Definition of an increasing function.
Function y=f(x) increases over the interval X, if for any and 
inequality holds. In other words, a larger argument value corresponds to a larger function value.
Definition of a decreasing function.
Function y=f(x) decreases on the interval X, if for any and 
inequality holds 
. In other words, a larger value of the argument corresponds to a smaller value of the function.
NOTE: if the function is defined and continuous at the ends of the increasing or decreasing interval (a;b), that is, when x=a And x=b, then these points are included in the interval of increasing or decreasing. This does not contradict the definitions of an increasing and decreasing function on the interval X.
For example, from the properties of basic elementary functions we know that y=sinx defined and continuous for all real values of the argument. Therefore, from the increase in the sine function on the interval, we can assert that it increases on the interval.
Extremum points, extrema of a function.
The point is called maximum point functions y=f(x), if for everyone x from its neighborhood the inequality is valid. The value of the function at the maximum point is called maximum of the function and denote .
The point is called minimum point functions y=f(x), if for everyone x from its neighborhood the inequality is valid. The value of the function at the minimum point is called minimum function and denote .
The neighborhood of a point is understood as the interval 
, where is a sufficiently small positive number.
The minimum and maximum points are called extremum points, and the values of the function corresponding to the extremum points are called extrema of the function.
Do not confuse the extrema of a function with the largest and smallest values of the function.
In the first figure, the largest value of the function on the segment is reached at the maximum point and is equal to the maximum of the function, and in the second figure - the highest value of the function is achieved at the point x=b, which is not a maximum point.
Sufficient conditions for increasing and decreasing functions.
Based on sufficient conditions (signs) for the increase and decrease of a function, intervals of increase and decrease of the function are found.
Here are the formulations of the signs of increasing and decreasing functions on an interval:
if the derivative of the function y=f(x) positive for anyone x from the interval X, then the function increases by X;
if the derivative of the function y=f(x) negative for anyone x from the interval X, then the function decreases by X.
Thus, to determine the intervals of increase and decrease of a function, it is necessary:
Let's consider an example of finding the intervals of increasing and decreasing functions to explain the algorithm.
Example.
Find the intervals of increasing and decreasing functions.
Solution.
The first step is to find the definition of the function. In our example, the expression in the denominator should not go to zero, therefore, .
Let's move on to finding the derivative of the function: 
To determine the intervals of increase and decrease of a function based on a sufficient criterion, we solve inequalities on the domain of definition. Let's use a generalization of the interval method. The only real root of the numerator is x = 2, and the denominator goes to zero at x=0. These points divide the domain of definition into intervals in which the derivative of the function retains its sign. Let's mark these points on the number line. We conventionally denote by pluses and minuses the intervals at which the derivative is positive or negative. The arrows below schematically show the increase or decrease of the function on the corresponding interval. 
Let f be continuous on an interval and differentiable at interior points of this interval. Then there is an internal point from this segment such that the tangent to the graph of the function, drawn at the point with abscissa c, is parallel to the chord AB, where A(a;f(x)) and B(b;f(x)). Or: on a smooth arc AB there is always a point c at which the tangent is parallel to the chord connecting the ends of the arc.
Let f be continuous on an interval and differentiable at interior points of this interval. Then there is an interior point from this segment such that
Corollary 1: if a function f is continuous on the segment and its derivative is equal to zero inside this segment, then the function f is constant on the segment.
Corollary 2: If the functions f and g are continuous on an interval and have the same derivatives inside this interval, then they differ by a constant term.
2. A sufficient sign of increasing function:
If f[/](x)>0 at each point of the interval I, then the function f increases on the interval I.
3. A sufficient sign of decreasing function:
If f[/](x)
Let us prove these signs using the Lagrange formula:
Let's take any two numbers from the interval. Let it be. According to Lagrange's formula, there is a number such that.
The number c belongs to the interval I, since the points belong to this interval. If f[/](x)>0 for, then f[/](c) >0, and therefore - this follows from formula (1), since ->0. This proves that the functions f increase on the interval I. If f[/](x) 0. The function f decreases on the interval I is proved.
Example 1. find the intervals of increasing and decreasing function
2. Find the derivative of the function and its critical points: or
3. Mark the points of extrema on the numerical axis and find the intervals of increase and decrease of the function
Answer: - the function increases
The function is decreasing
Example 2. Examine the increasing (decreasing) function:
2. Find the derivative and extrema points of the function:
3. Mark the critical point on the number axis and find the intervals of increase (decrease) of the function:
Answer: - the function is decreasing
Function increases
II. Critical points. Signs of finding the maximum and minimum of a function.
1. Critical points
Definition: critical points of a function are internal points of the function's domain of definition at which its derivative is zero or does not exist.
No. 1. Find the critical points of the function f: a) g(x) =
Answer: , where; , where b) g(x) =
2. Signs of finding the maximum and minimum of a function.
Sign of maximum functions:
If the function f is continuous at the point x0, and f[/](x)>0 on the interval (a;x0) and f[/](x)
Or: if at the point x0 the derivative changes sign from plus to minus, then x0 is the maximum point.
