How to understand whether a graph is decreasing or increasing. Sufficient signs of increasing and decreasing function

Let a rectangular coordinate system be specified on a certain plane. The graph of some function , (X-domain of definition) is the set of points of this plane with coordinates, where .

To construct a graph, you need to depict on a plane a set of points whose coordinates (x;y) are related by the relation.

Most often, the graph of a function is some kind of curve.

The simplest way to plot a graph is to plot by points.

A table is compiled in which the value of the argument is in one cell, and the value of the function from this argument is in the opposite cell. Then the resulting points are marked on the plane, and a curve is drawn through them.

An example of constructing a function graph using points:

Let's build a table.

Now let's build a graph.

But in this way it is not always possible to construct a sufficiently accurate graph - for accuracy you need to take a lot of points. Therefore, various methods of studying the function are used.

The full research scheme of the function is familiarized with in higher educational institutions. One of the points of studying a function is to find the intervals of increase (decrease) of the function.

A function is called increasing (decreasing) on a certain interval if, for any x 2 and x 1 from this interval, such that x 2 >x 1.

For example, a function whose graph is shown in the following figure, on intervals increases, and decreases in the interval (-5;3). That is, in the intervals The schedule is going uphill. And in the interval (-5;3) “downhill”.

Another point in the study of function is the study of function for periodicity.

A function is called periodic if there is a number T such that .

The number T is called the period of the function. For example, the function is periodic, here the period is 2P, so

Examples of graphs of periodic functions:

The period of the first function is 3, and the second is 4.

A function is called even if Example of an even function y=x 2 .

A function is called odd if Example of an odd function y=x 3 .

The graph of an even function is symmetrical about the op-amp axis (axial symmetry).

The graph of an odd function is symmetrical about the origin (central symmetry).

Examples of graphs of an even (left) and odd (right) function.

Monotone

A very important property of a function is its monotonicity. Knowing this property of various special functions, it is possible to determine the behavior of various physical, economic, social and many other processes.

The following types of monotony of functions are distinguished:

1) function increases, if on a certain interval, if for any two points and this interval such that . Those. a larger argument value corresponds to a larger function value;

2) function decreases, if on a certain interval, if for any two points and this interval such that . Those. a larger argument value corresponds to a smaller function value;

3) function non-decreasing, if on a certain interval, if for any two points and this interval such that ;

4) function does not increase, if on a certain interval, if for any two points and this interval such that .

2. For the first two cases, the term “strict monotonicity” is also used.

3. The last two cases are specific and are usually specified as a composition of several functions.

4. Separately, we note that the increase and decrease of the graph of a function should be considered from left to right and nothing else.

2. Even/odd.

The function is called odd, if when the sign of the argument changes, it changes its value to the opposite. The formula for this looks like this . This means that after substituting “minus x” values into the function in place of all x’s, the function will change its sign. The graph of such a function is symmetrical about the origin.

Examples of odd functions are etc.

For example, the graph actually has symmetry about the origin:

The function is called even, if when the sign of the argument changes, it does not change its value. The formula for this looks like this. This means that after substituting “minus x” values into the function in place of all x’s, the function will not change as a result. The graph of such a function is symmetrical about the axis.

Examples of even functions are etc.

For example, let’s show the symmetry of the graph about the axis:

If a function does not belong to any of the specified types, then it is called neither even nor odd or function general view . Such functions have no symmetry.

Such a function, for example, is the linear function we recently considered with a graph:

3. A special property of functions is periodicity.

The fact is that the periodic functions that are considered in the standard school curriculum are only trigonometric functions. We have already talked about them in detail when studying the relevant topic.

Periodic function is a function that does not change its values when a certain constant non-zero number is added to the argument.

This minimum number is called period of the function and are designated by the letter .

The formula for this looks like this: .

Let's look at this property using the example of a sine graph:

Let us remember that the period of the functions and is , and the period and is .

As we already know, trigonometric functions with complex arguments may have a non-standard period. We are talking about functions of the form:

Their period is equal. And about the functions:

Their period is equal.

As you can see, to calculate a new period, the standard period is simply divided by the factor in the argument. It does not depend on other modifications of the function.

Limitation.

Function y=f(x) is called bounded from below on the set X⊂D(f) if there is a number a such that for any xϵX the inequality f(x) holds< a.

Function y=f(x) is called bounded from above on the set X⊂D(f) if there is a number a such that for any хϵХ the inequality f(x) holds< a.

