This is a rather interesting section of mathematics that absolutely all graduate students and students face. However, not everyone likes matan. Some fail to understand even basic things like the seemingly standard function study. This article aims to correct this oversight. Want to learn more about function analysis? Would you like to know what extremum points are and how to find them? Then this article is for you.
Investigation of the graph of a function
To begin with, it is worth understanding why it is necessary to analyze the chart at all. There are simple functions that are easy to draw. A striking example of such a function is the parabola. It's not hard to draw her chart. All that is needed is, using a simple transformation, to find the numbers at which the function takes the value 0. And in principle, this is all you need to know in order to draw a parabola graph.
But what if the function we need to graph is much more complicated? Since the properties of complex functions are rather non-obvious, it is necessary to carry out a whole analysis. Only then can the function be represented graphically. How to do it? You can find the answer to this question in this article.
Function analysis plan
The first thing to do is to conduct a superficial study of the function, during which we will find the domain of definition. So, let's start in order. The domain of definition is the set of those values by which the function is defined. Simply put, these are the numbers that can be used in the function instead of x. In order to determine the scope, you just need to look at the record. For example, it is obvious that the function y (x) \u003d x 3 + x 2 - x + 43 has a domain of definition - the set of real numbers. Well, with a function like (x 2 - 2x) / x, everything is a little different. Since the number in the denominator should not be equal to 0, then the domain of this function will be all real numbers, except for zero.
Next, you need to find the so-called zeros of the function. These are the values of the argument for which the entire function takes the value zero. To do this, it is necessary to equate the function to zero, consider it in detail and perform some transformations. Let us take the already familiar function y(x) = (x 2 - 2x)/x. From the school course, we know that a fraction is 0 when the numerator is zero. Therefore, we discard the denominator and start working with the numerator, equating it to zero. We get x 2 - 2x \u003d 0 and take x out of brackets. Hence x (x - 2) \u003d 0. As a result, we find that our function is equal to zero when x is equal to 0 or 2.
During the study of the graph of a function, many are faced with a problem in the form of extremum points. And it's weird. After all, extremes are a rather simple topic. Don't believe? See for yourself by reading this part of the article, in which we will talk about the minimum and maximum points.
To begin with, it is worth understanding what an extremum is. An extremum is the limit value that a function reaches on a graph. From this it turns out that there are two extreme values - a maximum and a minimum. For clarity, you can look at the picture above. On the investigated area, point -1 is the maximum of the function y (x) \u003d x 5 - 5x, and point 1, respectively, is the minimum.
Also, do not confuse concepts with each other. The extremum points of a function are those arguments at which the given function acquires extreme values. In turn, the extremum is the value of the minima and maxima of the function. For example, consider the figure above again. -1 and 1 are the extremum points of the function, and 4 and -4 are the extremums themselves.
Finding extremum points
But how do you find the extremum points of a function? Everything is pretty simple. The first thing to do is to find the derivative of the equation. Let's say we got the task: "Find the extremum points of the function y (x), x is the argument. For clarity, let's take the function y (x) \u003d x 3 + 2x 2 + x + 54. Let's differentiate and get the following equation: 3x 2 + 4x + 1. As a result, we got a standard quadratic equation. All that needs to be done is to equate it to zero and find the roots. Since the discriminant is greater than zero (D \u003d 16 - 12 \u003d 4), this equation is determined by two roots. We find them and get two values: 1/3 and -1. These will be the extremum points of the function. However, how can you still determine who is who? Which point is the maximum and which is the minimum? To do this, you need to take a neighboring point and find out its value. For example , let's take the number -2, which is to the left along the coordinate line from -1. We substitute this value in our equation y (-2) = 12 - 8 + 1 = 5. As a result, we got a positive number. This means that on the interval from 1/3 to -1 the function increases, which, in turn, means that on the intervals from min from infinity to 1/3 and from -1 to plus infinity, the function decreases. Thus, we can conclude that the number 1/3 is the minimum point of the function on the investigated interval, and -1 is the maximum point.
It is also worth noting that the exam requires not only to find extremum points, but also to carry out some kind of operation with them (add, multiply, etc.). It is for this reason that it is worth paying special attention to the conditions of the problem. After all, due to inattention, you can lose points.
From this article, the reader will learn about what an extremum of functional value is, as well as about the features of its use in practice. The study of such a concept is extremely important for understanding the foundations of higher mathematics. This topic is fundamental to a deeper study of the course.
