Find the intervals of convexity of the function. Convexity and Concavity Intervals of a Function Plot
Using the online calculator, you can find inflection points and convexity intervals of the function graph with the design of the solution in Word. Whether a function of two variables f (x1, x2) is convex is solved using the Hesse matrix.
Function entry rules:
The direction of the convexity of the graph of the function. Inflection points
Definition: A curve y = f (x) is called convex downward in the interval (a; b) if it lies above the tangent at any point of this interval.Definition: A curve y = f (x) is called convex upward in the interval (a; b) if it lies below the tangent at any point of this interval.
Definition: The intervals in which the graph of a function is turned upward or downward, are called the intervals of the convexity of the graph of the function.
The downward or upward convexity of the curve, which is the graph of the function y = f (x), is characterized by the sign of its second derivative: if in some interval f ’’ (x)> 0, then the curve is convex downward in this interval; if f ’’ (x)< 0, то кривая выпукла вверх на этом промежутке.
Definition: The point of the graph of the function y = f (x), separating the intervals of the convexity of the opposite directions of this graph, is called the inflection point.
Only critical points of the second kind can serve as inflection points, i.e. points belonging to the domain of definition of the function y = f (x), at which the second derivative f '' (x) vanishes or has a discontinuity.
The rule for finding the inflection points of the graph of the function y = f (x)
- Find the second derivative f '' (x).
- Find critical points of the second kind of the function y = f (x), i.e. the point at which f '' (x) vanishes or breaks.
- Investigate the sign of the second derivative f '' (x) in the interval into which the found critical points divide the domain of definition of the function f (x). If in this case the critical point x 0 separates the intervals of the convexity of opposite directions, then x 0 is the abscissa of the inflection point of the graph of the function.
- Calculate the values of the function at the inflection points.
Example 1. Find the intervals of convexity and inflection points of the following curve: f (x) = 6x 2 –x 3.
Solution: Find f ’(x) = 12x - 3x 2, f’ ’(x) = 12 - 6x.
Find the critical points by the second derivative by solving the equation 12-6x = 0. x = 2.
f (2) = 6 * 2 2 - 2 3 = 16
Answer: The function is convex upward for x∈ (2; + ∞); the function is convex downward for x∈ (-∞; 2); inflection point (2; 16).
Example 2. Does the function have inflection points: f (x) = x 3 -6x 2 + 2x-1
Example 3. Find the intervals on which the graph of the function is convex and curved: f (x) = x 3 -6x 2 + 12x + 4
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 decrease (increase) on this interval.
Theorem 2. Let the function be defined and continuous together with its derivative on and has 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 a function.
Need. Let's take an arbitrary point. Expand the function near a point in a Taylor series
Equation of the tangent to a curve at a point with 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 a point. By virtue of continuity, if , then also for, belonging to a sufficiently small neighborhood of the point, and therefore, obviously, for any value other than the value belonging to the indicated neighborhood.
Hence, the graph of the function lies above the tangent line 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 a point into a Taylor series
The excess of the curve over the tangent to it at the point having the abscissa, determined by the expression is
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. Explore the convexity (concavity) function.
Its derivative increases on the entire number axis, which means that 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 the graph of a continuous function is called the point that separates the intervals in which the function is convex and concave.
It follows from this definition that the inflection points are the extremum points of the first derivative. This implies the following statements for the necessary and sufficient inflection conditions.
Theorem (necessary inflection condition)... 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 is equal to zero ( ) or did not exist.
Theorem (sufficient inflection condition). If the second derivative of a twice differentiable function changes sign when passing through a certain point, that 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 vicinity 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 areas of convexity and concavity.
3.4.2.3 Investigation of the function for convexity and the presence of inflection points
1. Find the second derivative.
2. Find the points at which the second derivative or does not exist.
Rice. 3.9.
3. Investigate 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. Examine 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 an interval, then the function is convex on that interval.
On the interval, then the function is concave on this interval.
3.4.2.4 General scheme of the study of functions and plotting
When examining a function and plotting its graph, it is recommended to use the following scheme:
- Find the domain of the function.
- Investigate the function for evenness - oddness. Recall that the graph of an even function is symmetric about the ordinate axis, and the graph of an odd function is symmetric about the origin.
- Find the 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 the points of intersection with the coordinate axes.
The study of the function is carried out simultaneously with the construction of its graph.
Example. Explore function and build her schedule.
1. Function definition area -.
2. The investigated function is even , therefore, its graph is symmetrical about the ordinate 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. Extrema and intervals of monotony. Find the first derivative
For, therefore, the function decreases in these intervals.
For, therefore, the function increases in these intervals.
When, therefore, the point is the critical point.
Find the second derivative
Since, the point is the minimum point of the function.
6. Intervals of convexity and points of inflection.
Function at , so on this interval the function is concave.
The function at, means on these intervals the function is convex.
The function does not vanish anywhere, so there are no inflection points.
7. Points of intersection with coordinate axes.
The equation has a solution, which means the point of intersection of the function graph with the ordinate axis (0, 1).
The equation has no solution, so 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 graph of a function is called a straight line, which has the property that the distance from a point () to this straight line tends to 0 with an unlimited distance from the origin of the graph point.
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Conclusion.
An important feature of the considered method is that it is based primarily on the detection and study of characteristic features in the behavior of the curve. The places where the function changes smoothly are not studied in particular detail, and there is no need for such a study. But those places where the function has any peculiarities in behavior are subject to complete research and the most accurate graphic representation. These features are the points of maximum, minimum, points of discontinuity of the function, etc.
Determination of the direction of concavity and bends, as well as the indicated method of finding the asymptotes, make it possible to study functions in even more detail and get a more accurate idea of their graphs.
Instructions
The inflection points of the function must belong to the domain of its definition, which must be found first. The graph of a function is a line that can be continuous or have discontinuities, decrease or increase monotonically, have minimum or maximum points (asymptotes), be convex or concave. An abrupt change in the last two states is called an inflection.
A necessary condition for the existence of an inflection of a function is the equality of the second to zero. Thus, by differentiating the function twice and equating the resulting expression to zero, one can find the abscissas of possible inflection points.
This condition follows from the definition of the properties of convexity and concavity of the graph of a function, i.e. negative and positive values of the second derivative. At the inflection point, there is a sharp change in these properties, which means that the derivative goes over the zero mark. However, equality to zero is still not enough to denote an inflection.
There are two sufficient that the abscissa found at the previous stage belongs to the inflection point: Through this point, you can draw a tangent to the function. The second derivative has different signs to the right and left of the assumed inflection point. Thus, its existence at the point itself is not necessary, it is enough to determine that it changes sign at it. The second derivative of the function is equal to zero, and the third is not.
The first sufficient condition is universal and is used more often than others. Consider an illustrative example: y = (3 x + 3) ∛ (x - 5).
Solution: Find the scope. 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 the appearance of the fraction. It follows from this that the range 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 the sufficiency of the inflection.
Determine the one-sided limits for the resulting expression as x → 5 - 0 and x → 5 + 0. They are -∞ and + ∞. You proved that a vertical tangent passes through the point x = 5. This point may turn out to be an inflection point, 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 the points x = 5 and x = 11 are inflection points. Analyze the behavior of the second derivative in their vicinity. It is obvious that at the point x = 5 it changes its sign from "+" to "-", and at the point x = 11 - vice versa. Conclusion: both points are inflection points. The first sufficient condition is satisfied.