Sufficient signs of increasing and decreasing functions. Intervals of increasing and decreasing

1. Find the domain of the function

2. Find the derivative of the function

3. Equate the derivative to zero and find the critical points of the function

4. Mark critical points on the definition area

5. Calculate the sign of the derivative in each of the resulting intervals

6. Find out the behavior of the function in each interval.

Example: Find the intervals of increasing and decreasing functionf(x) = and the number of zeros of this function on the interval .

Solution:

1.D( f) = R

2. f"(x) =

D( f") = D( f) = R

3. Find the critical points of the function by solving the equation f"(x) = 0.

x(x – 10) = 0

critical points of a function x= 0 and x = 10.

4. Let's determine the sign of the derivative.

f"(x) + – +

f(x) 0 10x

in the intervals (-∞; 0) and (10; +∞) the derivative of the function is positive and at the points x= 0 and x = 10 function f(x) is continuous, therefore, this function increases on the intervals: (-∞; 0]; .

Let us determine the sign of the function values ​​at the ends of the segment.

f(0) = 3, f(0) > 0

f(10) = , f(10) < 0.

Since the function decreases on the segment and the sign of the function values ​​changes, then there is one zero of the function on this segment.

Answer: the function f(x) increases on the intervals: (-∞; 0]; ;

on the interval the function has one function zero.

2. Extremum points of the function: maximum points and minimum points. Necessary and sufficient conditions for the existence of an extremum of a function. Rule for studying a function for extremum .

Definition 1:The points at which the derivative is equal to zero are called critical or stationary.

Definition 2. A point is called a minimum (maximum) point of a function if the value of the function at this point is less (greater than) the nearest values ​​of the function.

It should be kept in mind that the maximum and minimum in this case are local.

In Fig. 1. Local maxima and minima are shown.

Theorem 1.(a necessary sign of the existence of an extremum of a function). If a function differentiable at a point has a maximum or minimum at this point, then its derivative at vanishes, .

Theorem 2.(a sufficient sign of the existence of an extremum of the function). If a continuous function has a derivative at all points of some interval containing a critical point (with the possible exception of this point itself), and if the derivative, when the argument passes from left to right through the critical point, changes sign from plus to minus, then the function at this point has a maximum, and when the sign changes from minus to plus, it has a minimum.

Very important information about the behavior of a function is provided by the increasing and decreasing intervals. Finding them is part of the process of examining the function and plotting the graph. In addition, the extreme 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 the function on a certain interval.

In this article we will give the necessary definitions, formulate a sufficient criterion for the increase and decrease of a 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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Increasing and decreasing function on an interval.

Definition of an increasing function.

The function y=f(x) increases on the interval X if for any and inequality holds. In other words, a larger value of the argument corresponds to a larger value of the function.

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 basic 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.

Extremum points, extrema of a function.

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 smallest values ​​of the function.

In the first figure, the greatest value of the 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 greatest value of the function is achieved at the point x=b, which is not the 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) is positive for any x from the interval X, then the function increases by X;
• if the derivative of the function y=f(x) is negative for any x from the interval X, then the function decreases on 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 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] .

Sufficient conditions for the extremum of a function.

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.

The first sufficient condition for an extremum.

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.

• We find the domain of definition of the function.
• We find the derivative of the function on the domain of definition.
• We determine the zeros of the numerator, the zeros of the denominator of the derivative and the points of the domain of definition in which the derivative does not exist (all listed points are called points of possible extremum, passing through these points, the derivative can just change its sign).
• These points divide the domain of definition of the function into intervals in which the derivative retains its sign. We determine the signs of the derivative on each of the intervals (for example, by calculating the value of the derivative of a function at any point in a particular interval).
• We select points at which the function is continuous and, passing through which, the derivative changes sign - these are the extremum points.

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:

.

The second sign of an extremum of a function.

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.

The final work in the form of the Unified State Exam for 11th graders necessarily contains tasks on calculating limits, intervals of decreasing and increasing derivatives of a function, searching for extremum points and constructing graphs. Good knowledge of this topic allows you to correctly answer several exam questions and not experience difficulties in further professional training.

The fundamentals of differential calculus is one of the main topics of modern school mathematics. She studies the use of the derivative to study the dependencies of variables - it is through the derivative that one can analyze the increase and decrease of a function without resorting to a drawing.

Comprehensive preparation of graduates for passing the Unified State Exam on the Shkolkovo educational portal will help you deeply understand the principles of differentiation - understand the theory in detail, study examples of solving typical problems and try your hand at independent work. We will help you close gaps in knowledge - clarify your understanding of the lexical concepts of the topic and the dependencies of quantities. Students will be able to review how to find intervals of monotonicity, which means the derivative of a function rises or decreases on a certain segment when boundary points are and are not included in the intervals found.

