plotting

To determine the convexity (concavity) of a function on a certain interval, the following theorems can be used.

Theorem 1. Let the function be defined and continuous on the interval and have a finite derivative . For a function to be convex (concave) in , it is necessary and sufficient that its derivative decreases (increases) on this interval.

Theorem 2. Let the function be defined and continuous together with its derivative on and have a continuous second derivative inside . For the convexity (concavity) of the function in it is necessary and sufficient that inside

Let us prove Theorem 2 for the case of convexity of the function .

Need. Let's take an arbitrary point. We expand the function near the point in a Taylor series

The equation of a tangent to a curve at a point having an abscissa:

Then the excess of the curve over the tangent to it at the point is equal to

Thus, the remainder is equal to the excess of the curve over the tangent to it at the point . Due to continuity, if , then also for , belonging to a sufficiently small neighborhood of the point , and therefore, obviously, for any different from the value of , belonging to the indicated neighborhood.

This means that the graph of the function lies above the tangent and the curve is convex at an arbitrary point.

Adequacy. Let the curve be convex on the interval . Let's take an arbitrary point.

Similarly to the previous one, we expand the function near the point in a Taylor series

The excess of the curve over the tangent to it at the point having the abscissa , defined by the expression is equal to

Since the excess is positive for a sufficiently small neighborhood of the point , the second derivative is also positive. As we strive, we obtain that for an arbitrary point .

Example. Investigate for convexity (concavity) function .

Its derivative increases on the entire real axis, so by Theorem 1 the function is concave on .

Its second derivative , therefore, by Theorem 2, the function is concave on .

3.4.2.2 Inflection points

Definition. inflection point graphic arts continuous function is called the point separating the intervals in which the function is convex and concave.

It follows from this definition that the inflection points are the points of the extremum point of the first derivative. This implies the following assertions for the necessary and sufficient inflection conditions.

Theorem ( necessary condition inflection). In order for a point to be an inflection point of a twice differentiable function, it is necessary that its second derivative at this point be equal to zero ( ) or did not exist.

Theorem (sufficient condition for inflection). If the second derivative of a twice differentiable function changes sign when passing through a certain point, then there is an inflection point.

Note that the second derivative may not exist at the point itself.

The geometric interpretation of the inflection points is illustrated in fig. 3.9

In a neighborhood of a point, the function is convex and its graph lies below the tangent drawn at this point. In the neighborhood of a point, the function is concave and its graph lies above the tangent drawn at this point. At the inflection point, the tangent divides the graph of the function into regions of convexity and concavity.

3.4.2.3 Examining a function for convexity and the presence of inflection points

1. Find the second derivative.

2. Find points at which the second derivative or does not exist.

Rice. 3.9.

3. Examine the sign of the second derivative to the left and right of the found points and draw a conclusion about the intervals of convexity or concavity and the presence of inflection points.

Example. Investigate the function for convexity and the presence of inflection points.

2. The second derivative is equal to zero at .

3. The second derivative changes sign at , so the point is the inflection point.

On the interval , then the function is convex on this interval.

On the interval , then the function is concave on this interval.

3.4.2.4 General scheme for the study of functions and plotting

When studying a function and plotting its graph, it is recommended to use the following scheme:

Find the scope of the function.
Investigate the function for even - odd. Recall that the schedule even function is symmetric about the y-axis, and the graph of an odd function is symmetrical about the origin.
Find vertical asymptotes.
Explore the behavior of a function at infinity, find horizontal or oblique asymptotes.
Find extrema and intervals of monotonicity of the function.
Find the convexity intervals of the function and the inflection points.
Find points of intersection with coordinate axes.

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

Example. Explore Function and plot it.

1. Function scope - .

2. The function under study is even , so its graph is symmetrical about the y-axis.

3. The denominator of the function vanishes at , so the graph of the function has vertical asymptotes and .

The points are discontinuity points of the second kind, since the limits on the left and right at these points tend to .

4. Behavior of the function at infinity.

Therefore, the graph of the function has a horizontal asymptote.

5. Extremes and intervals of monotonicity. Finding the first derivative

For , therefore, the function decreases in these intervals.

For , therefore, the function increases in these intervals.

For , therefore, the point is a critical point.

Finding the second derivative

Since , then the point is the minimum point of the function .

6. Convexity intervals and inflection points.

Function at , so the function is concave on this interval.

The function at , means that the function is convex on these intervals.

The function never vanishes, so there are no inflection points.

7. Points of intersection with the coordinate axes.

The equation , has a solution , which means the point of intersection of the graph of the function with the y-axis (0, 1).

The equation has no solution, which means there are no points of intersection with the abscissa axis.

Taking into account the conducted research, it is possible to build a graph of the function

Schematic graph of a function shown in fig. 3.10.

Rice. 3.10.

3.4.2.5 Asymptotes of the graph of a function

Definition. Asymptote the function graph is called a straight line, which has the property that the distance from the point () to this straight line tends to 0 with an unlimited removal of the graph point from the origin.

When we plot a function, it is important to define the convex intervals and inflection points. We need them, along with the intervals of decreasing and increasing, for a clear representation of the function in a graphical form.

