When studying a function and constructing its graph, at one stage we determine inflection points and convexity intervals. These data, together with the intervals of increase and decrease, make it possible to schematically represent the graph of the function under study.
The further presentation assumes that you can do up to some order and different types.
Let's start studying the material with necessary definitions and concepts. Next, we will voice the connection between the value of the second derivative of a function on a certain interval and the direction of its convexity. After this, we will move on to the conditions that allow us to determine the inflection points of the function graph. According to the text we will give typical examples with detailed solutions.
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Convexity, concavity of a function, inflection point.
Definition.
convex down on the interval X if its graph is located not lower than the tangent to it at any point of the interval X.
Definition.
The function to be differentiated is called convex up on the interval X if its graph is located no higher than the tangent to it at any point in the interval X.
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 function graph y=f(x) if at a given point there is a tangent to the graph of the function (it can be parallel to the Oy axis) and there is a neighborhood of the point within which to the left and right of the point M the graph of the function has different directions of convexity.
In other words, point M is called an inflection point of the graph of a 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 some examples of inflection points (marked with red dots). Note that some functions may have no inflection points, while others may have one, several, or infinitely many inflection points.
Finding intervals of convexity of a function.
Let us formulate a theorem that allows us to determine the convexity intervals of a function.
Theorem.
If the function y=f(x) has a finite second derivative on the interval X and if the inequality holds 
(), then the graph of the function has a convexity directed downwards (upwards) by X.
This theorem allows you to find the 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 points at which the function y=f(x) is defined and the second derivative does not exist will be included in the concavity and convexity intervals.
Let's understand this with an example.
Example.
Find out the intervals on which the graph of the function 
has a convexity directed upward and a convexity directed downward.
Solution.
The domain of a function is the entire set 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, to find out the intervals of concavity and convexity, it is enough to solve and accordingly. 
Therefore, the function is convex downwards on the interval and convex upwards on the interval .
Graphic illustration.
The part of the function graph in the convex interval is shown in blue, and in the concavity interval – in red.
Now let's consider an example when the domain of definition of the second derivative does not coincide with the domain of definition of the function. In this case, as we have already noted, points of the domain of definition at which a finite second derivative does not exist should be included in the intervals of convexity and (or) concavity.
Example.
Find the intervals of convexity and concavity of the graph of the function.
Solution.
Let's start with the domain of the function:
Let's find the second derivative: 
The domain of definition of the second derivative is the set 
. As you can see, x=0 belongs to the domain of the original function, but does not belong to 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 inequalities on the domain of definition of the original function. Let's apply. Numerator of expression 
goes to zero at 
or 
, denominator – at x = 0 or x = 1. We schematically plot these points 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 as a shaded area on the lower number line). For a positive value we put a plus sign, for a negative value we put a minus sign. 
Thus,
And 
Therefore, by including the point x=0, we get the answer.
At 
the graph of the function has a convexity directed downward, with 
- convexity directed upward.
Graphic illustration.
The part of the graph of the function on the convexity interval is depicted in blue, on the concavity intervals - in red, the black dotted line is the vertical asymptote.
Necessary and sufficient conditions for inflection.
Necessary condition for inflection.
Let's formulate necessary condition for inflection function graphics.
Let the graph of the function y=f(x) have an inflection at a point and have a continuous second derivative, then the equality holds.
From this condition it follows that the abscissa of the inflection points should be sought among those at 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 abscissas of inflection points.
It should also be noted that the definition of an inflection point requires the existence of a tangent line, or a vertical one. What does this mean? And this means the following: the abscissas of inflection points can be everything from the domain of definition of the function for which 
And 
. These are usually the points at which the denominator of the first derivative vanishes.
The first sufficient condition for inflection.
After all that can be abscissas of inflection points have been found, you should use the first sufficient condition for inflection function graphics.
Let the function y=f(x) be continuous at the point, have a tangent (possibly vertical) at it, and let this function have a second derivative in some neighborhood of the point. Then, if within this neighborhood to the left and right of , the second derivative has different signs, then is the inflection point of the function graph.