Before you begin directly solving thematic problems, we recommend that you first go to the “Theoretical Background” section and repeat the definitions of concepts, rules and tabular formulas. Here you can read how to find and write down each interval of increasing and decreasing function on the derivative graph.

All information offered is presented in the most accessible form for understanding, practically from scratch. The site provides materials for perception and assimilation in several different forms - reading, video viewing and direct training under the guidance of experienced teachers. Professional teachers will tell you in detail how to find the intervals of increasing and decreasing derivatives of a function using analytical and graphical methods. During the webinars, you will be able to ask any question you are interested in, both on theory and on solving specific problems.

Having remembered the main points of the topic, look at examples of increasing the derivative of a function, similar to the tasks in the exam options. To consolidate what you have learned, take a look at the “Catalog” - here you will find practical exercises for independent work. The tasks in the section are selected at different levels of difficulty, taking into account the development of skills. For example, each of them is accompanied by solution algorithms and correct answers.

By choosing the “Constructor” section, students will be able to practice studying the increase and decrease of the derivative of a function on real versions of the Unified State Examination, constantly updated to take into account the latest changes and innovations.

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.

The 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

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

• 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 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 ..

The 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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Extrema of the function

Definition 2

A point $x_0$ is called a maximum point of a function $f(x)$ if there is a neighborhood of this point such that for all $x$ in this neighborhood the inequality $f(x)\le f(x_0)$ holds.

Definition 3

A point $x_0$ is called a maximum point of a function $f(x)$ if there is a neighborhood of this point such that for all $x$ in this neighborhood the inequality $f(x)\ge f(x_0)$ holds.

The concept of an extremum of a function is closely related to the concept of a critical point of a function. Let us introduce its definition.

Definition 4

$x_0$ is called a critical point of the function $f(x)$ if:

1) $x_0$ - internal point of the domain of definition;

2) $f"\left(x_0\right)=0$ or does not exist.

For the concept of extremum, we can formulate theorems on sufficient and necessary conditions for its existence.

Theorem 2

Sufficient condition for an extremum

Let the point $x_0$ be critical for the function $y=f(x)$ and lie in the interval $(a,b)$. Let on each interval $\left(a,x_0\right)\ and\ (x_0,b)$ the derivative $f"(x)$ exists and maintains a constant sign. Then:

1) If on the interval $(a,x_0)$ the derivative is $f"\left(x\right)>0$, and on the interval $(x_0,b)$ the derivative is $f"\left(x\right)

2) If on the interval $(a,x_0)$ the derivative $f"\left(x\right)0$, then the point $x_0$ is the minimum point for this function.

3) If both on the interval $(a,x_0)$ and on the interval $(x_0,b)$ the derivative $f"\left(x\right) >0$ or the derivative $f"\left(x\right)

This theorem is illustrated in Figure 1.

Figure 1. Sufficient condition for the existence of extrema

Examples of extremes (Fig. 2).

Figure 2. Examples of extreme points

Rule for studying a function for extremum

2) Find the derivative $f"(x)$;

7) Draw conclusions about the presence of maxima and minima on each interval, using Theorem 2.

Increasing and decreasing function

Let us first introduce the definitions of increasing and decreasing functions.

Definition 5

A function $y=f(x)$ defined on the interval $X$ is said to be increasing if for any points $x_1,x_2\in X$ at $x_1

Definition 6

A function $y=f(x)$ defined on the interval $X$ is said to be decreasing if for any points $x_1,x_2\in X$ for $x_1f(x_2)$.

Studying a function for increasing and decreasing

You can study increasing and decreasing functions using the derivative.

In order to examine a function for intervals of increasing and decreasing, you must do the following:

1) Find the domain of definition of the function $f(x)$;

2) Find the derivative $f"(x)$;

3) Find the points at which the equality $f"\left(x\right)=0$ holds;

4) Find the points at which $f"(x)$ does not exist;

5) Mark on the coordinate line all the points found and the domain of definition of this function;

6) Determine the sign of the derivative $f"(x)$ on each resulting interval;

7) Draw a conclusion: on intervals where $f"\left(x\right)0$ the function increases.

Examples of problems for studying functions for increasing, decreasing and the presence of extrema points

Example 1

Examine the function for increasing and decreasing, and the presence of maximum and minimum points: $f(x)=(2x)^3-15x^2+36x+1$

Since the first 6 points are the same, let’s carry them out first.

1) Domain of definition - all real numbers;

2) $f"\left(x\right)=6x^2-30x+36$;

3) $f"\left(x\right)=0$;

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4) $f"(x)$ exists at all points of the domain of definition;

5) Coordinate line:

Figure 3.

6) Determine the sign of the derivative $f"(x)$ on each interval:

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