Proof:
The derivative f[/](x)>0 on the interval (a;x0), and the function is continuous at the point x0, therefore the function f increases on the interval (a;x0], and therefore f(x)
On the interval [x0;c) the function decreases, and therefore f(x)
Signs of a minimum function:
If the function f is continuous at the point x0, and f[/](x) 0 on the interval (x0;b), then the point x0 is the minimum point of the function f.
Or: if at the point x0 the derivative changes sign from minus to plus, then x0 is the minimum point.
Proof:
Derivative f[/](x) f (x0) for all x from the interval (a; x0).
On the interval [x0;b), the function f increases, and therefore f(x) >f (x0) for everyone from the interval (a;b), that is, x0 is the minimum point of f.
III. Second derivative. Signs of convexity and concavity.
Let there be a second derivative at the point. Then, if, then the point is the minimum point, and if, then the point is the maximum point of the function.
If, then the bulge is directed downward. If, then the bulge is directed upward.
IV. Oblique asymptotes
Definition: A straight line is the slanted asymptote of the graph of a function, where and
Oblique asymptote equation
Vertical asymptotes equation of oblique asymptote
V. Function Study Design
1. Let's find the domain of definition of the function.
2. Examine the function for evenness (oddness).
3. Find the points of intersection of the graph with the coordinate axes and determine the intervals of constant sign of the function.
4. Find the derivative.
5. Find the extremum points of the function and the intervals of increase and decrease of the function.
6. Make a table.
7. Find the second derivative.
8. Find the inflection points of the function graph and establish the intervals of convexity and concavity of this graph.
9. Find asymptotes of the function graph, if necessary.
10. Construct a sketch of the graph of this function.
11. Find the set of function values.
VI. Examples for studying a function
2). It is impossible to talk about the parity of the function.
5) Find the extremum points of the function and the intervals of increase and decrease of the function:
Function increases
The function is decreasing
6) Let's make a table x
7) Find the second derivative
8) Find the inflection points: or
Bulge up
Bulge down
9) Let us find that oblique asymptotes do not exist. there are no oblique asymptotes.
10) Schedule
; x=2 - vertical asymptote
2). It is impossible to talk about the parity of the function
3) Find the points of intersection of the graph with the OX axis.
Let's find the points of intersection of the graph with the OU axis.
4) Find the derivative of the function:
5) Find the extremum points of the function and the points of increase and decrease of the function:
Function increases
The function is decreasing
6) Let's make a table x
7) Find the second derivative:
8) Find the inflection points: there are no inflection points
Bulge down
Bulge up
Oblique asymptote equation
10) Schedule
Vertical asymptote
2) we cannot talk about the parity of the function
There are no points of intersection with the OX axis.
Doesn't exist. There are no such points.
4) Find the derivative:
The function is decreasing
Function increases
6) Let's make a table:
7) Let's plot the function:
Vertical asymptote
2) - we cannot talk about the parity of the function
3) Find the points of intersection of the graph with the OX axis.
Let's find the points of intersection of the graph with the OY axis.
4) Find the derivative:
5) Find the extremum points of the function and the intervals of increase and decrease of the function.
There are no critical points.
There are no max and min points.
6) Let's make a table:
↘ 7) Find the second derivative:
8) Find the inflection points of the function graph and set the intervals of convexity and concavity:
There are no inflection points.
Bulge up
Bulge down
9) Find the oblique asymptotes:
Equation of horizontal asymptote, since k = 0.
10) Let's plot the function:
; - vertical asymptotes
2) - the function is odd, since. The graph is symmetrical about the origin.
3) Find the points of intersection of the graph with the OX axis.
Let's find the points of intersection of the graph with the OY axis.
4) Find the derivative:
5) Find the extremum points and intervals of increase and decrease of the function:
There is no solution.
The function is decreasing
Function increases
6) Let's make a table:
↘ Not noun.
↗ 7) Find the oblique asymptotes:
There are no oblique asymptotes.
8) Find the second derivative:
9) Find the inflection points: either or
Bulge down
Bulge up
10) Let's build a graph
VII. Historical information.
The end of the life of another creator of mathematical analysis, Gottfried Wilhelm Leibniz (1646 - 1716), was completely different. But first things first.
His ancestors came from Poland and bore the surname Lubenitz. After moving to Leipzig, their surname began to be pronounced in the German way. It is interesting to note that the very name of this city is also Slavic, it means >. Leibniz was born into the family of a professor of philosophy at the University of Leipzig. He lost his parents early: at the age of 6 he was left without a father, and at 17 - without a mother. During his school years, Leibniz amazed his teachers with his ability to compose poetry in Latin and Greek, his passion for philosophy and mathematics, he was very curious, he studied many subjects on his own, before meeting them at school, his memory was uneven. : he easily remembered complex things and worse - simple ones; he could not carry out calculations for a long time, but he gravitated towards generalizations and abstractions. And Leibniz retained such a memory and way of thinking throughout his life.
At the age of 15, Leibniz was a student at the Faculty of Philosophy at the University of Leipzig. This faculty was preparatory for law and theology. Having graduated brilliantly from the Faculty of Philosophy and then from the Faculty of Law, 20-year-old Leibniz was unable to get the desired position in his hometown. Conservative rules at the university created material barriers to obtaining a doctorate. He goes to Nuremberg and at the university there defends his legal dissertation for a doctorate with unprecedented success. The young scientist’s extraordinary talent was noticed. He is invited to the diplomatic service by the Elector (the prince who has the right to choose the king) of the city of Mainz, and later by the Duke of Hanover.