Understanding this topic requires knowing what the derivative of a function is and how to calculate it to some order, as well as being able to solve different types inequalities.

At the beginning of the article, the main concepts are defined. Then we will show what relationship exists between the direction of the convexity and the value of the second derivative over a certain interval. Next, we will indicate the conditions under which the inflection points of the graph can be determined. All reasoning will be illustrated by examples of problem solutions.

Definition 1

In a downward direction on a certain interval in the case when its graph is located not lower than the tangent to it at any point of this interval.

Definition 2

Differentiable function is convex upwards on a certain interval if the graph of this function is located no higher than the tangent to it at any point of this interval.

A downward convex function can also be called concave. Both definitions are clearly shown in the graph below:

Definition 3

Function inflection point is the point M (x 0 ; f (x 0)) at which there is a tangent to the graph of the function, provided that the derivative exists in the vicinity of the point x 0 , where the graph of the function takes different directions of convexity on the left and right sides.

Simply put, an inflection point is a place on a graph where there is a tangent, and the direction of the convexity of the graph when passing through this place will change the direction of the convexity. If you do not remember under what conditions the existence of a vertical and non-vertical tangent is possible, we advise you to repeat the section on the tangent of the graph of a function at a point.

Below is a graph of a function that has multiple inflection points highlighted in red. Let us clarify that the presence of inflection points is not mandatory. On the graph of one function, there can be one, two, several, infinitely many or none.

In this section, we will talk about a theorem with which you can determine the convexity intervals on the graph of a particular function.

Definition 4

The graph of the function will have a convexity in the direction downwards or upwards if the corresponding function y = f (x) has a second finite derivative on the specified interval x, provided that the inequality f "" (x) ≥ 0 ∀ x ∈ X (f "" (x) ≤ 0 ∀ x ∈ X) will be true.

Using this theorem, you can find the intervals of concavity and convexity on any graph of a function. To do this, you just need to solve the inequalities f "" (x) ≥ 0 and f "" (x) ≤ 0 on the domain of the corresponding function.

Let us clarify that those points where the second derivative does not exist, but the function y = f (x) is defined, will be included in the intervals of convexity and concavity.

Let's look at an example specific task how to properly apply this theorem.

Example 1

Condition: given a function y = x 3 6 - x 2 + 3 x - 1 . Determine at what intervals its graph will have convexity and concavity.

Solution

The domain of this function is the whole set real numbers. Let's start by calculating the second derivative.

y "= x 3 6 - x 2 + 3 x - 1" = x 2 2 - 2 x + 3 ⇒ y "" = x 2 2 - 2 x + 3 = x - 2

We see that the domain of the second derivative coincided with the domain of the function itself. Therefore, to identify the intervals of convexity, we need to solve the inequalities f "" (x) ≥ 0 and f "" (x) ≤ 0 .

y "" ≥ 0 ⇔ x - 2 ≥ 0 ⇔ x ≥ 2 y "" ≤ 0 ⇔ x - 2 ≤ 0 ⇔ x ≤ 2

We got that schedule given function will have concavity on the segment [ 2 ; + ∞) and convexity on the segment (- ∞ ; 2 ] .

For clarity, we will draw a graph of the function and mark the convex part in blue on it, and the concave part in red.

Answer: the graph of the given function will have a concavity on the segment [ 2 ; + ∞) and convexity on the segment (- ∞ ; 2 ] .

But what to do if the domain of the second derivative does not coincide with the domain of the function? Here the remark made above is useful to us: those points where the final second derivative does not exist, we will also include in the segments of concavity and convexity.

Example 2

Condition: given a function y = 8 x x - 1 . Determine in what intervals its graph will be concave, and in what intervals it will be convex.

Solution

First, let's find out the scope of the function.

x ≥ 0 x - 1 ≠ 0 ⇔ x ≥ 0 x ≠ 1 ⇔ x ∈ [ 0 ; 1) ∪ (1 ; + ∞)

Now we calculate the second derivative:

y "= 8 x x - 1" = 8 1 2 x (x - 1) - x 1 (x - 1) 2 = - 4 x + 1 x (x - 1) 2 y "" = - 4 x + 1 x (x - 1) 2 "= - 4 1 x x - 1 2 - (x + 1) x x - 1 2" x (x - 1) 4 = = - 4 1 x x - 1 2 - x + 1 1 2 x (x - 1) 2 + x 2 (x - 1) x x - 1 4 = = 2 3 x 2 + 6 x - 1 x 3 2 (x - 1) 3

The domain of the second derivative is the set x ∈ (0 ; 1) ∪ (1 ; + ∞) . We see that x equal to zero will be in the domain of the original function, but not in the domain of the second derivative. This point must be included in the segment of concavity or convexity.

After that, we need to solve the inequalities f "" (x) ≥ 0 and f "" (x) ≤ 0 on the domain of the given function. We use the interval method for this: at x \u003d - 1 - 2 3 3 ≈ - 2, 1547 or x \u003d - 1 + 2 3 3 ≈ 0, 1547 the numerator 2 (3 x 2 + 6 x - 1) x 2 3 x - 1 3 becomes 0 and the denominator is 0 when x is zero or one.