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What is an extreme?
In the school course, many definitions of the concept of "extremum" are given. This article is intended to give the deepest and clearest understanding of the term for those who are ignorant of the issue. So, the term is understood to what extent the functional interval acquires a minimum or maximum value on a particular set.
The extremum is both the minimum value of the function and the maximum at the same time. There is a minimum point and a maximum point, that is, the extreme values of the argument on the graph. The main sciences in which this concept is used:
- statistics;
- machine control;
- econometrics.
Extreme points play an important role in determining the sequence of a given function. The coordinate system on the graph at its best shows the change in extreme position depending on the change in functionality.
Extrema of the derivative function
There is also such a thing as a "derivative". It is necessary to determine the extremum point. It is important not to confuse the minimum or maximum points with the largest and smallest values. These are different concepts, although they may seem similar.
The value of the function is the main factor in determining how to find the maximum point. The derivative is not formed from the values, but exclusively from its extreme position in one order or another.
The derivative itself is determined based on the data of the extreme points, and not the largest or smallest value. In Russian schools, the line between these two concepts is not clearly drawn, which affects the understanding of this topic in general.
Let's now consider such a thing as a "sharp extremum". To date, there is an acute minimum value and an acute maximum value. The definition is given in accordance with the Russian classification of critical points of a function. The concept of an extremum point is the basis for finding critical points on a chart.
To define such a concept, Fermat's theorem is used. It is the most important in the study of extreme points and gives a clear idea of their existence in one form or another. To ensure extremeness, it is important to create certain conditions for decreasing or increasing on the chart.
To accurately answer the question "how to find the maximum point", you must follow these provisions:
- Finding the exact area of definition on the chart.
- Search for the derivative of a function and an extremum point.
- Solve standard inequalities for the domain of the argument.
- Be able to prove in which functions a point on a graph is defined and continuous.
Attention! The search for a critical point of a function is possible only if there is a derivative of at least the second order, which is ensured by a high proportion of the presence of an extremum point.
Necessary condition for the extremum of the function
In order for an extremum to exist, it is important that there are both minimum points and maximum points. If this rule is observed only partially, then the condition for the existence of an extremum is violated.
Each function in any position must be differentiated in order to identify its new meanings. It is important to understand that the case when a point vanishes is not the main principle of finding a differentiable point.
A sharp extremum, as well as a function minimum, is an extremely important aspect of solving a mathematical problem using extreme values. In order to better understand this component, it is important to refer to the tabular values for the assignment of the functional.
| A complete exploration of meaning | Plotting a Value |
| 1. Determination of points of increase and decrease of values. 2. Finding break points, extremum and intersection with coordinate axes. 3. The process of determining changes in position on the chart. 4. Determination of the index and direction of convexity and convexity, taking into account the presence of asymptotes. 5. Creation of a summary table of the study in terms of determining its coordinates. 6. Finding intervals of increase and decrease of extreme and acute points. 7. Determination of the convexity and concavity of the curve. 8. Building a graph based on the study allows you to find a minimum or maximum. |
The main element, when it is necessary to work with extremums, is the exact construction of its graph. School teachers do not often pay maximum attention to such an important aspect, which is a gross violation of the educational process. The graph is built only on the basis of the results of the study of functional data, the definition of sharp extrema, as well as points on the graph. Sharp extrema of the derivative of a function are displayed on a plot of exact values using the standard procedure for determining asymptotes. |
Consider the graph of a continuous function y=f(x) shown in the figure.
Function value at point x 1 will be greater than the values of the function at all neighboring points both to the left and to the right of x one . In this case, the function is said to have at the point x 1 max. At the point x The 3 function obviously also has a maximum. If we consider the point x 2 , then the value of the function in it is less than all neighboring values. In this case, the function is said to have at the point x 2 minimum. Similarly for the point x 4 .
Function y=f(x) at the point x 0 has maximum, if the value of the function at this point is greater than its values at all points of some interval containing the point x 0 , i.e. if there is such a neighborhood of the point x 0 , which is for everyone x≠x 0 , belonging to this neighborhood, we have the inequality f(x)<f(x 0 ) .
Function y=f(x) It has minimum at the point x 0 , if there is such a neighborhood of the point x 0 , what is for everyone x≠x 0 belonging to this neighborhood, we have the inequality f(x)>f(x0.