While in Paris on business for the Elector, Leibniz met with many famous scientists. Discussions of various problems awaken his interest in mathematics. Later, in a letter to I. Bernoulli, he recalled: >. After graduating from university (1666), Leibniz published a philosophical and mathematical work >, so when speaking about his >, he meant ignorance of the latest achievements of mathematics. To get acquainted with new results and ideas that arose at that time in mathematics, he turned to Huygens for help. He advises him to carefully study a number of works, and Leibniz gets down to business with enviable zeal: he studies the works of Saint-Vincent and Wallis, Descartes and Pascal, and engages in his own research.
But when he gets to London on diplomatic business and reports his results to English mathematicians, he is surprised to learn that many of these results are already known to them from Newton's manuscript, kept in the Royal Society. Leibniz, through the secretary of this society, Oldenburg (1615 - 1677), writes to Newton about his work. In the same letter he asks Newton to report his results. In response, he receives (again through Oldenburg) two letters in which Newton explains the operations of differentiation and integration using series.
Leibniz was in no hurry to publish his results in the field of new calculus, perhaps awaiting Newton's publications. But in 1683 Tschirnhauz published an article on the quadrature of algebraic curves. It does not mention the name of Leibniz, although Tschirnhaus owed a lot to him in solving these issues. To maintain the palm in this area, Leibniz publishes an article next year >, and a year later - >. The first of them contained the basics of differential calculus, the second - integral.
He based the new science on the concept of differential. Now the differential df(x0) of the function y=f(x) at the point x0 is given by the formula df(xo) = f"(xo)dx, where f"(xb) is the derivative calculated at the point xo, their is the increment of the argument. Leibniz defines the differential as one of the legs of the characteristic triangle, which was discussed in the previous chapter (section 9). From Figure 46 it can be seen that these definitions are equivalent.
Leibniz gives rules for calculating the differential of a sum, difference, product, quotient, degree, and solves differential equations. He defines the integral as the sum of differentiations, emphasizing the mutual inverse nature of the operations of differentiation and integration: >. Where do the properties of integrals and methods of calculating them come from? In subsequent articles, Leibniz developed a new analysis. He proved that any integrable function is bounded (a necessary condition for integrability), and developed an algorithm for calculating certain types of integrals, in particular, a method for integrating rational functions. The importance of this method cannot be overestimated, since with the help of various substitutions to integrals of rational functions a wide variety of integrals can be reduced. Let's look at this method in more detail.
To graphically solve the problem of integrating arbitrary functions, Leibniz invented (1693) a mechanical device - an integrator. If you move one pin of this device along the graph of the function, the other draws the graph of the antiderivative.
We still use the algorithms and notations developed by Leibniz, as well as most of the mathematical terms he introduced: function, variable, constant, coordinates, abscissa, algorithm, differential, etc. Many of these terms were used before, but did not have the specific meaning that Leibniz gave them.
At the beginning of the next century, a heated debate broke out about the priority of the invention of analysis. The reason for it was Leibniz's review (1704) of Newton's work, where he pointed out the ideological commonality of Newton and Fabry's interpretation of the infinitesimal. Such a comparison of the great Englishman with the little-known French mathematician O n o -re Fabry (1607 - 1688) caused the indignation of English scientists. (And Leibniz did not have any ulterior motives; Fabry’s book was simply one of the few that helped him eliminate > during the Parisian period.) They saw in this a belittlement of Newton’s merits, and so it began. In this dispute, Newton's rights were defended by English scientists, and Leibniz's by continental ones. The support of Leibniz by the majority of continental mathematicians was explained by the fact that his notations turned out to be so perfect, and the teaching itself so accessible, that they immediately found supporters among many scientists in Europe, which is extremely rare when a new theory appears.
Apparently, it was precisely this dispute that the wonderful Russian poet Valery Bryusov had in mind when he wrote the following lines:
O Leibniz, O sage, creator of prophetic books! You were above the world, like the ancient prophets. Your age, marveling at you, did not reach the prophecies And mixed crazy reproaches with flattery.
In fact, the claims of both sides were unfounded. Both scientists independently came to the creation of differential and integral calculus, and their approaches were completely different. Newton used the apparatus of power series, and Leibniz used the concept of differential. The heated dispute led to the fact that English mathematicians ignored everything that came from Leibniz and his school, and continental mathematicians ignored the work of the English. Since the continent relied on Leibniz’s symbolism, which was more advanced than Newton’s, and scientists were united by common ideas, published and accessible to everyone, continental mathematicians in the post-Newtonian period went far ahead in comparison with the English.
However, in the fate of Leibniz, the enmity between English and continental mathematicians played a fatal role. The Duke, for whom he served as a librarian, historian and biographer, having become the English king (1714), left for London. Leibniz could not follow him because of damaged relations with English mathematicians. In addition, the Duke was dissatisfied with his historiographer, believing that he did not pay enough attention to his direct official duties. Leibniz had to stay and work in the Duke's library. The disfavor of the newly crowned English king led to the fact that the scientist’s circle was greatly thinned out. Two years later he died, accompanied on his last journey only by his secretary and gravediggers. The offensive injustice of fate in relation to the great scientist, who did a lot.
