Very important information about the behavior of the function provide intervals of increasing and decreasing. Finding them is part of the process of examining the function and plotting the graph. In addition, the extremum points at which there is a change from increasing to decreasing or from decreasing to increasing are given Special attention when finding the largest and smallest values of a function on a certain interval.
In this article we will give necessary definitions, we formulate a sufficient sign of increasing and decreasing function on an interval and sufficient conditions for the existence of an extremum, and apply this entire theory to solving examples and problems.
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Definition of an increasing function.
The function y=f(x) increases on the interval X if for any and 
inequality holds. In other words, the larger value of the argument corresponds to higher value functions.
Definition of a decreasing function.
The 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, at x=a and x=b, then these points are included in the increasing or decreasing interval. This does not contradict the definitions of an increasing and decreasing function on the interval X.
For example, from the properties of the main elementary functions we know that y=sinx is 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.
The point is called maximum point function y=f(x) if the inequality is true for all x in its neighborhood. The value of the function at the maximum point is called maximum of the function and denote .
The point is called minimum point function y=f(x) if the inequality is true for all x in its neighborhood. 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 function values corresponding to the extremum points are called extrema of the function.
Do not confuse the extrema of a function with the largest and lowest value functions.
In the first picture highest value function on the segment is achieved at the maximum point and is equal to the maximum of the function, and in the second figure - the maximum value of the function is achieved at the point x=b, which is not the maximum point.
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:
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 function.
Solution.
The first step is to find the domain of 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. 
Thus, 
And 
.
At the point The x=2 function is defined and continuous, so it should be added to both the increasing and decreasing intervals. At the point x=0 the function is not defined, so we do not include this point in the required intervals.
We present a graph of the function to compare the results obtained with it.
Answer:
The function increases with 
, decreases on the interval (0;2] .
To find the maxima and minima of a function, you can use any of the three signs of extremum, of course, if the function satisfies their conditions. The most common and convenient is the first of them.
Let the function y=f(x) be differentiable in the -neighborhood of the point and continuous at the point itself.
In other words:
Algorithm for finding extremum points based on the first sign of extremum of a function.
There are too many words, let’s better look at a few examples of finding extremum points and extrema of a function using the first sufficient condition for the extremum of a function.
Example.
Find the extrema of the function.
Solution.
The domain of a function is the entire set of real numbers except x=2.
Finding the derivative: 
The zeros of the numerator are the points x=-1 and x=5, the denominator goes to zero at x=2. Mark these points on the number axis 
We determine the signs of the derivative at each interval; to do this, we calculate the value of the derivative at any of the points of each interval, for example, at the points x=-2, x=0, x=3 and x=6.
Therefore, on the interval the derivative is positive (in the figure we put a plus sign over this interval). Likewise 
Therefore, we put a minus above the second interval, a minus above the third, and a plus above the fourth.
It remains to select points at which the function is continuous and its derivative changes sign. These are the extremum points.
At the point x=-1 the function is continuous and the derivative changes sign from plus to minus, therefore, according to the first sign of extremum, x=-1 is the maximum point, the maximum of the function corresponds to it 
.
At the point x=5 the function is continuous and the derivative changes sign from minus to plus, therefore, x=-1 is the minimum point, the minimum of the function corresponds to it 
.
Graphic illustration.
Answer:
PLEASE NOTE: the first sufficient criterion for an extremum does not require differentiability of the function at the point itself.
Example.
Find extremum points and extrema of the function 
.
Solution.
The domain of a function is the entire set of real numbers. The function itself can be written as: 
Let's find the derivative of the function: 
At the point x=0 the derivative does not exist, since the values of the one-sided limits do not coincide when the argument tends to zero: 
At the same time, the original function is continuous at the point x=0 (see the section on studying the function for continuity): 
Let's find the value of the argument at which the derivative goes to zero: 
Let's mark all the obtained points on the number line and determine the sign of the derivative on each of the intervals. To do this, we calculate the values of the derivative at arbitrary points of each interval, for example, at x=-6, x=-4, x=-1, x=1, x=4, x=6.
That is, 
Thus, according to the first sign of an extremum, the minimum points are 
, the maximum points are 
.
We calculate the corresponding minima of the function 
We calculate the corresponding maxima of the function 
Graphic illustration.
Answer:
.
As you can see, this sign of an extremum of a function requires the existence of a derivative at least to the second order at the point.
Increasing and decreasing function function y = f(x) is called increasing on the interval [ a, b], if for any pair of points X And X", a ≤ x the inequality holds f(x) ≤
f (x"), and strictly increasing - if the inequality f (x) f(x"). Decreasing and strictly decreasing functions are defined similarly. For example, the function at = X 2 (rice.