Let's put the resulting points on the graph and determine the sign of the expression on all intervals that will be included in the domain of the original function. On the graph, this area is indicated by hatching. If the value is positive, mark the interval with a plus, if negative, then with a minus.

Consequently,

f "" (x) ≥ 0 x ∈ [ 0 ; 1) ∪ (1 ; + ∞) ⇔ x ∈ 0 ; - 1 + 2 3 3 ∪ (1 ; + ∞) , and f "" (x) ≤ 0 x ∈ [ 0 ; 1) ∪ (1 ; + ∞) ⇔ x ∈ [ - 1 + 2 3 3 ; one)

We turn on the previously marked point x = 0 and get the desired answer. The graph of the original function will have a downward bulge at 0 ; - 1 + 2 3 3 ∪ (1 ; + ∞) , and up - for x ∈ [ - 1 + 2 3 3 ; one) .

Let's draw a graph, marking the convex part in blue, and the concave in red. The vertical asymptote is marked with a black dotted line.

Answer: The graph of the original function will have a downward bulge at 0 ; - 1 + 2 3 3 ∪ (1 ; + ∞) , and up - for x ∈ [ - 1 + 2 3 3 ; one) .

Inflection Conditions for a Function Graph

Let's start with the formulation of the necessary condition for the inflection of the graph of some function.

Definition 5

Let's say we have a function y = f(x) whose graph has an inflection point. For x = x 0, it has a continuous second derivative, therefore, the equality f "" (x 0) = 0 will hold.

Considering this condition, we should look for inflection points among those in which the second derivative will go to 0 . This condition will not be sufficient: not all such points will suit us.

Also note that according to common definition, we will need a tangent line, vertical or non-vertical. In practice, this means that in order to find the inflection points, one should take those in which the second derivative of this function becomes 0. Therefore, to find the abscissas of the inflection points, we need to take all x 0 from the domain of the function, where lim x → x 0 - 0 f " (x) = ∞ and lim x → x 0 + 0 f " (x) = ∞ . Most often, these are points at which the denominator of the first derivative turns to 0.

The first sufficient condition for the existence of an inflection point of the function graph

We have found all x 0 values that can be taken as the abscissa of the inflection points. After that, we need to apply the first sufficient inflection condition.

Definition 6

Let's say we have a function y = f (x) that is continuous at the point M (x 0 ; f (x 0)) . Moreover, it has a tangent at this point, and the function itself has a second derivative in the vicinity of this point x 0 . In this case, if the second derivative acquires opposite signs on the left and right sides, then this point can be considered an inflection point.

We see that this condition does not require that the second derivative necessarily exist at this point, its presence in the neighborhood of the point x 0 is sufficient.

All of the above can be conveniently presented as a sequence of actions.

First you need to find all the abscissas x 0 of possible inflection points, where f "" (x 0) = 0, lim x → x 0 - 0 f "(x) = ∞, lim x → x 0 + 0 f "(x) = ∞ .
Find out at what points the derivative will change sign. These values are the abscissas of the inflection points, and the points M (x 0 ; f (x 0)) corresponding to them are the inflection points themselves.

For clarity, let's consider two problems.

Example 3

Condition: given a function y = 1 10 x 4 12 - x 3 6 - 3 x 2 + 2 x . Determine where the graph of this function will have inflection and bulge points.

Solution

This function is defined on the entire set of real numbers. We consider the first derivative:

y "= 1 10 x 4 12 - x 3 6 - 3 x 2 + 2 x" = 1 10 4 x 3 12 - 3 x 2 6 - 6 x + 2 = = 1 10 x 3 3 - x 2 2 - 6 x + 2

Now let's find the domain of the first derivative. It is also the set of all real numbers. Hence, the equalities lim x → x 0 - 0 f "(x) = ∞ and lim x → x 0 + 0 f" (x) = ∞ cannot be satisfied for any values of x 0 .

We calculate the second derivative:

y "" = = 1 10 x 3 3 - x 2 2 - 6 x + 2 " = 1 10 3 x 2 3 - 2 x 2 - 6 = 1 10 x 2 - x - 6

y "" = 0 ⇔ 1 10 (x 2 - x - 6) = 0 ⇔ x 2 - x - 6 = 0 D = (- 1) 2 - 4 1 (- 6) = 25 x 1 = 1 - 25 2 \u003d - 2, x 2 \u003d 1 + 25 2 \u003d 3

We found the abscissas of two likely inflection points - 2 and 3. All that remains for us to do is to check at what point the derivative changes its sign. Let's draw a numerical axis and plot these points on it, after which we will place the signs of the second derivative on the resulting intervals.

The arcs show the direction of the convexity of the graph in each interval.

The second derivative reverses sign (from plus to minus) at the point with abscissa 3 , passing through it from left to right, and does the same (from minus to plus) at the point with abscissa 3 . So, we can conclude that x = - 2 and x = 3 are the abscissas of the inflection points of the function graph. They will correspond to the points of the graph - 2; - 4 3 and 3 ; - 15 8 .