Despite his enormous busyness in compiling the history of the ducal house, which turned into the history of Western Europe, and other responsibilities distracting him from science, Leibniz left many works on mathematics, philosophy, biology, theory of knowledge, politics, law, and linguistics. A well-rounded scientist, he has made invaluable contributions to each of these areas. Ideas poured out of him as if from a cornucopia: every letter, every note or article contained something fundamentally new in the field of science under consideration, sometimes determining its further development. Much was done with his direct participation. In Berlin, he organized a scientific society, which was later transformed into the Berlin Academy of Sciences, and became its first president. He was the first foreign member of the Paris Academy of Sciences. Leibniz met repeatedly in Berlin with Peter I, for whom he developed a number of projects for the development of education and government in Russia, as well as the creation of the St. Petersburg Academy of Sciences.
But his most significant contribution was to mathematics. Having entered into it, he was able to completely transform it. After his work and the work of his closest associates, not only did mathematical analysis appear, but all of mathematics entered a new era.
1°. It is known that a constant function has a derivative equal to zero at each point of the segment. In full courses of analysis the opposite is proven, that a function f(x) is constant on the interval [a, b] if at each point of the interval its derivative f "(x) is equal to zero.
We illustrate this geometrically. If f" (x) = 0 at each point of the segment [a, b], then the tangent to the graph of the function y=f(x) in each of the points x (a ≤ x ≤ b) parallel to the axis Oh. When transitioning X from one value to its subsequent values dot M. the graph of the function, which is the point of contact of the tangent, is shifted to the right, but remains in the direction of the tangent drawn to the point M, since the tangent does not change its direction during this transition. As a result, on the segment [a, b]
graph of a function y=f(x) goes straight MN, parallel to the axis Oh, and the value of the function is equal to f(a), remains unchanged.
2°. If in between a
positive values. As a result, its limit is the derivative f "(x) - positive or equal to zero
f "(x) ≥ 0
If in between A<х function y=f(x) decreasing, then increasing X each subsequent value of the function is less than the previous one. Therefore, for any given value of x at the time the increment Δx positive, increment Δy negative, attitude Δy/Δx takes only negative values and when tending Δx to zero has as its limit a negative number or zero, i.e.
f "(x) ≤ 0.
Since the value of the derivative f "(x) equal to the slope of the tangent to the graph of the function y = f(x):
f "(x) = tanφ,
and for the increasing function f "(x) = tanφ ≥ 0, then the tangent to the graph of the increasing function forms with the axis Oh acute angle or parallel to axis Oh(drawing 106). In a decreasing function f "(x) = tanφ ≤ 0, the tangent to the graph forms with the axis Oh obtuse angle or parallel to the axis Oh(crap.).
In between a
Points on the graph of an increasing (or decreasing) function where the tangent is parallel to the axis Ox, are separate points in the sense that their abscissas do not form a segment. Fuck it. and damn. such points are R And R 1.
3°. In full courses of analysis the following sufficient signs of increasing and decreasing functions are proved:
function f(x) increases (or decreases) in the interval a
1) the derivative f "(x) is not negative (or not positive) in the interval a<х
f "(x) ≥ 0 (or f "(x) ≤ 0)
2) in this interval there is no segment a 1 ≤ x ≤ b 1 (a<а 1 .
4°. Example. Determine the intervals of increase and decrease of the function: y = x 3 - x 2 - 8x + 2.
Solution. To apply the signs of increasing and decreasing function, we find the derivative of this function and determine the values X, in which it is positive or negative:
y" = 3x 2 - 2x - 8.
Let us factorize the second-degree trinomial, since it is much easier to judge the sign of the product by the signs of the factors than the sign of the sum by the signs of the terms.
Trinomial roots:
|
y" =3(x+4/3)(x-2).
Multiplier x + 4/3 negative when X< - 4/3 и положителен при X> - 4/3. Factor X - 2 is negative at X< 2 и положителен при X> 2. The sign of the product will be one or the other depending on the location of the point X on the axis Oh relative to points -4/3 and 2.
Points -4/3 and 2 divide the entire axis into three spaces;
1) - ∞
To determine the sign of the derivative in each of the intervals, we make a table:
| Gap no. | Gap Characteristics | Sign x+4/3 | Sign x-2 | Sign f'(x) | This function |
| - ∞ < x< - 4/3 | - | - | + | increases | |
| -4/3 < x < 2 | + | - | - | decreases | |
| 2 < х < + ∞ | + | + | + | increases |
Consequently, this function increases in the intervals
- ∞
The graph of this function is shown in Fig.
5°.Function y = x 3(device) has a derivative y = 3x 2, which is positive for any value X, different from zero. At x = 0 derivative y" = 0. Function y = x 3 increases in the interval - ∞
Maximum and minimum functions
Problems of finding the largest and smallest values of quantities are important in technology and, as is clear from the examples, come down to finding the maximum and minimum of a function.
Definition. 1. The function f(x) has a maximum at x=c if its value at x=c is greater than at any other value of x taken in some neighborhood of the point x=c.
