Extremum of the derivative of a function. How to find the extremum (minimum and maximum points) of a function. Necessary condition for the extremum of the function

Passing in these inequalities to the limit as Δ x→ 0 and taking into account that the derivative f "(x 0) exists, and hence the limit on the left does not depend on how Δ x→ 0, we get: for Δ x → 0 – 0 f"(x 0) ≥ 0 and at Δ x → 0 + 0 f"(x 0) ≤ 0. Since f"(x 0) defines a number, then these two inequalities are compatible only if f"(x 0) = 0.

The proved theorem states that the maximum and minimum points can only be among those values ​​of the argument for which the derivative vanishes.

We have considered the case when a function has a derivative at all points of a certain segment. What happens when the derivative does not exist? Consider examples.

y=|x|.

The function does not have a derivative at a point x=0 (at this point, the graph of the function does not have a definite tangent), but at this point the function has a minimum, since y(0)=0, and for all x≠ 0y > 0.

Thus, from the given examples and the formulated theorem it is clear that the function can have an extremum only in two cases: 1) at the points where the derivative exists and is equal to zero; 2) at the point where the derivative does not exist.

However, if at some point x 0 we know that f"(x 0 ) =0, then it cannot be concluded from this that at the point x 0 the function has an extremum.

For example.

But point x=0 is not an extremum point, since to the left of this point the function values ​​are located below the axis Ox, and above on the right.

Values ​​of an argument from the domain of a function, for which the derivative of the function vanishes or does not exist, are called critical points.

It follows from the foregoing that the extremum points of a function are among the critical points, and, however, not every critical point is an extremum point. Therefore, to find the extremum of the function, you need to find all the critical points of the function, and then examine each of these points separately for maximum and minimum. For this, the following theorem serves.

Theorem 2. (A sufficient condition for the existence of an extremum.) Let the function be continuous on some interval containing the critical point x 0 , and is differentiable at all points of this interval (except, perhaps, the point itself x 0). If, when passing from left to right through this point, the derivative changes sign from plus to minus, then at the point x = x 0 the function has a maximum. If, when passing through x 0 from left to right, the derivative changes sign from minus to plus, then the function has a minimum at this point.

Thus, if

f"(x)>0 at x<x 0 and f"(x)< 0 at x > x 0 , then x 0 - maximum point;

Proof. Let us first assume that when passing through x 0, the derivative changes sign from plus to minus, i.e. for all x close to the point x 0 f "(x)> 0 for x< x 0 , f"(x)< 0 for x > x 0 . Let us apply the Lagrange theorem to the difference f(x) - f(x 0 ) = f "(c)(x- x 0), where c lies between x and x 0 .

Let be x< x 0 . Then c< x 0 and f "(c)> 0. So f "(c)(x-x 0)< 0 and, therefore,

f(x) - f(x 0 )< 0, i.e. f(x)< f(x 0 ).

Let be x > x 0 . Then c>x 0 and f"(c)< 0. Means f "(c)(x-x 0)< 0. So f(x) - f(x 0 ) <0,т.е.f(x)< f(x 0 ) .

Thus, for all values x close enough to x 0 f(x)< f(x 0 ) . And this means that at the point x 0 the function has a maximum.

The second part of the minimum theorem is proved similarly.

Let us illustrate the meaning of this theorem in the figure. Let be f"(x 1 ) =0 and for any x, close enough to x 1 , the inequalities

f"(x)< 0 at x< x 1 , f "(x)> 0 at x > x 1 .

Then to the left of the point x 1 the function is increasing, and decreasing on the right, therefore, when x = x 1 function goes from increasing to decreasing, that is, it has a maximum.

Similarly, one can consider the points x 2 and x 3 .

Schematically, all of the above can be depicted in the picture:

The rule for studying the function y=f(x) for an extremum

Find the scope of a function f(x).

Find the first derivative of a function f"(x).

Determine critical points, for this:

find the real roots of the equation f"(x)=0;

find all values x under which the derivative f"(x) does not exist.

Determine the sign of the derivative to the left and right of the critical point. Since the sign of the derivative remains constant between two critical points, it suffices to determine the sign of the derivative at any one point to the left and at one point to the right of the critical point.

Calculate the value of the function at the extremum points.

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 exploration and plotting. The extremum point is used when finding the largest and smallest values ​​of the function, since they increase or decrease the function from the interval.