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 neighborhood of the point.
Now let’s summarize all the information in the form of an algorithm.
Algorithm for finding inflection points of a function.
We find all the abscissas of possible inflection points of the function graph (or And ) and find out by passing through which the second derivative changes sign. Such values will be the abscissa of the inflection points, and the corresponding points will be the inflection points of the function graph.
Let's look at two examples of finding inflection points for clarification.
Example.
Find inflection points and intervals of convexity and concavity of the graph of a function 
.
Solution.
The domain of a function is the entire set of real numbers.
Let's find the first derivative: 
The domain of definition of the first derivative is also the entire set of real numbers, therefore the equalities And is not fulfilled for any .
Let's find the second derivative:
Let’s find out at what values of the argument x the second derivative goes to zero: 
Thus, the abscissas of possible inflection points are x=-2 and x=3.
Now it remains to check, using a sufficient sign of inflection, at which of these points the second derivative changes sign. To do this, plot the points x=-2 and x=3 on the number axis and, as in generalized interval method, we place the signs of the second derivative over each interval. Under each interval, the direction of convexity of the function graph is shown schematically with 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 abscissas of the inflection points of the function graph. They correspond to the graph points and .
Taking another look at the number line and the signs of the second derivative on its intervals, we can draw conclusions about the intervals of convexity and concavity. The graph of a function is convex on the interval and concave on the intervals and .
Graphic illustration.
The part of the function graph on the convex interval is shown in blue, on the concavity interval – in red, and inflection points are shown as black dots.
Example.
Find the abscissa of all inflection points of the function graph 
.
Solution.
The domain of definition 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 abscissa x=3 there is a vertical tangent to the graph of the original function. Thus, 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 obtained two more possible abscissas of inflection points. We mark all three points on the number line and determine the sign of the second derivative on each of the resulting intervals. 
The second derivative changes sign when passing through each of the points, therefore, they are all abscissas of inflection points.
Graph of a function y=f(x) called convex on the interval (a; b), if it is located below any of its tangents on this interval.
Graph of a function y=f(x) called concave on the interval (a; b), if it is located above any of its tangents on this interval.
The figure shows a curve that is convex at (a; b) and concave on (b;c).
Examples.
Let us consider a sufficient criterion that allows us to determine whether the graph of a function in a given interval will be convex or concave.
Theorem. Let y=f(x) differentiable on (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 – concave.
Proof. Let us assume for definiteness that f""(x) < 0 и докажем, что график функции будет выпуклым.
Let's take the functions on the graph y = f(x) arbitrary point M0 with abscissa x 0 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. at the same value x ordinate of curve 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 ordinate of the tangent corresponding to the abscissa x. Then . Consequently, the difference between the ordinates of the curve and the tangent for the same value x will .
Difference f(x) – f(x 0) transform according to Lagrange's theorem, where c between x And x 0.
Thus,
We again apply Lagrange’s theorem to the expression in square brackets: , where c 1 between c 0 And x 0. According to the conditions of the theorem f ""(x) < 0. Определим знак произведения второго и третьего сомножителей.
Thus, any point on the curve lies below the tangent to the curve for all values x And x 0 Î ( a; b), which means that the curve is convex. The second part of the theorem is proved in a similar way.
Examples.
Graph point continuous function, separating 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 determine sufficient conditions for the fact that given point the curve is the point of inflection.
Theorem. Let the curve be defined by the equation y = f(x). If f ""(x 0) = 0 or f ""(x 0) does not exist even when passing through the value x = x 0 derivative f ""(x) changes sign, then the point in the graph of the function with the abscissa x = x 0 there is an inflection point.
Proof. Let f ""(x) < 0 при x < x 0 And f ""(x) > 0 at x > x 0. Then at x < x 0 the curve is convex, and when x > x 0– concave. Therefore, the point A, lying on the curve, with abscissa x 0 there is an inflection point. The second case can be considered similarly, when f ""(x) > 0 at x < x 0 And f ""(x) < 0 при x > x 0.