If the interval X is not specified, then the function is considered to be limited over the entire domain of definition. A function bounded both above and below is called bounded.

The limitation of the function is easy to read from the graph. You can draw some line y=a, and if the function is higher than this line, then it is bounded from below.

If below, then accordingly above. Below is a graph of a function bounded below. Guys, try to draw a graph of a limited function yourself.

Topic: Properties of functions: intervals of increasing and decreasing; highest and lowest values; extremum points (local maximum and minimum), convexity of the function.

Intervals of increasing and decreasing.

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:

· find the domain of definition of the function;

· find the derivative of the function;

· solve inequalities on the domain of definition;

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 3

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 picture shows what you need to find highest value functions 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 4

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 ..

Definition 5

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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"Increasing and decreasing functions"

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.

Lesson progress

Today 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 the 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 the graph of a quadratic function, linear function, inverse proportionality, or 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.

Objective 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 the 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)

Derivative. If the derivative of a function is positive for any point in the interval, then the function increases; if it is negative, it decreases.

To find the intervals of increase and decrease of a function, you need to find its domain of definition, derivative, solve inequalities of the form F’(x) > 0 and F’(x)

Solution.

3. Solve the inequalities y’ > 0 and y’ 0;
(4 - x)/x³

Solution.
1. Let's find the domain of definition of the function. Obviously, the expression in the denominator must always be different from zero. Therefore, 0 is excluded from the domain of definition: the function is defined for x ∈ (-∞; 0)∪(0; +∞).

2. Calculate the derivative of the function:
y'(x) = ((3 x² + 2 x - 4)' x² – (3 x² + 2 x - 4) (x²)')/x^4 = ((6 x + 2) x² – (3 x² + 2 x - 4) 2 x)/x^4 = (6 x³ + 2 x² – 6 x³ – 4 x² + 8 x)/x^ 4 = (8 x – 2 x²)/x^4 = 2 (4 - x)/x³.

3. Solve the inequalities y’ > 0 and y’ 0;
(4 - x)/x³

4. The left side of the inequality has one real x = 4 and turns to at x = 0. Therefore, the value x = 4 is included in both the interval and the decreasing interval, and point 0 is not included.
So, the required function increases on the interval x ∈ (-∞; 0) ∪ .

Sources:

how to find decreasing intervals on a function

A function represents a strict dependence of one number on another, or the value of a function (y) on an argument (x). Each process (not only in mathematics) can be described by its own function, which will have characteristic features: intervals of decreasing and increasing, points of minimums and maximums, and so on.

You will need

- paper;
- pen.

Instructions

Example 2.
Find the intervals of decreasing f(x)=sinx +x.
The derivative of this function will be equal to: f’(x)=cosx+1.
Solving the inequality cosx+1

Interval monotony a function can be called an interval in which the function either only increases or only decreases. A number of specific actions will help to find such ranges for a function, which is often required in algebraic problems this kind.

Instructions

The first step in solving the problem of determining the intervals in which a function monotonically increases or decreases is to calculate this function. To do this, find out all the argument values (values along the x-axis) for which you can find the value of the function. Mark the points where discontinuities are observed. Find the derivative of the function. Once you have determined the expression that represents the derivative, set it equal to zero. After this, you should find the roots of the resulting . Not about the area of permissible.

The points at which the function or at which its derivative is equal to zero represent the boundaries of the intervals monotony. These ranges, as well as the points separating them, should be sequentially entered into the table. Find the sign of the derivative of the function in the resulting intervals. To do this, substitute any argument from the interval into the expression corresponding to the derivative. If the result is positive, the function in this range increases; otherwise, it decreases. The results are entered into the table.

In the line denoting the derivative of the function f’(x), the corresponding values of the arguments are written: “+” - if the derivative is positive, “-” - negative or “0” - equal to zero. In the next line, note the monotony of the original expression itself. An up arrow corresponds to an increase, and a down arrow corresponds to a decrease. Check the functions. These are the points at which the derivative is zero. An extremum can be either a maximum point or a minimum point. If the previous section of the function increased and the current one decreased, this is the maximum point. If the function was decreasing before a given point, but is now increasing, this is the minimum point. Enter the values of the function at the extremum points into the table.

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what is the definition of monotony

The behavior of a function that has a complex dependence on the argument is studied using the derivative. By the nature of the change in the derivative, you can find critical points and areas of growth or decrease of the function.