2. The function f(x) has a minimum at x=c if its value at x=c is less than at any other value of x taken in some neighborhood of the point x=c.
The terms “maximum” and “minimum” are combined into one common term “extremum”.
The argument value that gives the maximum (or minimum) of the function is called a maximum (minimum) point, or an extremum point.
A function can only have a maximum, for example the function y = 60x- 2x 2(Fig. 111), or only the minimum, for example the function y = 2x+72/x(Fig. 112), or have
maximum and minimum, such as the function y = x 3 - - x 2 - 8x+2(drawing 108). A function can have several maxima and minima (Fig. 113), and in this case the maxima and minima alternate. A function may have neither a maximum nor a minimum. For example, functions y = x 3, y = ctgx, y = a x have neither maximum nor minimum, since with increasing X from - ∞ to +∞ the first and third functions increase, and the second only decreases.
The maximum (minimum) of a function may not be its largest (smallest) value. So, depicted in the devil. 113 function has at point With. value greater than maximum with 1 M 1 And with 3 M 2, and at the point from 0 value less than minimum c 2 m 1, And c 4 m 2, minimum c 4 m 2 more than maximum with 1 M 1. The maximum (minimum) of a function at a given point is generally the largest (smallest) value of the function compared to its values at points lying to the left and right of the extremum point only in sufficient proximity to it.
To determine the nature of a function and talk about its behavior, it is necessary to find intervals of increase and decrease. This process is called function research and graphing. The extremum point is used when finding the largest and smallest values of a function, since at them the function increases or decreases from the interval.
This article reveals the definitions, formulates a sufficient sign of increase and decrease on the interval and a condition for the existence of an extremum. This applies to solving examples and problems. The section on differentiating functions should be repeated, because the solution will need to use finding the derivative.
Yandex.RTB R-A-339285-1 Definition 1
The function y = f (x) will increase on the interval x when, for any x 1 ∈ X and x 2 ∈ X, x 2 > x 1, the inequality f (x 2) > f (x 1) is satisfied. In other words, a larger value of the argument corresponds to a larger value of the function.
Definition 2
The function y = f (x) is considered to be decreasing on the interval x when, for any x 1 ∈ X, x 2 ∈ X, x 2 > x 1, the equality f (x 2) > f (x 1) is considered true. In other words, a larger function value corresponds to a smaller argument value. Consider the figure below.
Comment: When the function is definite and continuous at the ends of the interval of increasing and decreasing, that is (a; b), where x = a, x = b, the points are included in the interval of increasing and decreasing. This does not contradict the definition; it means that it takes place on the interval x.
The main properties of elementary functions of the type y = sin x are certainty and continuity for real values of the arguments. From here we get that the sine increases over the interval - π 2; π 2, then the increase on the segment has the form - π 2; π 2.
Definition 3The point x 0 is called maximum point for the function y = f (x), when for all values of x the inequality f (x 0) ≥ f (x) is valid. Maximum function is the value of the function at a point, and is denoted by y m a x .
The point x 0 is called the minimum point for the function y = f (x), when for all values of x the inequality f (x 0) ≤ f (x) is valid. Minimum functions is the value of the function at a point, and has a designation of the form y m i n .
Neighborhoods of the point x 0 are considered extremum points, and the value of the function that corresponds to the extremum points. Consider the figure below.
Extrema of a function with the largest and smallest value of the function. Consider the figure below.
The first figure says that it is necessary to find the largest value of the function from the segment [a; b ] . It is found using maximum points and is equal to the maximum value of the function, and the second figure is more like finding the maximum point at x = b.
Sufficient conditions for a function to increase and decrease
To find the maxima and minima of a function, it is necessary to apply signs of extremum in the case when the function satisfies these conditions. The first sign is considered the most frequently used.
The first sufficient condition for an extremum
Definition 4Let a function y = f (x) be given, which is differentiable in an ε neighborhood of the point x 0, and has continuity at the given point x 0. From here we get that
- when f " (x) > 0 with x ∈ (x 0 - ε ; x 0) and f " (x)< 0 при x ∈ (x 0 ; x 0 + ε) , тогда x 0 является точкой максимума;
- when f "(x)< 0 с x ∈ (x 0 - ε ; x 0) и f " (x) >0 for x ∈ (x 0 ; x 0 + ε), then x 0 is the minimum point.
In other words, we obtain their conditions for setting the sign:
- when the function is continuous at the point x 0, then it has a derivative with a changing sign, that is, from + to -, which means the point is called a maximum;
- when the function is continuous at the point x 0, then it has a derivative with a changing sign from - to +, which means the point is called a minimum.
To correctly determine the maximum and minimum points of a function, you must follow the algorithm for finding them:
- find the domain of definition;
- find the derivative of the function on this area;
- identify zeros and points where the function does not exist;
- determining the sign of the derivative on intervals;
- select points where the function changes sign.
Let's consider the algorithm by solving several examples of finding extrema of a function.
Example 1
Find the maximum and minimum points of the given function y = 2 (x + 1) 2 x - 2 .