The points at which the function reaches its maximum and minimum are called extremum points, and the values of the function at these points are the extrema of the function.
Let us pay attention to the fact that a function defined on a segment can reach its maximum and minimum only at points contained within the segment under consideration.
Note that if a function has a maximum at a point, this does not mean that at this point the function has the maximum value in the entire domain. In the figure discussed above, the function at the point x 1 has a maximum, although there are points at which the values of the function are greater than at the point x 1 . In particular, f(x 1) < f(x 4) i.e. the minimum of the function is greater than the maximum. From the definition of the maximum, it only follows that this is the largest value of the function at points sufficiently close to the maximum point.
Theorem 1. (A necessary condition for the existence of an extremum.) If the differentiable function y=f(x) has at the point x=x 0 extremum, then its derivative at this point vanishes.
Proof. Let, for definiteness, at the point x 0 the function has a maximum. Then for sufficiently small increments Δ x we have f(x 0 + Δ x)
Passing in these inequalities to the limit as Δ x→ 0 and taking into account that the derivative f "(x 0) exists, and hence the limit on the left does not depend on how Δ x→ 0, we get: for Δ x → 0 – 0 f"(x 0) ≥ 0 and at Δ x → 0 + 0 f"(x 0) ≤ 0. Since f"(x 0) defines a number, then these two inequalities are compatible only if f"(x 0) = 0.
The proved theorem states that the maximum and minimum points can only be among those values of the argument for which the derivative vanishes.
We have considered the case when a function has a derivative at all points of a certain segment. What happens when the derivative does not exist? Consider examples.
Examples.
- y=|x|.
The function does not have a derivative at a point x=0 (at this point, the graph of the function does not have a definite tangent), but at this point the function has a minimum, since y(0)=0, and for all x≠ 0y > 0.
- Let x< x
0 . Then c< x
0 and f "(c)> 0.
That's why f "(c)(x-x 0)<
0 and, therefore,
f(x) - f(x 0 )< 0, i.e. f(x)< f(x 0 ).
- Let x > x 0 . Then c>x 0 and f"(c)< 0. Means f "(c)(x-x 0)< 0. That's why f(x) - f(x 0 ) <0,т.е.f(x)< f(x 0 ) .
- Find the scope of a function f(x).
- Find the first derivative of a function f"(x).
- Determine critical points, for this:
- find the real roots of the equation f"(x)=0;
- find all values x under which the derivative f"(x) does not exist.
- Determine the sign of the derivative to the left and right of the critical point. Since the sign of the derivative remains constant between two critical points, it suffices to determine the sign of the derivative at any one point to the left and at one point to the right of the critical point.
- Calculate the value of the function at the extremum points.
- Find all critical points of a function in the interval ( a, b) and calculate the function values at these points.
- Calculate the values of the function at the ends of the segment for x=a, x=b.
- Of all the obtained values, choose the largest and smallest.
The function has no derivative at x=0, since it goes to infinity when x=0. But at this point, the function has a maximum.
The function has no derivative at x=0 because 
at x→0. At this point, the function has neither a maximum nor a minimum. Really, f(x)=0 and at x<0f(x)<0, а при x>0f(x)>0.
Thus, from the given examples and the formulated theorem it is clear that the function can have an extremum only in two cases: 1) at the points where the derivative exists and is equal to zero; 2) at the point where the derivative does not exist.
However, if at some point x 0 we know that f"(x 0 ) =0, then it cannot be concluded from this that at the point x 0 the function has an extremum.
For example. 
.
But point x=0 is not an extremum point, since to the left of this point the function values are located below the axis Ox, and above on the right.
Values of an argument from the domain of a function, for which the derivative of the function vanishes or does not exist, are called critical points.
It follows from the foregoing that the extremum points of a function are among the critical points, and, however, not every critical point is an extremum point. Therefore, to find the extremum of the function, you need to find all the critical points of the function, and then examine each of these points separately for maximum and minimum. For this, the following theorem serves.
Theorem 2. (A sufficient condition for the existence of an extremum.) Let the function be continuous on some interval containing the critical point x 0 , and is differentiable at all points of this interval (except, perhaps, the point itself x 0). If, when passing from left to right through this point, the derivative changes sign from plus to minus, then at the point x = x 0 the function has a maximum. If, when passing through x 0 from left to right, the derivative changes sign from minus to plus, then the function has a minimum at this point.
