Extreme function online solution. How to find the minimum and maximum points of a function: features, methods and examples. Problems of Finding the Extremum of a Function

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

This article reveals the definitions, we formulate a sufficient indicator of an increase and decrease in an 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 in the solution 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 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 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 satisfiable. In other words, the larger the value of the function, the smaller the value of the argument. Consider the figure below.

Comment: When the function is definite and continuous at the ends of the increasing and decreasing interval, that is, (a; b), where x = a, x = b, the points are included in the increasing and decreasing interval. This does not contradict the definition, which means that there is a place to be 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. Hence, we find 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.

Point x 0 is called maximum point for the function y = f (x), when the inequality f (x 0) ≥ f (x) is valid for all values ​​of x. Maximum function Is the value of the function at the point, and 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. Function minimum Is the value of the function at the point, and has a designation of the form y m i n.

The neighborhoods of the point x 0 are considered extremum points, and the value of the function, which corresponds to the extremum points. Consider the figure below.

Extrema of the function with the largest and smallest function value. 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 the 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 INCREASE AND DECREASE OF A FUNCTION

To find the maxima and minima of a function, it is necessary to apply the extremum criteria in the case when the function satisfies these conditions. The first sign is considered to be the most frequently 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 a 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 a 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 that the point is called the maximum;
• when the function is continuous at the point x 0, then it has a derivative with an alternating 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 in this area;
• define zeros and points where the function does not exist;
• determination of the sign of the derivative on intervals;
• select the points where the function changes sign.

Let us consider the algorithm by the example of solving several examples for finding the 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 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

Hence we see that the zeros of the function are x = - 1, x = 5, x = 2, that is, each parenthesis must be equated to zero. Let's mark on the number axis and get:

Now let us 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) = 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. In a similar way, 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 the continuity, it is necessary to pay attention to the sign of the derivative, if it changes, then this is the 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 criterion, we have that x = - 1 is a maximum point, so 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 changes sign from - to +. Hence, x = -1 is a 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 noting that the use of the first sufficient criterion for an extremum does not require differentiability of the function with 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.

Solution.

The scope 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 yx → 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 yx → 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 yx → 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 yx → 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 equal to 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 should 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 the 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 line looks like

Hence, we come to the conclusion 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:

ymin = y - 4 - 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = - 4 - 2 3 3 = - 8 27 3 ymin = y (0) = 1 6 x 3 - 2 2 + 22 3 x - 8 x = 0 = - 8 ymin = 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

ymax = y - 4 + 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = - 4 + 2 3 3 = 8 27 3 ymax = 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:

ymin = y - 4 - 2 3 3 = - 8 27 3 ymin = y (0) = - 8 ymin = y 4 + 2 3 3 = - 8 27 3 ymax = y - 4 + 2 3 3 = 8 27 3 ymax = y 4 - 2 3 3 = 8 27 3

If a function f "(x 0) = 0 is given, then for its 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 xx + 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 = 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 an extremum, we obtain that x = 1 is a maximum point. Otherwise, the record 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 n-th order in the ε neighborhood of the given point x 0 and the derivative up to the n + 1-th 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 whole rational, it follows 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 vanish at x 1 = - 1, x 2 = 5 7, x 3 = 3. That is, the points can be points of a possible extreme. It is necessary to apply the third sufficient condition for an extremum. Finding the second derivative allows us to accurately determine the presence of the maximum and minimum of the 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 3 sufficient criterion, we find that for n = 1 and f (n + 1) 5 7< 0 .

It is necessary to determine the nature of the points x 1 = - 1, x 3 = 3. To do this, you need to find the third derivative, 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, 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 = 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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Function y = f (x) is called increasing (diminishing) in some interval, if for x 1< x 2 выполняется неравенство (f (x 1) < f (x 2) (f (x 1) >f (x 2)).

If a differentiable function y = f (x) increases (decreases) on an interval, then its derivative on this interval f " (x)> 0

(f "(x)< 0).

Point x O called point of local maximum (minimum) of the function f (x) if there exists a neighborhood of the point x about, for all points of which the inequality f (x)≤ f (x o) (f (x)≥ f (x o)).

The maximum and minimum points are called extremum points, and the values ​​of the function at these points are its extrema.

Extremum points

Necessary conditions for an extremum ... If point x O is the extremum point of the function f (x), then either f " (x о) = 0, or f(x o) does not exist. Such points are called critical, moreover, the function itself is defined at the critical point. The extrema of a function should be sought among its critical points.