Thus, inflection points should be sought only among those points where the second derivative vanishes or does not exist.
Examples. Find inflection points and determine the intervals of convexity and concavity of curves.
ASYMPTOTES OF THE GRAPH OF THE FUNCTION
When studying a function, it is important to establish the shape of its graph at an unlimited distance 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.
The straight line is called asymptote function graphics y = f(x), if the distance from the variable point M graphics to this line when removing a point M to infinity tends to zero, i.e. a point on the graph of a function, as it tends to infinity, must indefinitely approach the asymptote.
A curve can approach its asymptote while remaining on one side of it or on different sides, infinite set once crossing the asymptote and moving from one side to the other.
If we denote by d the distance from the point M curve to the asymptote, then it is clear that d tends to zero as the point moves away M to infinity.
We will further distinguish between vertical and oblique asymptotes.
VERTICAL ASYMPTOTES
Let at x→ x 0 from any side function y = f(x) increases unlimitedly in absolute value, i.e. or or 
. Then from the definition of an asymptote it follows that the straight line x = x 0 is an asymptote. The opposite is also obvious, if the line x = x 0 is an asymptote, i.e. .
Thus, the vertical asymptote of the graph of the function y = f(x) is called a straight line if f(x)→ ∞ under at least one of the conditions x→ x 0– 0 or x → x 0 + 0, x = x 0
Therefore, to find the vertical asymptotes of the graph of the function y = f(x) need to find those values x = x 0, at which the function goes to infinity (suffers an infinite discontinuity). Then the vertical asymptote has the equation x = x 0.
Examples.
SLANT ASYMPTOTES
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) then and only when 
. A similar statement is true for x → –∞.
Proof. Let MP– length of the segment, equal to the distance from point M to asymptote. By condition . Let us 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 we graph a function, it is important to identify the convexity intervals and inflection points. We need them, along with the intervals of decrease and increase, to clearly represent the function in graphical form.
Understanding this topic requires knowledge of what the derivative of a function is and how to evaluate it to some order, as well as the ability to solve different types inequalities
At the beginning of the article, basic 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 arguments will be illustrated with examples of problem solutions.
Yandex.RTB R-A-339285-1 Definition 1
In the downward direction over a certain interval in the case when its graph is located not lower than the tangent to it at any point in this interval.
Definition 2
The function to be differentiated is convex upward over a certain interval if the graph of a given function is located no higher than the tangent to it at any point in this interval.
A downward convex function can also be called a concave function. Both definitions are clearly shown in the graph below:
Definition 3
Inflection point of a function– this is a point M (x 0 ; f (x 0)), at which there is a tangent to the graph of the function, subject to the existence of a derivative in the vicinity of the point x 0, where from the left and right side the graph of the function takes different directions of convexity.
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 recommend repeating the section on the tangent of the graph of a function at a point.
Below is a graph of a function that has several inflection points, which are 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 a function will be convex downward or upward 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 the function. To do this, you simply need to solve the inequalities f "" (x) ≥ 0 and f "" (x) ≤ 0 on the domain of definition of the corresponding function.
Let us clarify that those points at which the second derivative does not exist, but the function y = f (x) is defined, will be included in the convexity and concavity intervals.
Let's look at an example specific task how to correctly apply this theorem.
Example 1
Condition: given the 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 definition of this function is the entire set of 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 definition of the second derivative coincides with the domain of the function itself. This means that 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 a concavity on the segment [ 2 ; + ∞) and convexity on the segment (- ∞; 2 ] .
For clarity, let’s draw a graph of the function and mark the convex part in blue 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 definition of the second derivative does not coincide with the domain of definition of the function? Here the remark made above will be useful to us: we will also include those points where the finite second derivative does not exist in the concavity and convex segments.
Example 2
Condition: given the function y = 8 x x - 1 . Determine in which intervals its graph will be concave and in which it will be convex.