Let's look again at the image of the numerical axis and the resulting signs on the intervals in order to draw conclusions about the places of concavity and convexity. It turns out that the bulge will be located on the segment - 2; 3 , and concavity on segments (- ∞ ; - 2 ] and [ 3 ; + ∞) .

The solution to the problem is clearly shown on the graph: blue color - convexity, red - concavity, black color means inflection points.

Answer: the bulge will be located on the segment - 2; 3 , and concavity on segments (- ∞ ; - 2 ] and [ 3 ; + ∞) .

Example 4

Condition: calculate the abscissas of all inflection points of the graph of the function y = 1 8 · x 2 + 3 x + 2 · x - 3 3 5 .

Solution

The domain of the given function is the set of all real numbers. We calculate the derivative:

y "= 1 8 (x 2 + 3 x + 2) x - 3 3 5" = = 1 8 x 2 + 3 x + 2 " (x - 3) 3 5 + (x 2 + 3 x + 2) x - 3 3 5 " = = 1 8 2 x + 3 (x - 3) 3 5 + (x 2 + 3 x + 2) 3 5 x - 3 - 2 5 = 13 x 2 - 6 x - 39 40 (x - 3) 2 5

Unlike a function, its first derivative will not be determined at an x value of 3 , but:

lim x → 3 - 0 y "(x) = 13 (3 - 0) 2 - 6 (3 - 0) - 39 40 3 - 0 - 3 2 5 = + ∞ lim x → 3 + 0 y " (x) = 13 (3 + 0) 2 - 6 (3 + 0) - 39 40 3 + 0 - 3 2 5 = + ∞

This means that a vertical tangent to the graph will pass through this point. Therefore, 3 can be the abscissa of the inflection point.

We calculate the second derivative. We also find the area of its definition and the points at which it turns to 0:

y "" = 13 x 2 - 6 x - 39 40 x - 3 2 5 " = = 1 40 13 x 2 - 6 x - 39 " (x - 3) 2 5 - 13 x 2 - 6 x - 39 x - 3 2 5 " (x - 3) 4 5 = = 1 25 13 x 2 - 51 x + 21 (x - 3) 7 5 , x ∈ (- ∞ ; 3) ∪ (3 ; + ∞ ) y "" (x) = 0 ⇔ 13 x 2 - 51 x + 21 = 0 D = (- 51) 2 - 4 13 21 = 1509 x 1 = 51 + 1509 26 ≈ 3 , 4556 , x 2 = 51 - 1509 26 ≈ 0.4675

We have two more possible inflection points. We put them all on a number line and mark the resulting intervals with signs:

The change of sign will occur when passing through each specified point, which means that they are all inflection points.

Answer: Let's draw a graph of the function, marking concavities in red, convexities in blue, and inflection points in black:

Knowing the first sufficient inflection condition, we can determine the necessary points where the presence of the second derivative is not necessary. Based on this, the first condition can be considered the most universal and suitable for solving various types of problems.

Note that there are two more inflection conditions, but they can only be applied when there is a finite derivative at the specified point.

If we have f "" (x 0) = 0 and f """ (x 0) ≠ 0 , then x 0 will be the abscissa of the inflection point of the graph y = f (x) .

Example 5

Condition: the function y = 1 60 x 3 - 3 20 x 2 + 7 10 x - 2 5 is given. Determine if the function graph will have an inflection at point 3 ; 4 5 .

Solution

The first thing to do is to make sure that given point will generally belong to the graph of this function.

y (3) = 1 60 3 3 - 3 20 3 2 - 2 5 = 27 60 - 27 20 + 21 10 - 2 5 = 9 - 27 + 42 - 8 20 = 4 5

The specified function is defined for all arguments that are real numbers. We calculate the first and second derivatives:

y "= 1 60 x 3 - 3 20 x 2 + 7 10 x - 2 5" = 1 20 x 2 - 3 10 x + 7 10 y "" = 1 20 x 2 - 3 10 x + 7 10" = 1 10 x - 3 10 = 1 10 (x - 3)

We got that the second derivative will go to 0 if x is equal to 0 . This means that the necessary inflection condition for this point will be satisfied. Now we use the second condition: we find the third derivative and find out if it will turn to 0 at 3:

y " " " = 1 10 (x - 3) " = 1 10

The third derivative will not vanish for any value of x. Therefore, we can conclude that this point will be the inflection point of the graph of the function.

Answer: Let's show the solution in the illustration:

Let's say that f "(x 0) = 0, f "" (x 0) = 0, . . . , f (n) (x 0) = 0 and f (n + 1) (x 0) ≠ 0 . In this case, for even n, we get that x 0 is the abscissa of the inflection point of the graph y \u003d f (x) .

Example 6

Condition: given a function y = (x - 3) 5 + 1 . Calculate the inflection points of its graph.

Solution

This function is defined on the entire set of real numbers. Calculate the derivative: y " = ((x - 3) 5 + 1) " = 5 x - 3 4 . Since it will also be defined for all real values of the argument, then at any point in its graph there will be a non-vertical tangent.