This article reveals the definitions, we formulate a sufficient sign of increase and decrease on the interval and the condition for the existence of an extremum. This applies to solving examples and problems. The section on differentiation of functions should be repeated, because when solving it will be necessary to use finding the derivative.

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) will be feasible. 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 feasible. In other words, a larger function value corresponds to a smaller argument value. Consider the figure below.

Comment: When the function is defined and continuous at the ends of the ascending and descending interval, i.e. (a; b) where x = a, x = b, the points are included in the ascending and descending interval. This does not contradict the definition, which means that it takes place on the interval x.

The main properties of elementary functions of the type y = sin x are definiteness and continuity for real values ​​of the arguments. From here we get that the increase in the sine occurs on 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 a function y = f (x) when for all values ​​of x the inequality f (x 0) ≥ f (x) is true. Feature maximum is the value of the function at the point, and is denoted by y m a x .

The point x 0 is called the minimum point for the function y \u003d f (x) when for all values ​​of x the inequality f (x 0) ≤ f (x) is true. Feature Minimum is the value of the function at the point, and has the notation of the form y m i n .

The 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 the 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 equals the maximum value of the function, and the second figure is more like finding a maximum point at x = b.

Sufficient conditions for increasing and decreasing functions

To find the maxima and minima of a function, it is necessary to apply the signs of an extremum in the case when the function satisfies these conditions. The first feature is the most commonly used.

The first sufficient condition for an extremum

Let a function y = f (x) be given, which is differentiable in the ε neighborhood of the point x 0 , and has continuity at the given point x 0 . Hence 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 sign setting conditions:

• 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 that the point is called the maximum;
• when the function is continuous at the point x 0, then it has a derivative with a changing sign from - to +, which means that the point is called a minimum.

To correctly determine the maximum and minimum points of the 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 the points where the function changes sign.

Consider the algorithm on the example of solving several examples of finding the extrema of the function.

Example 1

Find the maximum and minimum points of the given function y = 2 (x + 1) 2 x - 2 .

Decision

The domain of this function is all real numbers except x = 2. First, we 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 \u003d - 1, x \u003d 5, x \u003d 2, that is, each bracket must be equated to zero. Mark on the number line and get:

Now we determine the signs of the derivative from each interval. It is necessary to select a point included in the interval, substitute it into the expression. For example, points x = - 2, x = 0, x = 3, x = 6.

We get that

y "(- 2) \u003d 2 (x + 1) (x - 5) (x - 2) 2 x \u003d - 2 \u003d 2 (- 2 + 1) (- 2 - 5) (- 2 - 2) 2 \u003d 2 7 16 \u003d 7 8 > 0, therefore, the interval - ∞; - 1 has a positive derivative. Similarly, we obtain 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 segment will be negative. The third with a minus, the fourth with a plus. To determine continuity, it is necessary to pay attention to the sign of the derivative, if it changes, then this is an extremum point.

We get 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 the 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 +. Hence, x=-1 is the minimum point, and its finding 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 sign of an extremum does not require the function to be differentiable from the point x 0 , and 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 .

Decision.

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 has no 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

Calculations must be made 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 points obtained must be marked on the 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) \u003d - 1 2 x 2 - 4 x - 22 3 x \u003d - 6 \u003d - 1 2 - 6 2 - 4 (- 6) - 22 3 \u003d - 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 a straight line has the form

So, we come to the point that it is necessary to resort to the first sign of an extremum. We calculate and get 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 us 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 the function f "(x 0) = 0 is given, then with its f "" (x 0) > 0 we get that x 0 is the 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 .

Decision

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

When x = 1, the derivative becomes equal to zero, which means that the point is a possible extremum. For clarification, it is necessary to find the second derivative and calculate the value at x \u003d 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

Hence, using the 2 sufficient condition for the extremum, we obtain that x = 1 is the maximum point. Otherwise, the entry is 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 the 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)(x0)< 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 .

Decision

The original function is an entire rational one, hence it follows that the domain of definition is all real numbers. The function needs to be differentiated. 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 points of a possible extremum. It is necessary to apply the third sufficient extremum condition. 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 \u003d 5 7 is the maximum point. Applying 3 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, 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

Hence, 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 the above, we conclude that x 3 \u003d 3 is the minimum point of the function.

Graphic image

Answer: x 2 \u003d 5 7 is the maximum point, x 3 \u003d 3 - the minimum point of the given function.

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Lesson on the topic: "Finding extremum points of functions. Examples"

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What will we study:
1. Introduction.
2. Points of minimum and maximum.