Solution
First, let's find out the domain of definition 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 definition of the second derivative is the set x ∈ (0 ; 1) ∪ (1 ; + ∞) . We see that x equal to zero will belong to the domain of the original function, but not to the domain of the second derivative. This point must be included in the concavity or convex segment.
After this, we need to solve the inequalities f "" (x) ≥ 0 and f "" (x) ≤ 0 on the domain of definition of the given function. We use the interval method for this: with x = - 1 - 2 3 3 ≈ - 2, 1547 or x = - 1 + 2 3 3 ≈ 0, 1547 numerator 2 · (3 x 2 + 6 x - 1) x 2 3 · x - 1 3 becomes 0, and the denominator is 0 at x, equal to zero or unit.
Let's plot the resulting points on the graph and determine the sign of the expression on all intervals that will be included in the domain of definition of the original function. This area is indicated by shading on the graph. If the value is positive, we mark the interval with a plus, if negative, then with a minus.
Hence,
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 ; 1)
We include the previously marked point x = 0 and get the desired answer. The graph of the original function will be convex downwards at 0; - 1 + 2 3 3 ∪ (1 ; + ∞) , and upward – for x ∈ [ - 1 + 2 3 3 ; 1) .
Let's draw a graph, marking the convex part in blue and the concave part in red. Vertical asymptote marked with a black dotted line.
Answer: The graph of the original function will be convex downwards at 0; - 1 + 2 3 3 ∪ (1 ; + ∞) , and upward – for x ∈ [ - 1 + 2 3 3 ; 1) .
Conditions for inflection of a function graph
Let's start by formulating the necessary condition for the inflection of the graph of a certain function.
Definition 5
Let's say that we have a function y = f (x), the graph of which has an inflection point. At 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 at which the second derivative will turn to 0. This condition will not be sufficient: not all such points are suitable for us.
Also note that, according to general definition, we will need a tangent line, vertical or non-vertical. In practice, this means that to find inflection points, you should take those at which the second derivative of a given function turns to 0. Therefore, to find the abscissa of the inflection points, we need to take all x 0 from the domain of definition 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 becomes 0.
The first sufficient condition for the existence of an inflection point in the graph of a function
We have found all the values of x 0 that can be taken as abscissas of inflection points. After this, we need to apply the first sufficient inflection condition.
Definition 6
Let's say that we have a function y = f (x) which 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 on the left and right sides the second derivative acquires opposite signs, then this point can be considered an inflection point.
We see that this condition does not require that a second derivative necessarily exist at this point; its presence in the vicinity of the point x 0 is sufficient.
It is convenient to present everything said above in the form of 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) = ∞ .
- Let's 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, we will analyze two problems.
Example 3
Condition: given the 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 points and convexity points.
Solution
The specified function is defined on the entire set of real numbers. We calculate 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 definition of the first derivative. It is also the set of all real numbers. This means that 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 = - 2, x 2 = 1 + 25 2 = 3
We found the abscissa of two possible 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 number line 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 changes sign to the opposite (from plus to minus) at the point with abscissa 3, passing through it from left to right, and also does this (from minus to plus) at the point with abscissa 3. This means that we can conclude that x = - 2 and x = 3 are the abscissas of the inflection points of the function graph. They will correspond to graph points - 2; - 4 3 and 3; - 15 8 .
Let's take a look again at the image of the number axis and the resulting signs at the intervals in order to draw conclusions about the places of concavity and convexity. It turns out that the convexity will be located on the segment - 2; 3, and the concavity on the segments (- ∞; - 2 ] and [ 3; + ∞).
The solution to the problem is clearly depicted in the graph: Blue colour– convexity, red – concavity, black color means inflection points.
Answer: the convexity will be located on the segment - 2; 3, and the concavity on the segments (- ∞; - 2 ] and [ 3; + ∞).
Example 4
Condition: calculate the abscissa 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 definition of a 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 defined at a value of x equal to 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 may be the abscissa of the inflection point.