Now let's calculate for what values the second derivative will turn to 0:

y "" = 5 (x - 3) 4 " = 20 x - 3 3 y "" = 0 ⇔ x - 3 = 0 ⇔ x = 3

We have found that for x = 3 the graph of the function may have an inflection point. We use the third condition to confirm this:

y " " " = 20 (x - 3) 3 " = 60 x - 3 2 , y " " " (3) = 60 3 - 3 2 = 0 y (4) = 60 (x - 3) 2 " = 120 (x - 3) , y (4) (3) = 120 (3 - 3) = 0 y (5) = 120 (x - 3) " = 120 , y (5) (3 ) = 120 ≠ 0

We have n = 4 in the third sufficient condition. This is an even number, so x \u003d 3 will be the abscissa of the inflection point and the point of the graph of the function (3; 1) corresponds to it.

Answer: Here is a graph of this function with the convexity, concavity and inflection point marked:

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Function Graph y=f(x) called convex on the interval (a;b), if it is located below any of its tangents on this interval.

Function Graph y=f(x) called concave on the interval (a;b), if it is located above any of its tangents in this interval.

The figure shows a curve convex on (a;b) and concave to (b;c).

Examples.

Consider a sufficient sign that allows you to determine whether the graph of a function in a given interval will be convex or concave.

Theorem. Let y=f(x) differentiable by (a;b). If at all points of the interval (a;b) second derivative of the function y = f(x) negative, i.e. f ""(x) < 0, то график функции на этом интервале выпуклый, если же f""(x) > 0 is concave.

Proof. Assume for definiteness that f""(x) < 0 и докажем, что график функции будет выпуклым.

Take on the function graph y = f(x) arbitrary point M0 with abscissa x0 Î ( a; b) and draw through the point M0 tangent. Her equation. We must show that the graph of the function on (a;b) lies below this tangent, i.e. with the same value x curve ordinate y = f(x) will be less than the ordinate of the tangent.

So the equation of the curve is y = f(x). Let us denote the tangent ordinate corresponding to the abscissa x. Then . Therefore, the difference between the ordinates of the curve and the tangent at the same value x will be .

Difference f(x) – f(x0) transform according to the Lagrange theorem, where c between x and x0.

In this way,

We again apply the Lagrange theorem to the expression in square brackets: , where c 1 between c 0 and x0. According to the theorem f ""(x) < 0. Определим знак произведения второго и третьего сомножителей.

Thus, any point of the curve lies below the tangent to the curve for all values x and x0 Î ( a; b), which means that the curve is convex. The second part of the theorem is proved similarly.

Examples.

The point on the graph of a continuous function that separates its convex part from the concave part is called inflection point.

Obviously, at the inflection point, the tangent, if it exists, intersects the curve, because on one side of this point, the curve lies under the tangent, and on the other side, above it.

Let us define sufficient conditions for a given point of the curve to be an inflection point.

Theorem. Let the curve be defined by the equation y = f(x). If a f ""(x 0) = 0 or f ""(x 0) does not exist and when passing through the value x = x0 derivative f ""(x) changes sign, then the point of the graph of the function with the abscissa x = x0 there is an inflection point.

Proof. Let f ""(x) < 0 при x < x0 and f ""(x) > 0 at x > x0. Then at x < x0 the curve is convex, and x > x0- concave. Hence the point A, lying on the curve, with abscissa x0 there is an inflection point. Similarly, we can consider the second case, when f ""(x) > 0 at x < x0 and f ""(x) < 0 при x > x0.

Thus, inflection points should be sought only among those points where the second derivative vanishes or does not exist.

Examples. Find the inflection points and determine the intervals of convexity and concavity of the curves.

ASYMPTOTS OF THE GRAPH OF A FUNCTION

When investigating a function, it is important to establish the shape of its graph with an unlimited removal of the graph point from the origin.

Of particular interest is the case when the graph of a function, when its variable point is removed to infinity, indefinitely approaches a certain straight line.

Direct called asymptote function graph y = f(x) if the distance from the variable point M graph to this line when the point is removed M to infinity tends to zero, i.e. the point of the graph of the function, as it tends to infinity, must approach the asymptote indefinitely.

The curve can approach its asymptote, remaining on one side of it or on different sides, intersecting the asymptote an infinite number of times and moving from one side to the other.

If we denote by d the distance from the point M curve to the asymptote, it is clear that d tends to zero as the point is removed M to infinity.

We will further distinguish between vertical and oblique asymptotes.

VERTICAL ASYMPTOTS

Let at x→ x0 either side of the function y = f(x) indefinitely increases in absolute value, i.e. or or . Then it follows from the definition of the asymptote that the line x = x0 is an asymptote. The converse is also obvious if the line x = x0 is an asymptote, so .

Thus, the vertical asymptote of the graph of the function y = f(x) is called a line if f(x)→ ∞ under at least one of the conditions x→ x0– 0 or x → x0 + 0, x = x0

Therefore, to find the vertical asymptotes of the graph of the function y = f(x) need to find those values x = x0, at which the function goes to infinity (suffers an infinite discontinuity). Then the vertical asymptote has the equation x = x0.

Examples.

SLANT ASYMPTOTS

Since the asymptote is a straight line, then if the curve y = f(x) has an oblique asymptote, then its equation will be y = kx + b. Our task is to find the coefficients k and b.