, a) strictly increases on the segment , and (rice.
, b) strictly decreases on this segment. Increasing functions are designated f (x), and decreasing f (x)↓. In order for a differentiable function f (x) was increasing on the segment [ A, b], it is necessary and sufficient that its derivative f"(x) was non-negative on [ A, b]. Along with the increase and decrease of a function on a segment, we consider the increase and decrease of a function at a point. Function at = f (x) is called increasing at the point x 0 if there is an interval (α, β) containing the point x 0, which for any point X from (α, β), x> x 0 , the inequality holds f (x 0) ≤
f (x), and for any point X from (α, β), x 0 , the inequality holds f (x) ≤ f (x 0). The strict increase of a function at the point is defined similarly x 0 . If f"(x 0) >
0, then the function f(x) strictly increases at the point x 0 . If f (x) increases at each point of the interval ( a, b), then it increases over this interval. S. B. Stechkin.
Big Soviet encyclopedia. - M.: Soviet Encyclopedia. 1969-1978 .
Concepts mathematical analysis. The function f(x) is called the ratio of the numbers of different age groups of the population that increases on the segment AGE STRUCTURE OF THE POPULATION. Depends on birth and death rates, life expectancy of people... Big Encyclopedic Dictionary
Concepts of mathematical analysis. A function f(x) is said to be increasing on the segment if for any pair of points x1 and x2, a≤x1 ... encyclopedic Dictionary
Concepts of mathematics. analysis. Function f(x) is called. increasing on the segment [a, b], if for any pair of points x1 and x2, and<или=х1 <х<или=b, выполняется неравенство f(x1)
A branch of mathematics that studies derivatives and differentials of functions and their applications to the study of functions. Design of D. and. into an independent mathematical discipline is associated with the names of I. Newton and G. Leibniz (second half of 17 ... Great Soviet Encyclopedia
A branch of mathematics in which the concepts of derivative and differential and how they are applied to the study of functions are studied. Development of D. and. closely related to the development of integral calculus. Their content is also inseparable. Together they form the basis... ... Mathematical Encyclopedia
This term has other meanings, see function. The "Display" request is redirected here; see also other meanings... Wikipedia
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I Heart The heart (Latin cor, Greek cardia) is a hollow fibromuscular organ that, functioning as a pump, ensures the movement of blood in the circulatory system. Anatomy The heart is located in the anterior mediastinum (Mediastinum) in the Pericardium between... ... Medical encyclopedia
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"Increasing and decreasing function"
Lesson objectives:
1. Learn to find periods of monotony.
2. Development of thinking abilities that ensure analysis of the situation and development of adequate methods of action (analysis, synthesis, comparison).
3. Forming interest in the subject.
During the classesToday we continue to study the application of the derivative and consider the question of its application to the study of functions. Front work
Now let’s give some definitions to the properties of the “Brainstorming” function.
1. What is a function called?
2. What is the name of the variable X?
3. What is the name of the variable Y?
4. What is the domain of a function?
5. What is the value set of a function?
6. Which function is called even?
7. Which function is called odd?
8. What can you say about the graph of an even function?
9. What can you say about the graph of an odd function?
10. What function is called increasing?
11. Which function is called decreasing?
12. Which function is called periodic?
Mathematics is the study of mathematical models. One of the most important mathematical models is a function. There are different ways to describe functions. Which one is the most obvious?
– Graphic.
– How to build a graph?
- Point by point.
This method is suitable if you know in advance what the graph approximately looks like. For example, what is a graph quadratic function, linear function, inverse proportionality, function y = sinx? (The corresponding formulas are demonstrated, students name the curves that are graphs.)
But what if you need to plot a graph of a function or even more complex one? You can find multiple points, but how does the function behave between these points?
Place two dots on the board and ask students to show what the graph “between them” might look like:
Its derivative helps you figure out how a function behaves.
Open your notebooks, write down the number, great job.
The purpose of the lesson: learn how the graph of a function is related to the graph of its derivative, and learn to solve two types of problems:
1. Using the derivative graph, find the intervals of increase and decrease of the function itself, as well as the extremum points of the function;
2. Using the scheme of derivative signs on intervals, find the intervals of increase and decrease of the function itself, as well as the extremum points of the function.
Similar tasks are not in our textbooks, but are found in the tests of the unified state exam (parts A and B).
Today in the lesson we will look at a small element of the work of the second stage of studying the process, the study of one of the properties of the function - determining the intervals of monotonicity
To solve this problem, we need to recall some issues discussed earlier.
So, let's write down the topic of today's lesson: Signs of increasing and decreasing functions.
Signs of increasing and decreasing function:
If the derivative of a given function is positive for all values of x in the interval (a; b), i.e. f"(x) > 0, then the function increases in this interval.