4. How to calculate extremums?
5. Examples.

Introduction to extrema of functions

Guys, let's look at the graph of some function:

Note that the behavior of our function y=f (x) is largely determined by the two points x1 and x2. Let's take a closer look at the graph of the function at and around these points. Up to the point x2, the function increases, at the point x2 there is an inflection, and immediately after this point, the function decreases to the point x1. At the point x1, the function bends again, and after that it increases again. Points x1 and x2 will be called inflection points for the time being. Let's draw tangents at these points:

The tangents at our points are parallel to the x-axis, which means that the slope of the tangent is zero. This means that the derivative of our function at these points is zero.

Let's look at the graph of this function:

Tangents at points x2 and x1 cannot be drawn. Hence, the derivative at these points does not exist. Now let's look again at our points on the two charts. The point x2 is the point where the function reaches its maximum value in some area (near the point x2). The point x1 is the point at which the function reaches its smallest value in some area (near the point x1).

High and low points

Definition: The point x= x0 is called the minimum point of the function y=f(x) if there is a neighborhood of the point x0 where the following inequality is true: f(x) ≥ f(x0).

Definition: The point x=x0 is called the maximum point of the function y=f(x) if there is a neighborhood of the point x0 where the following inequality is true: f(x) ≤ f(x0).

Guys, what is the neighborhood?

Definition: The neighborhood of a point is a set of points containing our point and close to it.

We can define the neighborhood ourselves. For example, for a point x=2, we can define the neighborhood as points 1 and 3.

Let's return to our graphs, look at the point x2, it is larger than all other points from some neighborhood, then by definition it is a maximum point. Now let's look at the point x1, it is less than all other points from some neighborhood, then by definition it is a minimum point.

Guys, let's introduce the notation:

Ymin - minimum point,
ymax - maximum point.

Important! Guys, do not confuse the maximum and minimum points with the smallest and largest value of the function. The smallest and largest values ​​are sought over the entire domain of definition of the given function, and the minimum and maximum points are sought in some neighborhood.

Function extremes

There is a common term for minimum and maximum points - extremum points.

Extremum (lat. extremum - extreme) - the maximum or minimum value of a function on a given set. The point at which the extremum is reached is called the extremum point.

Accordingly, if the minimum is reached, the extremum point is called the minimum point, and if the maximum is reached, the maximum point.

How to find extrema of a function?

Let's get back to our charts. At our points, the derivative either vanishes (on the first graph) or does not exist (on the second graph).

Then we can make an important statement: If the function y= f(x) has an extremum at the point x=x0, then at this point the derivative of the function is either equal to zero or does not exist.

The points where the derivative is equal to zero are called stationary.

Points where the derivative of a function does not exist are called critical.

How to calculate extremes?

Guys, let's go back to the first graph of the function:

Analyzing this graph, we said: up to the point x2 the function increases, at the point x2 there is an inflection, and after this point the function decreases to the point x1. At the point x1, the function bends again, and after that the function increases again.

Based on such reasoning, we can conclude that the function at the extremum points changes the nature of monotonicity, and hence the derivative function changes sign. Recall that if the function is decreasing, then the derivative is less than or equal to zero, and if the function is increasing, then the derivative is greater than or equal to zero.

Let's generalize the obtained knowledge by the statement:

Theorem: Sufficient extremum condition: let the function y=f(x) be continuous on some interval X and have a stationary or critical point x= x0 inside the interval. Then:

• If this point has a neighborhood in which f’(x)>0 is satisfied for x x0, then the point x0 is the minimum point of the function y= f(x).
• If this point has such a neighborhood in which for x 0, and for x> x0 f'(x) there is no extremum.

To solve problems, remember the following rules: If the signs of derivatives are defined then:

Algorithm for studying the continuous function y= f(x) for monotonicity and extrema:

• Find the derivative y'.
• Find stationary (the derivative is zero) and critical points (the derivative does not exist).
• Mark the stationary and critical points on the number line and determine the signs of the derivative on the resulting intervals.
• Based on the above statements, draw a conclusion about the nature of the extremum points.

Examples of finding extremum points

1) Find the extremum points of the function and determine their nature: y= 7+ 12*x - x 3

Solution: Our function is continuous, then we will use our algorithm:
a) y "= 12 - 3x 2,
b) y"= 0, at x= ±2,

The point x= -2 is the minimum point of the function, the point x= 2 is the maximum point of the function.
Answer: x= -2 - function minimum point, x= 2 - function maximum point.

2) Find the extremum points of the function and determine their nature.