Solution
The domain of definition of this function is all real numbers except x = 2. First, let's find the derivative of the function and get:
y " = 2 x + 1 2 x - 2 " = 2 x + 1 2 " (x - 2) - (x + 1) 2 (x - 2) " (x - 2) 2 = = 2 2 (x + 1) (x + 1) " (x - 2) - (x + 1) 2 1 (x - 2) 2 = 2 2 (x + 1) (x - 2 ) - (x + 2) 2 (x - 2) 2 = = 2 · (x + 1) · (x - 5) (x - 2) 2
From here we see that the zeros of the function are x = - 1, x = 5, x = 2, that is, each bracket must be equated to zero. Let's mark it on the number axis and get:
Now we determine the signs of the derivative from each interval. It is necessary to select a point included in the interval and substitute it into the expression. For example, points x = - 2, x = 0, x = 3, x = 6.
We get that
y " (- 2) = 2 · (x + 1) · (x - 5) (x - 2) 2 x = - 2 = 2 · (- 2 + 1) · (- 2 - 5) (- 2 - 2) 2 = 2 · 7 16 = 7 8 > 0, which means that the interval - ∞ - 1 has a positive derivative. Similarly, we find that.
y " (0) = 2 · (0 + 1) · 0 - 5 0 - 2 2 = 2 · - 5 4 = - 5 2< 0 y " (3) = 2 · (3 + 1) · (3 - 5) (3 - 2) 2 = 2 · - 8 1 = - 16 < 0 y " (6) = 2 · (6 + 1) · (6 - 5) (6 - 2) 2 = 2 · 7 16 = 7 8 > 0
Since the second interval turned out to be less than zero, it means that the derivative on the interval will be negative. The third with a minus, the fourth with a plus. To determine continuity, you need to pay attention to the sign of the derivative; if it changes, then this is an extremum point.
We find that at the point x = - 1 the function will be continuous, which means that the derivative will change sign from + to -. According to the first sign, we have that x = - 1 is a maximum point, which means we get
y m a x = y (- 1) = 2 (x + 1) 2 x - 2 x = - 1 = 2 (- 1 + 1) 2 - 1 - 2 = 0
The point x = 5 indicates that the function is continuous, and the derivative will change sign from – to +. This means that x = -1 is the minimum point, and its determination has the form
y m i n = y (5) = 2 (x + 1) 2 x - 2 x = 5 = 2 (5 + 1) 2 5 - 2 = 24
Graphic representation
Answer: y m a x = y (- 1) = 0, y m i n = y (5) = 24.
It is worth paying attention to the fact that the use of the first sufficient criterion for an extremum does not require the differentiability of the function at the point x 0, this simplifies the calculation.
Example 2
Find the maximum and minimum points of the function y = 1 6 x 3 = 2 x 2 + 22 3 x - 8.
Solution.
The domain of a function is all real numbers. This can be written as a system of equations of the form:
1 6 x 3 - 2 x 2 - 22 3 x - 8 , x< 0 1 6 x 3 - 2 x 2 + 22 3 x - 8 , x ≥ 0
Then you need to find the derivative:
y " = 1 6 x 3 - 2 x 2 - 22 3 x - 8 " , x< 0 1 6 x 3 - 2 x 2 + 22 3 x - 8 " , x >0 y " = - 1 2 x 2 - 4 x - 22 3 , x< 0 1 2 x 2 - 4 x + 22 3 , x > 0
The point x = 0 does not have a derivative, because the values of the one-sided limits are different. We get that:
lim y " x → 0 - 0 = lim y x → 0 - 0 - 1 2 x 2 - 4 x - 22 3 = - 1 2 (0 - 0) 2 - 4 (0 - 0) - 22 3 = - 22 3 lim y "x → 0 + 0 = lim y x → 0 - 0 1 2 x 2 - 4 x + 22 3 = 1 2 (0 + 0) 2 - 4 (0 + 0) + 22 3 = + 22 3
It follows that the function is continuous at the point x = 0, then we calculate
lim y x → 0 - 0 = lim x → 0 - 0 - 1 6 x 3 - 2 x 2 - 22 3 x - 8 = = - 1 6 · (0 - 0) 3 - 2 · (0 - 0) 2 - 22 3 (0 - 0) - 8 = - 8 lim y x → 0 + 0 = lim x → 0 - 0 1 6 x 3 - 2 x 2 + 22 3 x - 8 = = 1 6 (0 + 0) 3 - 2 · (0 + 0) 2 + 22 3 · (0 + 0) - 8 = - 8 y (0) = 1 6 x 3 - 2 x 2 + 22 3 x - 8 x = 0 = 1 6 · 0 3 - 2 0 2 + 22 3 0 - 8 = - 8
It is necessary to perform calculations to find the value of the argument when the derivative becomes zero:
1 2 x 2 - 4 x - 22 3 , x< 0 D = (- 4) 2 - 4 · - 1 2 · - 22 3 = 4 3 x 1 = 4 + 4 3 2 · - 1 2 = - 4 - 2 3 3 < 0 x 2 = 4 - 4 3 2 · - 1 2 = - 4 + 2 3 3 < 0
1 2 x 2 - 4 x + 22 3 , x > 0 D = (- 4) 2 - 4 1 2 22 3 = 4 3 x 3 = 4 + 4 3 2 1 2 = 4 + 2 3 3 > 0 x 4 = 4 - 4 3 2 1 2 = 4 - 2 3 3 > 0