Theorem. Straight y = kx + b serves as an oblique asymptote at x→ +∞ for the graph of the function y = f(x) if and only if . A similar statement is true for x → –∞.

Proof. Let MP- the length of the segment equal to the distance from the point M to the asymptote. By condition . Denote by φ the angle of inclination of the asymptote to the axis Ox. Then from ΔMNP follows that . Since φ is a constant angle (φ ≠ π/2), then , but

When examining a function and constructing its graph, at one of the stages we determine the inflection points and convexity intervals. These data, together with the intervals of increase and decrease, allow us to schematically present the graph of the function under study.

What follows assumes that you know up to a certain order and different types.

Let's start the study of the material with the necessary definitions and concepts. Next, we voice the relationship between the value of the second derivative of a function on a certain interval and the direction of its convexity. After that, let's move on to the conditions that allow us to determine the inflection points of the function graph. In the text, we will give typical examples with detailed solutions.

Page navigation.

Convexity, concavity of a function, inflection point.

Definition.

convex down on the X interval, if its graph is located not lower than the tangent to it at any point of the X interval.

Definition.

The differentiable function is called convex up on the X interval, if its graph is located no higher than the tangent to it at any point of the X interval.

An upward convex function is often called convex, and convex down - concave.

Look at the drawing illustrating these definitions.

Definition.

The point is called inflection point of the graph of the function y \u003d f (x) if at a given point there is a tangent to the function graph (it can be parallel to the Oy axis) and there is such a neighborhood of the point , within which the graph of the function has different directions of convexity to the left and right of the point M.

In other words, the point M is called the inflection point of the graph of the function, if there is a tangent at this point and the graph of the function changes the direction of the convexity, passing through it.

If necessary, refer to the section to recall the conditions for the existence of a non-vertical and vertical tangent.

The figure below shows several examples of inflection points (marked with red dots). Note that some functions may not have inflection points, while others may have one, several, or infinitely many inflection points.

Finding the convexity intervals of a function.

We formulate a theorem that allows us to determine the intervals of convexity of a function.

Theorem.

If the function y=f(x) has a finite second derivative on the interval X and if the inequality (), then the graph of the function has a convexity directed down (up) on X.

This theorem allows you to find intervals of concavity and convexity of a function, you only need to solve the inequalities and , respectively, on the domain of definition of the original function.

It should be noted that the points at which the function y=f(x) is defined and the second derivative does not exist will be included in the intervals of concavity and convexity.

Let's deal with this with an example.

Example.

Find the intervals at which the graph of the function has a convexity directed upwards and a convexity directed downwards.

Solution.

The domain of a function is the entire set of real numbers.

Let's find the second derivative.

The domain of definition of the second derivative coincides with the domain of definition of the original function, therefore, in order to find out the intervals of concavity and convexity, it is enough to solve and respectively.

Therefore, the function is downward convex on the interval and upward convex on the interval .

Graphic illustration.

Part of the graph of the function on the convex interval is shown in blue, on the concavity interval - in red.

Now consider an example where the domain of the second derivative does not coincide with the domain of the function. In this case, as we have already noted, the points of the domain where there is no finite second derivative should be included in the intervals of convexity and (or) concavity.

Example.

Find the intervals of convexity and concavity of the function graph.

Solution.

Let's start with the scope of the function:

Let's find the second derivative:

The domain of the second derivative is the set . As you can see, x=0 is in the domain of the original function, but not in the domain of the second derivative. Do not forget about this point, it will need to be included in the interval of convexity and (or) concavity.

Now we solve the inequalities and on the domain of the original function. Applicable . Expression numerator goes to zero at or , denominator - at x = 0 or x = 1 . We plot these points schematically on the number line and find out the sign of the expression on each of the intervals included in the domain of definition of the original function (it is shown by the shaded area on the bottom number line). A positive value is a plus sign, a negative value is a minus sign.

In this way,

and

Therefore, by including the point x=0 , we get the answer.

At the graph of the function has a convexity directed downwards, with - bulge directed upwards.

Graphic illustration.

Part of the graph of the function on the convex interval is shown in blue, on the concavity intervals - in red, the black dotted line is the vertical asymptote.

Necessary and sufficient conditions for an inflection.

Necessary condition for an inflection.

Let's formulate necessary condition for inflection function graph.

Let the graph of the function y=f(x) have an inflection at a point and have a continuous second derivative for , then the equality is true.

It follows from this condition that the abscissas of the inflection points should be sought among those in which the second derivative of the function vanishes. BUT, this condition is not sufficient, that is, not all values in which the second derivative is equal to zero are the abscissas of the inflection points.

It should also be noted that, by definition of the inflection point, the existence of a tangent line is required, it can also be vertical. What does this mean? And this means the following: the abscissas of the inflection points can be everything from the domain of the function, for which and . Usually these are the points at which the denominator of the first derivative vanishes.

The first sufficient condition for an inflection.

After all are found that can be abscissas of inflection points, you should use the first sufficient condition for the inflection function graph.

Let the function y=f(x) be continuous at the point , have a tangent at it (possibly vertical) and this function have a second derivative in some neighborhood of the point . Then, if within this neighborhood to the left and to the right of , the second derivative has different signs, then it is the inflection point of the graph of the function.