3) Find the extremum points of the function y= x - 2cos(x) and determine their character, for -π ≤ x ≤ π.

Solution: Our function is continuous, let's use our algorithm:
a) y"= 1 + 2sin(x),
b) find the values ​​in which the derivative is equal to zero: 1 + 2sin(x)= 0, sin(x)= -1/2,
because -π ≤ x ≤ π, then: x= -π/6, -5π/6,
c) mark the stationary points on the real line and determine the signs of the derivative: d) look at our figure, which shows the rules for determining extremums.
The point x= -5π/6 is the maximum point of the function.
The point x= -π/6 is the minimum point of the function.
Answer: x= -5π/6 - maximum point of the function, x= -π/6 - minimum point of the function.

4) Find the extremum points of the function and determine their nature:

Tasks for independent solution

The extremum point of a function is the point in the function's domain where the value of the function takes on a minimum or maximum value. The function values ​​at these points are called extrema (minimum and maximum) of the function.

Definition. Dot x1 function scope f(x) is called maximum point of the function , if the value of the function at this point is greater than the values ​​of the function at points close enough to it, located to the right and left of it (that is, the inequality f(x0 ) > f(x 0 + Δ x) x1 maximum.

Definition. Dot x2 function scope f(x) is called minimum point of the function, if the value of the function at this point is less than the values ​​of the function at points close enough to it, located to the right and left of it (that is, the inequality f(x0 ) < f(x 0 + Δ x) ). In this case, the function is said to have at the point x2 minimum.

Let's say the point x1 - maximum point of the function f(x) . Then in the interval up to x1 function increases, so the derivative of the function is greater than zero ( f "(x) > 0 ), and in the interval after x1 the function is decreasing, so function derivative less than zero ( f "(x) < 0 ). Тогда в точке x1

Let us also assume that the point x2 - minimum point of the function f(x) . Then in the interval up to x2 the function is decreasing and the derivative of the function is less than zero ( f "(x) < 0 ), а в интервале после x2 the function is increasing and the derivative of the function is greater than zero ( f "(x) > 0 ). In this case also at the point x2 the derivative of the function is zero or does not exist.

Fermat's theorem (a necessary criterion for the existence of an extremum of a function). If point x0 - extremum point of the function f(x) , then at this point the derivative of the function is equal to zero ( f "(x) = 0 ) or does not exist.

Definition. The points at which the derivative of a function is equal to zero or does not exist are called critical points .

Example 1 Let's consider a function.

At the point x= 0 the derivative of the function is equal to zero, therefore, the point x= 0 is the critical point. However, as can be seen on the graph of the function, it increases in the entire domain of definition, so the point x= 0 is not an extremum point of this function.

Thus, the conditions that the derivative of a function at a point is equal to zero or does not exist are necessary conditions for an extremum, but not sufficient, since other examples of functions can be given for which these conditions are satisfied, but the function does not have an extremum at the corresponding point. So must have sufficient indications, which make it possible to judge whether there is an extremum at a particular critical point and which one - a maximum or a minimum.

Theorem (the first sufficient criterion for the existence of an extremum of a function). Critical point x0 f(x) , if the derivative of the function changes sign when passing through this point, and if the sign changes from "plus" to "minus", then the maximum point, and if from "minus" to "plus", then the minimum point.

If near the point x0 , to the left and to the right of it, the derivative retains its sign, this means that the function either only decreases or only increases in some neighborhood of the point x0 . In this case, at the point x0 there is no extremum.

So, to determine the extremum points of the function, you need to do the following :

1. Find the derivative of a function.
2. Equate the derivative to zero and determine the critical points.
3. Mentally or on paper, mark the critical points on the numerical axis and determine the signs of the derivative of the function in the resulting intervals. If the sign of the derivative changes from "plus" to "minus", then the critical point is the maximum point, and if from "minus" to "plus", then the critical point is the minimum point.
4. Calculate the value of the function at the extremum points.

Example 2 Find extrema of a function .

Decision. Let's find the derivative of the function:

Equate the derivative to zero to find the critical points:

.

Since for any values ​​\u200b\u200bof "x" the denominator is not equal to zero, then we equate the numerator to zero:

Got one critical point x= 3 . We determine the sign of the derivative in the intervals delimited by this point:

in the range from minus infinity to 3 - minus sign, that is, the function decreases,

in the range from 3 to plus infinity - a plus sign, that is, the function increases.

That is, point x= 3 is the minimum point.