Thus, if
Proof. Let us first assume that when passing through x 0, the derivative changes sign from plus to minus, i.e. for all x close to the point x 0 f "(x)> 0 for x< x 0 , f"(x)< 0 for x > x 0 . Let us apply the Lagrange theorem to the difference f(x) - f(x 0 ) = f "(c)(x- x 0), where c lies between x and x 0 .
Thus, for all values x close enough to x 0 f(x)< f(x 0 ) . And this means that at the point x 0 the function has a maximum.
The second part of the minimum theorem is proved similarly.
Let us illustrate the meaning of this theorem in the figure. Let f"(x 1 ) =0 and for any x, close enough to x 1 , the inequalities
f"(x)< 0 at x< x 1 , f "(x)> 0 at x > x 1 .
Then to the left of the point x 1 the function is increasing, and decreasing on the right, therefore, when x = x 1 function goes from increasing to decreasing, that is, it has a maximum.
Similarly, one can consider the points x 2 and x 3 .
Schematically, all of the above can be depicted in the picture:
The rule for studying the function y=f(x) for an extremum
Examples. Explore functions for minimum and maximum.
THE GREATEST AND MINIMUM FUNCTION VALUES ON THE INTERCEPT
the largest the value of a function on a segment is the largest of all its values on this segment, and least is the smallest of all its values.
Consider the function y=f(x) continuous on the segment [ a, b]. As is known, such a function reaches its maximum and minimum values, either on the boundary of the segment, or inside it. If the maximum or minimum value of the function is reached at the internal point of the segment, then this value is the maximum or minimum of the function, that is, it is reached at critical points.
Thus, we get the following the rule for finding the largest and smallest values of a function on the segment [ a, b] :
>> Extremes
Function extremum
Definition of extremum
Function y = f(x) is called increasing (waning) in some interval if for x 1< x 2 выполняется неравенство (f (x 1) < f (x 2) (f (x 1) >f(x2)).
If a differentiable function y \u003d f (x) on a segment increases (decreases), then its derivative on this segment f " (x )> 0
(f"(x)< 0).
Dot x about called local maximum point (minimum) of the function f (x ) if there is a neighborhood of the point x o, for all points of which the inequality f (x)≤ f (x o) (f (x)≥ f (x o )).
The maximum and minimum points are called extremum points, and the values of the function at these points are its extrema.
extremum points
Necessary conditions for an extremum . If point x about is an extremum point of the function f (x), then either f " (x o ) = 0, or f(x o ) does not exist. Such points are called critical, where the function itself is defined at the critical point. The extrema of a function should be sought among its critical points.
The first sufficient condition. Let x about - critical point. If f" (x ) when passing through the point x about changes the plus sign to minus, then at the point x o the function has a maximum, otherwise it has a minimum. If the derivative does not change sign when passing through a critical point, then at the point x about there is no extremum.
The second sufficient condition.
Let the function f(x) have
f" (x ) in the vicinity of the point x
about
and the second derivative at the very point x o. If f"(x o) = 0, >0 ( <0), то точка x o is a local minimum (maximum) point of the function f(x). If =0, then one must either use the first sufficient condition, or involve higher ones.
On a segment, the function y \u003d f (x) can reach the smallest or largest value either at critical points or at the ends of the segment.
Example 3.22.
Solution. Because f " (
Tasks for finding the extremum of a function
Example 3.23. a
Solution. x and y y
0
≤ x≤
> 0, while x >a /4 S " < 0, значит, в точке x=a
/4 функция S имеет максимум. Значение
functions sq..
units).
Example 3.24. p ≈
Solution. pp
S"
R = 2, H = 16/4 = 4.
Example 3.22.Find the extrema of the function f (x) = 2x 3 - 15x 2 + 36x - 14.
Solution. Because f " (x) \u003d 6x 2 - 30x +36 \u003d 6 (x -2) (x - 3), then the critical points of the function x 1 \u003d 2 and x 2 \u003d 3. Extreme points can only be at these points. Since when passing through the point x 1 \u003d 2, the derivative changes sign from plus to minus, then at this point the function has a maximum. When passing through the point x 2 \u003d 3, the derivative changes sign from minus to plus, therefore, at the point x 2 \u003d 3, the function has a minimum. Calculating the values of the function in points
x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f (2) = 14 and minimum f (3) = 13.