We calculate the second derivative. We also find the domain 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 now have two more possible inflection points. Let's plot them all on the number line and mark the resulting intervals with signs:
The sign will change when passing through each indicated 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 condition for the inflection, we can determine the necessary points at which 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 different types tasks.
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 whether the graph of the function will have an inflection point at point 3; 4 5 .
Solution
The first thing to do is to make sure that this 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 given function is defined for all arguments that are real numbers. Let's 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 found 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: find the third derivative and find out whether 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 function graph.
Answer: Let's show the solution in the illustration:
Let us assume 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 obtain that x 0 is the abscissa of the inflection point of the graph y = f (x).
Example 6
Condition: given the 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. We calculate the derivative: y " = ((x - 3) 5 + 1) " = 5 x - 3 4 . Since it will also be determined for everyone real values argument, then at any point in its graph there will be a non-vertical tangent.
Now let’s calculate at 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 found that at x = 3 the graph of the function may have an inflection point. Let's 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 by the third sufficient condition. This even number, which means that x = 3 will be the abscissa of the inflection point and the graph point of the function (3; 1) corresponds to it.
Answer: Here is a graph of this function with the convexities, concavities and inflection point marked:
If you notice an error in the text, please highlight it and press Ctrl+Enter
It remains to consider convexity, concavity and kinks of the graph. Let's start with the sites visitors love so much physical exercise. Please stand up and lean forward or backward. This is a bulge. Now stretch your arms out in front of you, palms up, and imagine that you are holding a large log on your chest... ...well, if you don’t like the log, let something/someone else do it =) This is concavity. A number of sources contain synonymous terms bulge up And bulge down, but I'm a fan of short titles.
! Attention
: some authors determine convexity and concavity exactly the opposite. This is also mathematically and logically correct, but often completely incorrect from a substantive point of view, including at the level of our layman’s understanding of the terms. So, for example, a lens with tubercles is called a biconvex lens, but not with depressions (biconcave).
And, say, a “concave” bed - it still clearly does not “stick up” =) (however, if you climb under it, then we will already talk about convexity; =)) I adhere to an approach that corresponds to natural human associations.
The formal definition of convexity and concavity of a graph is quite difficult for a teapot, so we will limit ourselves to a geometric interpretation of the concept on specific examples. Consider the graph of a function that continuous on the entire number line: 
It's easy to build with geometric transformations, and, probably, many readers are aware of how it is obtained from a cubic parabola.
Let's call chord line connecting two various points graphic arts.
The graph of a function is convex on some interval, if it is located not less any chord of a given interval. The experimental line is convex on , and, obviously, here any part of the graph is located ABOVE its chord. To illustrate the definition, I drew three black lines.
Graph functions are concave on the interval, if it is located not higher any chord of this interval. In the example under consideration, the patient is concave at the interval . A pair of brown segments convincingly demonstrates that here any piece of the graph is located UNDER its chord.
The point on the graph at which it changes from convex to concave or concavity to convexity is called inflection point. We have it in a single copy (the first case), and, in practice, by the inflection point we can mean both the green point belonging to the line itself and the “X” value.
IMPORTANT! The kinks of the graph should be drawn carefully and very smooth. All kinds of “irregularities” and “roughness” are unacceptable. It just takes a little training.
The second approach to determining convexity/concavity in theory is given through tangents:
Convex on the interval the graph is located not higher tangent drawn to it at arbitrary point of this interval. Concave on the interval graph – not less any tangent on this interval.
The hyperbola is concave on the interval and convex on: 
When passing through the origin of coordinates, the concavity changes to convexity, but the point DO NOT COUNT inflection point, since the function not determined in it.
More rigorous statements and theorems on the topic can be found in the textbook, and we move on to the intense practical part:
How to find convexity intervals, concavity intervals
and inflection points of the graph?
The material is simple, stencilled and structurally repeats study of a function for an extremum.
The convexity/concavity of the graph characterizes second derivative functions.