All obtained points must be marked on a straight line to determine the sign of each interval. Therefore, it is necessary to calculate the derivative at arbitrary points for each interval. For example, we can take points with values x = - 6, x = - 4, x = - 1, x = 1, x = 4, x = 6. We get that
y " (- 6) = - 1 2 x 2 - 4 x - 22 3 x = - 6 = - 1 2 · - 6 2 - 4 · (- 6) - 22 3 = - 4 3< 0 y " (- 4) = - 1 2 x 2 - 4 x - 22 3 x = - 4 = - 1 2 · (- 4) 2 - 4 · (- 4) - 22 3 = 2 3 >0 y "(- 1) = - 1 2 x 2 - 4 x - 22 3 x = - 1 = - 1 2 (- 1) 2 - 4 (- 1) - 22 3 = 23 6< 0 y " (1) = 1 2 x 2 - 4 x + 22 3 x = 1 = 1 2 · 1 2 - 4 · 1 + 22 3 = 23 6 >0 y "(4) = 1 2 x 2 - 4 x + 22 3 x = 4 = 1 2 4 2 - 4 4 + 22 3 = - 2 3< 0 y " (6) = 1 2 x 2 - 4 x + 22 3 x = 6 = 1 2 · 6 2 - 4 · 6 + 22 3 = 4 3 > 0
The image on the straight line looks like
This means that we come to the conclusion that it is necessary to resort to the first sign of an extremum. Let us calculate and find that
x = - 4 - 2 3 3 , x = 0 , x = 4 + 2 3 3 , then from here the maximum points have the values x = - 4 + 2 3 3 , x = 4 - 2 3 3
Let's move on to calculating the minimums:
y m i n = y - 4 - 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = - 4 - 2 3 3 = - 8 27 3 y m i n = y (0) = 1 6 x 3 - 2 2 + 22 3 x - 8 x = 0 = - 8 y m i n = y 4 + 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = 4 + 2 3 3 = - 8 27 3
Let's calculate the maxima of the function. We get that
y m a x = y - 4 + 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = - 4 + 2 3 3 = 8 27 3 y m a x = y 4 - 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = 4 - 2 3 3 = 8 27 3
Graphic representation
Answer:
y m i n = y - 4 - 2 3 3 = - 8 27 3 y m i n = y (0) = - 8 y m i n = y 4 + 2 3 3 = - 8 27 3 y m a x = y - 4 + 2 3 3 = 8 27 3 y m a x = y 4 - 2 3 3 = 8 27 3
If a function f " (x 0) = 0 is given, then if f "" (x 0) > 0, we obtain that x 0 is a minimum point if f "" (x 0)< 0 , то точкой максимума. Признак связан с нахождением производной в точке x 0 .
Example 3
Find the maxima and minima of the function y = 8 x x + 1.
Solution
First, we find the domain of definition. We get that
D(y) : x ≥ 0 x ≠ - 1 ⇔ x ≥ 0
It is necessary to differentiate the function, after which we get
y " = 8 x x + 1 " = 8 x " (x + 1) - x (x + 1) " (x + 1) 2 = = 8 1 2 x (x + 1) - x 1 (x + 1) 2 = 4 x + 1 - 2 x (x + 1) 2 x = 4 - x + 1 (x + 1) 2 x
At x = 1, the derivative becomes zero, which means that the point is a possible extremum. To clarify, it is necessary to find the second derivative and calculate the value at x = 1. We get:
y "" = 4 - x + 1 (x + 1) 2 x " = = 4 (- x + 1) " (x + 1) 2 x - (- x + 1) x + 1 2 x " (x + 1) 4 x = = 4 (- 1) (x + 1) 2 x - (- x + 1) x + 1 2 " x + (x + 1) 2 x " (x + 1) 4 x = = 4 - (x + 1) 2 x - (- x + 1) 2 x + 1 (x + 1) " x + (x + 1) 2 2 x (x + 1) 4 x = = - (x + 1) 2 x - (- x + 1) x + 1 2 x + x + 1 2 x (x + 1) 4 x = = 2 3 x 2 - 6 x - 1 x + 1 3 x 3 ⇒ y "" (1) = 2 3 1 2 - 6 1 - 1 (1 + 1) 3 (1) 3 = 2 · - 4 8 = - 1< 0
This means that using the 2 sufficient condition for an extremum, we obtain that x = 1 is a maximum point. Otherwise, the entry looks like y m a x = y (1) = 8 1 1 + 1 = 4.
Graphic representation
Answer: y m a x = y (1) = 4 ..
Definition 5The function y = f (x) has its derivative up to the nth order in the ε neighborhood of a given point x 0 and its derivative up to the n + 1st order at the point x 0 . Then f " (x 0) = f "" (x 0) = f " " " (x 0) = . . . = f n (x 0) = 0 .
It follows that when n is an even number, then x 0 is considered an inflection point, when n is an odd number, then x 0 is an extremum point, and f (n + 1) (x 0) > 0, then x 0 is a minimum point, f (n + 1) (x 0)< 0 , тогда x 0 является точкой максимума.
Example 4
Find the maximum and minimum points of the function y y = 1 16 (x + 1) 3 (x - 3) 4.