If the derivative of a given function is negative for all values of x in the interval (a; b), i.e. f"(x)< 0, то функция в этом интервале убывает
The order of finding intervals of monotonicity:
Find the domain of definition of the function.
1. Find the first derivative of the function.
2. decide for yourself on the board
Find critical points, investigate the sign of the first derivative in the intervals into which the found critical points divide the domain of definition of the function. Find intervals of monotonicity of functions:
a) domain of definition,
b) find the first derivative:
c) find the critical points: ; , And
3. Let us examine the sign of the derivative in the resulting intervals and present the solution in the form of a table.
point to extremum points
Let's look at several examples of studying functions for increasing and decreasing.
A sufficient condition for the existence of a maximum is to change the sign of the derivative when passing through the critical point from “+” to “-”, and for a minimum from “-” to “+”. If, when passing through the critical point, the sign of the derivative does not change, then there is no extremum at this point
1. Find D(f).
2. Find f"(x).
3. Find stationary points, i.e. points where f"(x) = 0 or f"(x) does not exist.
(The derivative is 0 at the zeros of the numerator, the derivative does not exist at the zeros of the denominator)
4. Place D(f) and these points on the coordinate line.
5. Determine the signs of the derivative on each of the intervals
6. Apply signs.
7. Write down the answer.
Consolidation of new material.
Students work in pairs and write down the solution in their notebooks.
a) y = x³ - 6 x² + 9 x - 9;
b) y = 3 x² - 5x + 4.
Two people are working at the board.
a) y = 2 x³ – 3 x² – 36 x + 40
b) y = x4-2 x³
3. Lesson summary
Homework: test (differentiated)
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.
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.
Let 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
In other words, we obtain their conditions for setting the sign:
To correctly determine the maximum and minimum points of a function, you must follow the algorithm for finding them:
Let's consider the algorithm by solving several examples of finding extrema of a function.
Example 1
Find maximum and minimum points 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 less than zero, which means the derivative on the segment 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 image
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 image
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 image
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 obtain 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 image
Answer: x 2 = 5 7 is the maximum point, x 3 = 3 is the minimum point of the given function.
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Finding from the graph the intervals of increasing and decreasing quadratic function xy 0 11 The function is decreasing on the interval if a larger value of x corresponds to a smaller value of y, i.e., when moving from left to right, the graph goes down (click to view) The function is increasing on the interval if a larger x value corresponds to a larger y value, i.e., when moving from left to right, the graph goes up (click to view)
8 y x0 11 Find from the graph and write down the intervals of increase and decrease of the quadratic function. Please note that the graph of the quadratic function consists of two branches. The branches are connected to each other by the vertex of a parabola. When recording intervals of increasing and decreasing, the most main role the abscissa (x) of the vertices of the parabola will play Example 1. Consider the movement along each branch of the parabola separately: along the left branch, when moving from left to right, the graph goes down, which means the function decreases; along the right branch - the graph goes up, which means the function is increasing. Answer: decreasing interval (- ∞; -1 ]; increasing interval [ -1; +∞)
8 y x0 11 Find from the graph and write down the intervals of increase and decrease of the quadratic function Example 2. Consider the movement along each branch of the parabola separately: along the left branch, when moving from left to right, the graph goes up, which means the function increases; along the right branch - the graph goes down, which means the function is decreasing. Answer: interval of increase (- ∞; 3 ]; interval of decrease [ 3; +∞).
Tasks for independent decision(complete in notebook) Task 1 Task 2 Task 3 Task 4 Appendix
increasing interval (- ∞; -1 ]; decreasing interval [ -1; +∞). check the answer. Find from the graph and write down the intervals of increasing and decreasing quadratic function 88 y x0 1 11 watch the animation write the answer yourself
“decreasing interval (- ∞; 3 ]; increasing interval [ 3; +∞). Find from the graph and write down the intervals of increasing and decreasing quadratic function y x 11 0 8 2 watch the animation write down the answer check the answer yourself
Find from the graph and write down the intervals of increase and decrease of the quadratic function 8 y 0 1 1 x3 view the animation write down the answer yourself the interval of decrease (- ∞; 0 ]; interval of increase [ 0; +∞). check the answer
“Find from the graph and write down the intervals of increase and decrease of the quadratic function 8 1 y 01 x4 view the animation write down the answer yourself the interval of increase (- ∞; - 0. 5 ]; interval of decrease [ - 0. 5; + ∞). check the answer
Appendix The boundary point of the intervals of increasing and decreasing is the abscissa of the vertex of the parabola. The boundary point of the intervals of increasing and decreasing is always written in the answer with a square bracket, since the quadratic function is continuous