As you can see, the first sufficient condition does not require the existence of the second derivative at the point itself, but requires its existence in the vicinity of the point.

Now we summarize all the information in the form of an algorithm.

Algorithm for finding the inflection points of a function.

We find all the abscissas of the possible inflection points of the graph of the function (or and ) and find out, passing through which the second derivative changes sign. Such values will be the abscissas of the inflection points, and the points corresponding to them will be the inflection points of the function graph.

Consider two examples of finding inflection points for clarification.

Example.

Find the inflection points and the intervals of convexity and concavity of the function graph.

Solution.

The domain of the function is the entire set of real numbers.

Let's find the first derivative:

The domain of the first derivative is also the entire set of real numbers, so the equalities and is not executed for any .

Let's find the second derivative:

Let us find out at what values of the argument x the second derivative vanishes:

So the abscissas of the possible inflection points are x=-2 and x=3 .

Now it remains to check, by a sufficient inflection criterion, at which of these points the second derivative changes sign. To do this, put the points x=-2 and x=3 on the real axis and, as in generalized interval method, we place the signs of the second derivative over each interval. Under each interval, the direction of the convexity of the graph of the function is shown schematically by arcs.

The second derivative changes sign from plus to minus passing through the point x=-2 from left to right, and changes sign from minus to plus passing through x=3 . Therefore, both x=-2 and x=3 are the abscissas of the inflection points of the function graph. They correspond to the graph points and .

Looking again at the real axis and the signs of the second derivative on its intervals, we can conclude about the intervals of convexity and concavity. The graph of the function is convex on the interval and concave on the intervals and .

Graphic illustration.

Part of the graph of the function on the convex interval is shown in blue, on the concavity intervals - in red, the inflection points are shown as black dots.

Example.

Find the abscissas of all inflection points of a function graph .

Solution.

The domain of this function is the entire set of real numbers.

Let's find the derivative.

The first derivative, unlike the original function, is not defined at x=3 . But and . Therefore, at the point with the abscissa x=3, there is a vertical tangent to the graph of the original function. So x=3 can be the abscissa of the inflection point of the function graph.

We find the second derivative, its domain of definition and the points at which it vanishes:

We got two more possible abscissas of the inflection points. We mark all three points on the number line and determine the sign of the second derivative on each of the obtained intervals.

The second derivative changes sign, passing through each of the points, therefore, they are all abscissas of the inflection points.

Graphic illustration.

Parts of the graph of the function at convex intervals are shown in blue, at concavity intervals - in red, inflection points are shown as black dots.

The first sufficient condition for the inflection of the function graph allows one to determine the inflection points and does not require the existence of a second derivative at them. Therefore, the first sufficient condition can be considered universal and most used.

Now we formulate two more sufficient conditions for inflection, but they are applicable only if there is a finite derivative at the inflection point up to a certain order.

The second sufficient condition for an inflection.

If , a , then is the abscissa of the inflection point of the graph of the function y=f(x) x=3 different from zero.

Obviously, the value of the third derivative is non-zero for any x , including x=3 . Therefore, according to the second sufficient condition for the inflection of the function graph, the point is an inflection point.

Graphic illustration.

The third sufficient condition for an inflection.

Let , and , then if n is an even number, then is the abscissa of the inflection point of the graph of the function y=f(x) .

Example.

Find the inflection points of the graph of a function .

Solution.

The function is defined on the entire set of real numbers.

Let's find its derivative: . Obviously, it is also defined for all real x , so there is a non-vertical tangent at any point on its graph.

Let us determine the values of x at which the second derivative vanishes.

Thus, at the point with the abscissa x=3 there may be an inflection in the graph of the function. To make sure that x=3 is indeed the abscissa of the inflection point, we use the third sufficient condition.

According to the third sufficient condition for the inflection of the function graph, we have n=4 (the fifth derivative vanishes) - even, therefore x=3 is the abscissa of the inflection point and it corresponds to the point of the function graph (3;1) .

Graphic illustration.

Part of the graph of the function on the convex interval is shown in blue, on the concavity interval - in red, the inflection point is shown as a black dot.

The concept of convexity of a function

Consider the function \(y = f\left(x \right),\) which is assumed to be continuous on the segment \(\left[ (a,b) \right].\) The function \(y = f\left(x \right),\) )\) is called convex down (or simply convex) if for any points \((x_1)\) and \((x_2)\) from \(\left[ (a,b) \right]\) x_1),(x_2) \in \left[ (a,b) \right],\) such that \((x_1) \ne (x_2),\) then the function \(f\left(x \right) \) are called strictly convex down

An upward convex function is defined similarly. The function \(f\left(x \right)\) is called convex up (or concave) if for any points \((x_1)\) and \((x_2)\) of the segment \(\left[ (a,b) \right]\) the inequality \ If this inequality is strict for any \(( x_1),(x_2) \in \left[ (a,b) \right],\) such that \((x_1) \ne (x_2),\) then the function \(f\left(x \right) \) are called strictly convex upwards on the segment \(\left[ (a,b) \right].\)

Geometric interpretation of the convexity of a function

The introduced definitions of a convex function have a simple geometric interpretation.