Solution
The original function is a rational entire function, which means that the domain of definition is all real numbers. It is necessary to differentiate the function. We get that
y " = 1 16 x + 1 3 " (x - 3) 4 + (x + 1) 3 x - 3 4 " = = 1 16 (3 (x + 1) 2 (x - 3) 4 + (x + 1) 3 4 (x - 3) 3) = = 1 16 (x + 1) 2 (x - 3) 3 (3 x - 9 + 4 x + 4) = 1 16 (x + 1) 2 (x - 3) 3 (7 x - 5)
This derivative will go to zero at x 1 = - 1, x 2 = 5 7, x 3 = 3. That is, the points can be possible extremum points. It is necessary to apply the third sufficient condition for the extremum. Finding the second derivative allows you to accurately determine the presence of a maximum and minimum of a function. The second derivative is calculated at the points of its possible extremum. We get that
y "" = 1 16 x + 1 2 (x - 3) 3 (7 x - 5) " = 1 8 (x + 1) (x - 3) 2 (21 x 2 - 30 x - 3) y "" (- 1) = 0 y "" 5 7 = - 36864 2401< 0 y "" (3) = 0
This means that x 2 = 5 7 is the maximum point. Applying the 3rd sufficient criterion, we find that for n = 1 and f (n + 1) 5 7< 0 .
It is necessary to determine the nature of the points x 1 = - 1, x 3 = 3. To do this, you need to find the third derivative and calculate the values at these points. We get that
y " " " = 1 8 (x + 1) (x - 3) 2 (21 x 2 - 30 x - 3) " = = 1 8 (x - 3) (105 x 3 - 225 x 2 - 45 x + 93) y " " " (- 1) = 96 ≠ 0 y " " " (3) = 0
This means that x 1 = - 1 is the inflection point of the function, since for n = 2 and f (n + 1) (- 1) ≠ 0. It is necessary to investigate the point x 3 = 3. To do this, we find the 4th derivative and perform calculations at this point:
y (4) = 1 8 (x - 3) (105 x 3 - 225 x 2 - 45 x + 93) " = = 1 2 (105 x 3 - 405 x 2 + 315 x + 57) y (4) ( 3) = 96 > 0
From what was decided above, we conclude that x 3 = 3 is the minimum point of the function.
Graphic representation
Answer: x 2 = 5 7 is the maximum point, x 3 = 3 is the minimum point of the given function.
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Signs of local increase and decrease of a function.
One of the main tasks of studying a function is to find the intervals of its increase and decrease. Such a study can be easily carried out using the derivative. Let us formulate the corresponding statements.
A sufficient sign of increasing function.
If f’(x) > 0 at each point of the interval I, then the function f increases by I.
A sufficient sign of decreasing function.
If f'(x)< 0 в каждой точке интервала I, то функция f убывает на I.
The proof of these signs is carried out on the basis of the Lagrange formula (see paragraph 19). Take any two numbers x 1 and x 2 from the interval. Let x 1
(1)
The number c belongs to the interval I, since the points x 1 and x 2 belong to I. If f"(x)>0 for x∈I then f’(c)>0, and therefore F(x 1 )
The visual meaning of the signs is clear from physical reasoning (for definiteness, let us consider the sign of increase).
Let a point moving along the ordinate axis at time t have an ordinate y = f(t). Then the speed of this point at time t is equal to f"(t) (see. Instantaneous speed ). If f’ (t)>0 at each moment of time from the interval t, then the point moves in the positive direction of the ordinate axis, i.e. if t 1
Note 1.
If the function f is continuous at any end of the increasing (decreasing) interval, then this point is attached to this interval.
Note 2.
To solve the inequalities f" (x)>0 and f" (x)<0 удобно пользоваться обобщением метода интервалов (теоремой Дарбу) : точки, в которых производная равна 0 или не существует, разбивают область определения функции f на промежутки, в каждом из которых f" сохраняет постоянный знак. (Этот факт доказывается в курсах математического анализа.) Знак можно определить, вычислив значение f" в какой-нибудь точке промежутка.
Necessary and sufficient conditions for the existence of an extremum of a function at a point.
Necessary condition for extremum
The function g(x) at a point has an extremum (maximum or minimum) if the function is defined in a two-sided neighborhood of the point and for all points x of some region: , the inequality is satisfied accordingly
(in case of maximum) or (in case of minimum).
The extremum of the function can be found from the condition: if the derivative exists, i.e. we equate the first derivative of the function to zero.
Sufficient condition for an extremum
1) First sufficient condition:
a) f(x) is a continuous function and is defined in some neighborhood of a point such that the first derivative at this point is equal to zero or does not exist.
b) f(x) has a finite derivative in the vicinity of the specification and continuity of the function
c) the derivative retains a certain sign to the right of a point and to the left of the same point, then the point can be characterized as follows
This condition is not very convenient, since you need to check many conditions and memorize the table, but if nothing is said about derivatives of higher orders, then this is the only way to find the extremum of the function.
2) Second sufficient condition
If the function g(x) has a second derivative and at some point the first derivative is equal to zero, and the second derivative is different from zero. Then point extremum of the function g(x), and if , then the point is a maximum; if , then the point is a minimum.