First sufficient condition. Let be x O - critical point. If f " (x) when passing through the point x O changes the plus sign to minus, then at the point x about the function has a maximum, otherwise it has a minimum. If the derivative does not change sign when passing through the critical point, then at the point x O there is no extreme.

Second sufficient condition. Let the function f (x) have
f " (x) in the vicinity of the point x O and the second derivative f "" (x 0) at the very point x about... If f "(x about) = 0, f "" (x 0)> 0 (f "" (x 0)<0), то точка x about is the point of the local minimum (maximum) of the function f (x). If f "" (x 0) = 0, then either use the first sufficient condition or involve higher ones.

On a segment, the function y = f (x) can reach the smallest or largest value either at critical points or at the ends of the segment.

Example 3.22.

Solution. Because f " (

Problems of Finding the Extremum of a Function

Example 3.23. a

Solution. x and y y
0 ≤ x≤
> 0, and for x> a / 4 S " < 0, значит, в точке x=a /4 функция S имеет максимум. Значение functions sq.. units).

Example 3.24. p ≈

Solution. p p
S "
R = 2, H = 16/4 = 4.

Example 3.22.Find the extrema of the function f (x) = 2x 3 - 15x 2 + 36x - 14.

Solution. Because f " (x) = 6x 2 - 30x +36 = 6 (x -2) (x - 3), then the critical points of the function x 1 = 2 and x 2 = 3. Extrema can be only at these points. Since when passing through the point x 1 = 2 the derivative changes the sign plus to minus, then at this point the function has a maximum. When passing through the point x 2 = 3, the derivative changes its minus sign to plus, therefore, at the point x 2 = 3, the function has a minimum. Calculating the values ​​of the function in points
x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f (2) = 14 and minimum f (3) = 13.

Example 3.23.It is necessary to build a rectangular area near the stone wall so that on three sides it is fenced off with a wire mesh, and on the fourth side it is adjacent to the wall. For this there is a running meters of mesh. At what aspect ratio will the site have the largest area?

Solution.We denote the sides of the site by x and y... The area of ​​the site is S = xy. Let be y is the length of the side adjacent to the wall. Then, by condition, the equality 2x + y = a must be fulfilled. Therefore, y = a - 2x and S = x (a - 2x), where
0 ≤ x≤ a / 2 (the length and width of the pad cannot be negative). S "= a - 4x, a - 4x = 0 for x = a / 4, whence
y = a - 2 × a / 4 = a / 2. Insofar as x = a / 4 is the only critical point, let us check whether the sign of the derivative changes when passing through this point. For x a / 4 S "> 0, and for x> a / 4 S " < 0, значит, в точке x=a /4 функция S имеет максимум. Значение functions S (a / 4) = a / 4 (a - a / 2) = a 2/8 (sq.. units). Since S is continuous on and its values ​​at the ends S (0) and S (a / 2) are equal to zero, the found value will be the largest value of the function. Thus, the most advantageous aspect ratio of the site under the given conditions of the problem is y = 2x.

Example 3.24.It is required to make a closed cylindrical tank with a capacity of V = 16 p ≈ 50 m 3. What are the dimensions of the tank (radius R and height H) so that the least amount of material is used to make it?

Solution.The total surface area of ​​the cylinder is S = 2 p R (R + H). We know the volume of the cylinder V = p R 2 H Þ H = V / p R 2 = 16 p / p R 2 = 16 / R 2. Hence, S (R) = 2 p (R 2 + 16 / R). Find the derivative of this function:
S "(R) = 2 p (2R- 16 / R 2) = 4 p (R- 8 / R 2). S " (R) = 0 for R 3 = 8, therefore
R = 2, H = 16/4 = 4.

We can also say that at these points the direction of movement of the function changes: if the function stops falling and begins to grow, this is the point of the minimum, on the contrary - the maximum.

Lows and Highs are collectively referred to as extrema of the function.

In other words, all five points highlighted on the chart above are extreme.

This makes it easy to find these points, even if you don't have a function graph.

Attention! When they write extremes or the maxima / minima mean the value of the function i.e. \ (y \). When they write extremum points or the high / low points mean the x at which the highs / lows are reached. For example, in the picture above, \ (- 5 \) is the minimum point (or extremum point), and \ (1 \) is the minimum (or extremum).

How to find the extremum points of a function according to the derivative graph (7th task of the exam)?