Example 3.23.It is necessary to build a rectangular area near the stone wall so that it is fenced off with wire mesh on three sides, and adjoins the wall on the fourth side. For this there is a linear meters of the grid. At what aspect ratio will the site have the largest area?
Solution.Denote the sides of the site through x and y. The area of the site is equal to S = xy. Let y is the length of the side adjacent to the wall. Then, by condition, the equality 2x + y = a must hold. Therefore y = a - 2x and S = x (a - 2x), where
0
≤ x≤ a /2 (the length and width of the pad cannot be negative). S "= a - 4x, a - 4x = 0 for x = a/4, whence
y \u003d a - 2 × a / 4 \u003d a / 2. Because the x = a /4 is the only critical point, let's check whether the sign of the derivative changes when passing through this point. For x a /4 S "> 0, while x >a /4 S " < 0, значит, в точке x=a
/4 функция S имеет максимум. Значение
functions S(a/4) = a/4(a - a/2) = a 2 /8 (sq..
units).
Since S is continuous on and its values at the ends of S(0) and S(a /2) are equal to zero, then the value found will be the largest value of the function. Thus, the most favorable aspect ratio of the site under the given conditions of the problem is y = 2x.
Example 3.24.It is required to make a closed cylindrical tank with a capacity of V=16 p ≈ 50 m 3. What should be the dimensions of the tank (radius R and height H) in order to use the least amount of material for its manufacture?
Solution.The total surface area of the cylinder is S = 2 p R(R+H). We know the volume of the cylinder V = p R 2 N Þ N \u003d V / p R 2 \u003d 16 p / p R 2 \u003d 16 / R 2. So S(R) = 2 p (R2+16/R). We find the derivative of this function:
S"(R) \u003d 2 p (2R- 16 / R 2) \u003d 4 p (R- 8 / R 2). S" (R) = 0 for R 3 = 8, therefore,
R = 2, H = 16/4 = 4.
Functions, it is not at all necessary to know about the presence of the first and second derivatives and understand their physical meaning. First you need to understand the following:
- the extrema of the function maximize or, conversely, minimize the value of the function in an arbitrarily small neighborhood;
- at the extremum point there should not be a discontinuity of the function.
And now the same thing, only in simple terms. Look at the tip of a ballpoint pen. If the pen is placed vertically, with the writing end up, then the very middle of the ball will be the extreme point - the highest point. In this case, we talk about the maximum. Now, if you turn the pen with the writing end down, then at the middle of the ball there will already be a minimum of the function. With the help of the figure given here, you can imagine the listed manipulations for a stationery pencil. So, the extrema of a function are always critical points: its maxima or minima. The adjacent section of the chart can be arbitrarily sharp or smooth, but it must exist on both sides, only in this case the point is an extremum. If the chart is present only on one side, this point will not be an extremum even if the extremum conditions are met on one of its sides. Now let's study the extrema of the function from a scientific point of view. In order for a point to be considered an extremum, it is necessary and sufficient that:
- the first derivative was equal to zero or did not exist at the point;
- the first derivative changes sign at this point.
The condition is interpreted somewhat differently from the point of view of higher-order derivatives: for a function differentiable at a point, it is sufficient that there is an odd-order derivative that is not equal to zero, while all lower-order derivatives must exist and be equal to zero. This is the simplest interpretation of theorems from textbooks. But for the most ordinary people, it is worth explaining this point with an example. The basis is an ordinary parabola. Immediately make a reservation, at the zero point it has a minimum. Just a little math:
- first derivative (X 2) | = 2X, for zero point 2X = 0;
- second derivative (2X) | = 2, for zero point 2 = 2.
In this simple way, the conditions that determine the extrema of the function both for first-order derivatives and for higher-order derivatives are illustrated. We can add to this that the second derivative is just the same derivative of an odd order, unequal to zero, which was discussed a little higher. When it comes to extrema of a function of two variables, the conditions must be met for both arguments. When generalization occurs, then partial derivatives come into play. That is, it is necessary for the presence of an extremum at a point that both first-order derivatives be equal to zero, or at least one of them does not exist. For the sufficiency of the presence of an extremum, an expression is investigated, which is the difference between the product of second-order derivatives and the square of the mixed second-order derivative of the function. If this expression is greater than zero, then there is an extremum, and if there is an equality to zero, then the question remains open, and additional research is needed.