For the function, convex down (drawing \(1\)), the midpoint \(B\) of any chord \((A_1)(A_2)\) lies above

Similarly, for the function convex up (drawing \(2\)), the midpoint \(B\) of any chord \((A_1)(A_2)\) lies below corresponding point \((A_0)\) of the graph of the function or coincides with this point.

Convex functions have another visual property, which is related to the location tangent to the graph of the function. The function \(f\left(x \right)\) is convex down on the segment \(\left[ (a,b) \right]\) if and only if its graph lies not lower than the tangent drawn to it at any point \((x_0)\) of the segment \(\left[ (a ,b) \right]\) (figure \(3\)).

Accordingly, the function \(f\left(x \right)\) is convex up on the segment \(\left[ (a,b) \right]\) if and only if its graph lies no higher than the tangent drawn to it at any point \((x_0)\) of the segment \(\left[ (a ,b) \right]\) (figure \(4\)). These properties are a theorem and can be proved using the definition of convexity of a function.

Sufficient conditions for convexity

Let for the function \(f\left(x \right)\) the first derivative \(f"\left(x \right)\) exist on the segment \(\left[ (a,b) \right],\) and the second derivative \(f""\left(x \right)\) − on the interval \(\left((a,b) \right).\) Then the following sufficient criteria for convexity hold:

If \(f""\left(x \right) \ge 0\) for all \(x \in \left((a,b) \right),\) then the function \(f\left(x \right )\) convex down on the segment \(\left[ (a,b) \right];\)

If \(f""\left(x \right) \le 0\) for all \(x \in \left((a,b) \right),\) then the function \(f\left(x \right )\) convex up on the segment \(\left[ (a,b) \right].\)

In cases where the second derivative is strictly greater than (less than) zero, one speaks, respectively, of strict convexity down (or up ).

Let us prove the above theorem for the case of a downward convex function. Let the function \(f\left(x \right)\) have a non-negative second derivative on the interval \(\left((a,b) \right):\) \(f""\left(x \right) \ge 0.\) Denote by \((x_0)\) the midpoint of the segment \(\left[ ((x_1),(x_2)) \right].\) Assume that the length of this segment is equal to \(2h.\) Then the coordinates \((x_1)\) and \((x_2)\) can be written as: \[(x_1) = (x_0) - h,\;\;(x_2) = (x_0) + h.\] Expand the function \(f\left(x \right)\) at the point \((x_0)\) into a Taylor series with a remainder term in the Lagrange form. We get the following expressions: \[ (f\left(((x_1)) \right) = f\left(((x_0) - h) \right) ) = (f\left(((x_0)) \right) - f"\left(((x_0)) \right)h + \frac((f""\left(((\xi _1)) \right)(h^2)))((2},} \] \[ {f\left({{x_2}} \right) = f\left({{x_0} + h} \right) } = {f\left({{x_0}} \right) + f"\left({{x_0}} \right)h + \frac{{f""\left({{\xi _2}} \right){h^2}}}{{2!}},} \] где \({x_0} - h !}
Add both equalities: \[ (f\left(((x_1)) \right) + f\left(((x_2)) \right) ) = (2f\left(((x_0)) \right) + \frac (((h^2)))(2)\left[ (f""\left(((\xi _1)) \right) + f""\left(((\xi _2)) \right)) \right].) \] Since \((\xi _1),(\xi _2) \in \left((a,b) \right),\) the second derivatives on the right-hand side are non-negative. Therefore, \ or \ that is, according to the definition, the function \(f\left(x \right)\) convex down .

Note that the necessary convexity condition for a function (i.e., a direct theorem in which, for example, from the convexity condition it follows that \(f""\left(x \right) \ge 0\)) is satisfied only for a nonstrict inequalities. In the case of strict convexity, the necessary condition is generally not satisfied. For example, the function \(f\left(x \right) = (x^4)\) is strictly downward convex. However, at the point \(x = 0\) its second derivative is equal to zero, i.e. the strict inequality \(f""\left(x \right) \gt 0\) is not satisfied in this case.

Properties of convex functions

We list some properties of convex functions, assuming that all functions are defined and continuous on the segment \(\left[ (a,b) \right].\)

If the functions \(f\) and \(g\) are downward (upward) convex, then any of them linear combination \(af + bg,\) where \(a\), \(b\) are positive real numbers, also convex downwards (upwards).

If the function \(u = g\left(x \right)\) is downward convex and the function \(y = f\left(u \right)\) is downward convex and nondecreasing, then complex function \(y = f\left((g\left(x \right)) \right)\) will also convex down.

If the function \(u = g\left(x \right)\) is upward convex and the function \(y = f\left(u \right)\) is downward convex and nonincreasing, then complex function \(y = f\left((g\left(x \right)) \right)\) will convex down.

Local maximum convex upward function defined on the segment \(\left[ (a,b) \right],\) is simultaneously its highest value on this segment.

Local minimum downward convex function defined on the segment \(\left[ (a,b) \right],\) is simultaneously its the smallest value on this segment.