Find the value of the function at the minimum point:

Thus, the extremum point of the function is found: (3; 0) , and it is the minimum point.

Theorem (the second sufficient criterion for the existence of an extremum of a function). Critical point x0 is the extremum point of the function f(x) , if the second derivative of the function at this point is not equal to zero ( f ""(x) ≠ 0 ), moreover, if the second derivative is greater than zero ( f ""(x) > 0 ), then the maximum point, and if the second derivative is less than zero ( f ""(x) < 0 ), то точкой минимума.

Remark 1. If at a point x0 both the first and second derivatives vanish, then at this point it is impossible to judge the presence of an extremum on the basis of the second sufficient sign. In this case, you need to use the first sufficient criterion for the extremum of the function.

Remark 2. The second sufficient criterion for the extremum of a function is also inapplicable when the first derivative does not exist at the stationary point (then the second derivative does not exist either). In this case, it is also necessary to use the first sufficient criterion for the extremum of the function.

The local nature of the extrema of the function

From the above definitions it follows that the extremum of a function is of a local nature - this is the largest and smallest value of the function compared to the nearest values.

Suppose you consider your earnings in a time span of one year. If in May you earned 45,000 rubles, and in April 42,000 rubles and in June 39,000 rubles, then the May earnings are the maximum of the earnings function compared to the nearest values. But in October you earned 71,000 rubles, in September 75,000 rubles, and in November 74,000 rubles, so the October earnings are the minimum of the earnings function compared to nearby values. And you can easily see that the maximum among the values ​​of April-May-June is less than the minimum of September-October-November.

Generally speaking, a function may have several extrema on an interval, and it may turn out that any minimum of the function is greater than any maximum. So, for the function shown in the figure above, .

That is, one should not think that the maximum and minimum of the function are, respectively, its maximum and minimum values ​​on the entire segment under consideration. At the maximum point, the function has the greatest value only in comparison with those values ​​that it has at all points sufficiently close to the maximum point, and at the minimum point, the smallest value only in comparison with those values ​​that it has at all points sufficiently close to the minimum point.

Therefore, we can refine the above concept of extremum points of a function and call the minimum points local minimum points, and the maximum points - local maximum points.

We are looking for the extrema of the function together

Example 3

Solution. The function is defined and continuous on the whole number line. Its derivative also exists on the entire number line. Therefore, in this case only those at which , i.e., , whence and . Critical points and divide the entire domain of the function into three intervals of monotonicity: . We select one control point in each of them and find the sign of the derivative at this point.

For the interval, the reference point can be : we find . Taking a point in the interval, we get , and taking a point in the interval, we have . So, in the intervals and , and in the interval . According to the first sufficient sign of an extremum, there is no extremum at the point (since the derivative retains its sign in the interval ), and the function has a minimum at the point (since the derivative changes sign from minus to plus when passing through this point). Find the corresponding values ​​of the function: , and . In the interval, the function decreases, since in this interval , and in the interval it increases, since in this interval.

To clarify the construction of the graph, we find the points of intersection of it with the coordinate axes. When we obtain an equation whose roots and , i.e., two points (0; 0) and (4; 0) of the graph of the function are found. Using all the information received, we build a graph (see at the beginning of the example).

For self-checking during calculations, you can use online derivatives calculator .

Example 4 Find the extrema of the function and build its graph.

The domain of the function is the entire number line, except for the point, i.e. .

To shorten the study, we can use the fact that this function is even, since . Therefore, its graph is symmetrical about the axis Oy and the study can only be performed for the interval .

Finding the derivative and critical points of the function:

1) ;

2) ,

but the function suffers a break at this point, so it cannot be an extremum point.

Thus, the given function has two critical points: and . Taking into account the parity of the function, we check only the point by the second sufficient sign of the extremum. To do this, we find the second derivative and determine its sign at : we get . Since and , then is the minimum point of the function, while .

To get a more complete picture of the graph of the function, let's find out its behavior on the boundaries of the domain of definition:

(here the symbol indicates the desire x to zero on the right, and x remains positive; similarly means aspiration x to zero on the left, and x remains negative). Thus, if , then . Next, we find

,

those. if , then .

The graph of the function has no points of intersection with the axes. The picture is at the beginning of the example.

For self-checking during calculations, you can use online derivatives calculator .

We continue to search for extremums of the function together

Example 8 Find the extrema of the function .

Decision. Find the domain of the function. Since the inequality must hold, we obtain from .

Let's find the first derivative of the function.