Let the function be twice differentiable on some interval. Then:
– if the second derivative is on an interval, then the graph of the function is convex on this interval;
– if the second derivative is on an interval, then the graph of the function is concave on this interval.
Regarding the signs of the second derivative with respect to spaces educational institutions a prehistoric association is walking around: “–” shows that “you cannot pour water into the graph of a function” (convexity),
and “+” – “gives such an opportunity” (concavity).
Necessary condition of inflection
If at a point there is an inflection point in the graph of the function, That:
or the value does not exist(let's sort it out, read!).
This phrase implies that the function continuous at a point and in the case – is twice differentiable in some neighborhood of it.
The necessity of the condition suggests that the converse is not always true. That is, from equality (or non-existence of value) shouldn't yet the existence of an inflection in the graph of a function at point . But in both situations they call critical point of the second derivative.
Sufficient condition for inflection
If the second derivative changes sign when passing through a point, then at this point there is an inflection in the graph of the function.
There may be no inflection points (an example has already been met) at all, and in this sense some elementary examples are indicative. Let's analyze the second derivative of the function: 
A positive constant function is obtained, that is for any value of "x". Facts lying on the surface: the parabola is concave throughout domain of definition, there are no inflection points. It is easy to notice that the negative coefficient at “inverts” the parabola and makes it convex (as the second derivative, a negative constant function, will tell us).
Exponential function also concave at: 
for any value of "x".
Of course, the graph has no inflection points.
We examine the graph for convexity/concavity logarithmic function :
Thus, the branch of the logarithm is convex on the interval. The second derivative is also defined on the interval, but consider it IT IS FORBIDDEN, because the given interval not included in domain functions The requirement is obvious - since there is no logarithm graph there, then, naturally, there is no talk about any convexity/concavity/inflections.
As you can see, everything is really very reminiscent of the story with increasing, decreasing and extrema of the function. Similar to myself algorithm for studying the graph of a functionfor convexity, concavity and the presence of kinks:
2) Looking for critical values. To do this, take the second derivative and solve the equation. Points at which there is no 2nd derivative, but which are included in the domain of definition of the function itself, are also considered critical!
3) Mark on the number line all the found discontinuity points and critical points (there may be neither one nor the other - then there is no need to draw anything (as in too simple case), it is enough to limit yourself to a written comment). Interval method determine the signs on the resulting intervals. As just explained, one should consider only those intervals that are included in the domain of definition of the function. We draw conclusions about the convexity/concavity and inflection points of the function graph. We give the answer.
Try to verbally apply the algorithm to functions 
. In the second case, by the way, there is an example when there is no inflection point in the graph at the critical point. However, let's start with a little more difficult tasks:
Example 1
Solution:
1) The function is defined and continuous on the entire number line. Very good.
2) Let's find the second derivative. You can first perform cube construction, but it is much more profitable to use rule for differentiation of complex functions:
Please note that 
, which means the function is non-decreasing. Although this is not relevant to the task, it is always advisable to pay attention to such facts.
Let's find the critical points of the second derivative:
- critical point
3) Let's check the execution sufficient condition inflection. Let us determine the signs of the second derivative on the resulting intervals.
Attention! Now we are working with the second derivative (and not with a function!)
As a result, one critical point was obtained: .
3) Mark two discontinuity points on the number line, a critical point, and determine the signs of the second derivative on the resulting intervals:
I remind you important technique interval method, allowing you to significantly speed up the solution. Second derivative 
turned out to be very cumbersome, so it is not necessary to calculate its values, it is enough to make an “estimate” at each interval. Let us choose, for example, a point belonging to the left interval,
and perform the substitution: 
Now let's analyze the multipliers: 
Two “minus” and “plus” give “plus”, therefore, which means that the second derivative is positive over the entire interval.
The commented actions are easy to perform verbally. In addition, it is advantageous to ignore the factor altogether - it is positive for any “x” and does not affect the signs of our second derivative.
So, what information did you provide us with?
Answer: The graph of the function is concave at 
and convex on 
. At the origin (it's clear that ) there is an inflection point in the graph.
When passing through points, the second derivative also changes sign, but they are not considered inflection points, since the function suffers at them endless breaks.
In the analyzed example, the first derivative 
informs us about the growth of the function throughout domain of definition. There would always be such a freebie =) In addition, it is obvious that there are three asymptote. A lot of data has been obtained, which allows high degree present reliability appearance graphic arts. To the heap, the function is also odd. Based on the established facts, try to make a rough sketch. Picture at the end of the lesson.
Assignment for independent decision:
Example 6
Examine the graph of a function for convexity, concavity and find inflection points of the graph, if they exist.
There is no drawing in the sample, but it is not forbidden to put forward a hypothesis;)
We grind the material without numbering the points of the algorithm:
Example 7
Examine the graph of a function for convexity, concavity and find inflection points, if they exist.
Solution: function tolerates endless gap at point .
As usual, everything is fine with us: 
Derivatives are not the most difficult, the main thing is to be careful with their “hairstyle”.
In the induced marathon, two critical points of the second derivative are revealed: 
Let us determine the signs on the resulting intervals:
There is an inflection point in the graph at a point; let’s find the ordinate of the point: 
When passing through a point, the second derivative does not change sign, therefore, there is NO inflection in the graph.
Answer: convexity intervals: 
; concavity interval: ; inflection point: .
Let's consider final examples with additional bells and whistles:
Example 8
Find the intervals of convexity, concavity and inflection points of the graph
Solution: with finding domain of definition There are no special problems:
, while the function suffers discontinuities at points.
Let's go down the beaten path: 
- critical point.
Let's define the signs and consider the intervals only from the function domain:
There is an inflection point in the graph at a point; let’s calculate the ordinate: 
Instructions
Points inflection functions must belong to the domain of its definition, which must be found first. Schedule functions is a line that can be continuous or have breaks, monotonically decrease or increase, have minimum or maximum points(asymptotes), be convex or concave. A sharp change in the last two states is called an inflection point.
Prerequisite existence inflection functions consists in the equality of the second to zero. Thus, by differentiating the function twice and equating the resulting expression to zero, we can find the abscissa of possible points inflection.
This condition follows from the definition of the properties of convexity and concavity of the graph functions, i.e. negative and positive value second derivative. At the point inflection abrupt change these properties, which means that the derivative passes the zero mark. However, being equal to zero is not yet enough to indicate an inflection.
There are two sufficient conditions that the abscissa found at the previous stage belongs to the point inflection:Through this point you can draw a tangent to functions. The second derivative has different signs to the right and left of the expected one points inflection. Thus, its existence at the point itself is not necessary; it is enough to determine that at it it changes sign. Second derivative functions is equal to zero, and the third is not.
Solution: Find . IN in this case there are no restrictions, therefore, it is the entire space of real numbers. Calculate the first derivative: y’ = 3 ∛(x - 5) + (3 x + 3)/∛(x - 5)².
Pay attention to . It follows from this that the domain of definition of the derivative is limited. The point x = 5 is punctured, which means that a tangent can pass through it, which partly corresponds to the first sign of sufficiency inflection.
Determine the resulting expression for x → 5 – 0 and x → 5 + 0. They are equal to -∞ and +∞. You have proven that a vertical tangent passes through the point x=5. This point may turn out to be a point inflection, but first calculate the second derivative: Y'' = 1/∛(x - 5)² + 3/∛(x - 5)² – 2/3 (3 x + 3)/∛(x - 5)^5 = (2 x – 22)/∛(x - 5)^5.
Omit the denominator since you have already taken into account the point x = 5. Solve the equation 2 x – 22 = 0. It has a single root x = 11. The last step is to confirm that points x=5 and x=11 are points inflection. Analyze the behavior of the second derivative in their vicinity. Obviously, at the point x = 5 it changes sign from “+” to “-”, and at the point x = 11 - vice versa. Conclusion: both points are points inflection. The first sufficient condition is satisfied.