Let's find together the number of extremum points of the function from the graph of the derivative using an example:

We have a graph, which means we are looking for at what points on the graph the derivative is equal to zero. Obviously, these are the points \ (- 13 \), \ (- 11 \), \ (- 9 \), \ (- 7 \) and \ (3 \). The number of extremum points of the function is \ (5 \).

Attention! If a schedule is given derivative functions, but you need to find extremum points of the function, we do not count the highs and lows of the derivative! We count the points at which the derivative of the function vanishes (i.e., crosses the \ (x \) axis).

How to find the maximum or minimum points of a function according to the derivative graph (7th task of the exam)?

To answer this question, you need to remember two more important rules:

- The derivative is positive where the function increases.
- The derivative is negative where the function decreases.

Using these rules, let's find the minimum and maximum points of the function on the graph of the derivative.

It is clear that minimums and maximums should be sought among the extremum points, i.e. among \ (- 13 \), \ (- 11 \), \ (- 9 \), \ (- 7 \) and \ (3 \).

To make it easier to solve the problem, let us first place the plus and minus signs in the figure, denoting the sign of the derivative. Then the arrows - denoting the increase, decrease of the function.

Let's start with \ (- 13 \): up to \ (- 13 \) the derivative is positive i.e. the function grows, after the derivative is negative i.e. function falls. If we imagine this, it becomes clear that \ (- 13 \) is a maximum point.

\ (- 11 \): the derivative is first positive and then negative, which means the function increases and then decreases. Again, try to draw this mentally and it will become obvious to you that \ (- 11 \) is the minimum.

\ (- 9 \): the function increases and then decreases - the maximum.

\ (- 7 \): minimum.

\ (3 \): maximum.

All of the above can be summarized by the following conclusions:

- The function has a maximum where the derivative is zero and changes sign from plus to minus.
- The function has a minimum where the derivative is zero and changes sign from minus to plus.

How to find the points of maxima and minima if the formula of the function is known (12 task of the exam)?

To answer this question, you need to do the same as in the previous paragraph: find where the derivative is positive, where is negative and where is zero. To make it clearer, I will write an algorithm with an example of a solution:

1. Find the derivative of the function \ (f "(x) \).
2. Find the roots of the equation \ (f "(x) = 0 \).
3. Draw the \ (x \) axis and mark the points obtained in step 2 on it, draw with arcs the intervals into which the axis is divided. Label above the \ (f "(x) \) axis, and below the \ (f (x) \) axis.
4. Determine the sign of the derivative in each interval (using the interval method).
5. Place the sign of the derivative in each interval (above the axis), and use the arrow to indicate the increase (↗) or decrease (↘) of the function (below the axis).
6. Determine how the sign of the derivative changed when passing through the points obtained in step 2:
   - if \ (f ’(x) \) changed the sign from“ \ (+ \) ”to“ \ (- \) ”, then \ (x_1 \) is a maximum point;
   - if \ (f '(x) \) changed the sign from "\ (- \)" to "\ (+ \)", then \ (x_3 \) is a minimum point;
   - if \ (f '(x) \) has not changed sign, then \ (x_2 \) - can be an inflection point.

Everything! Highs and lows are found.

Plotting points on the axis at which the derivative is equal to zero, the scale can be ignored. The behavior of the function can be shown as shown in the figure below. So it will be more obvious where is the maximum and where is the minimum.

Example(Exam)... Find the maximum point of the function \ (y = 3x ^ 5-20x ^ 3-54 \).
Solution:
1. Find the derivative of the function: \ (y "= 15x ^ 4-60x ^ 2 \).
2. Let us equate it to zero and solve the equation:

\ (15x ^ 4-60x ^ 2 = 0 \) \ (|: 15 \)
\ (x ^ 4-4x ^ 2 = 0 \)
\ (x ^ 2 (x ^ 2-4) = 0 \)
\ (x = 0 \) \ (x ^ 2-4 = 0 \)
\ (x = ± 2 \)

3. - 6. Draw points on the numerical axis and determine how the sign of the derivative changes and how the function moves:

It is now obvious that the maximum point is \ (- 2 \).

Answer. \(-2\).

Point x 0 is called maximum point(minimum) of the function f (x) if the inequality f (x) ≤f (x 0) (f (x) ≥f (x 0)) holds in some neighborhood of the point x 0.

The value of the function at this point is called accordingly maximum or minimum functions. The maximum and minimum functions are united by a common name. extremum functions.

The extremum of a function in this sense is often called local extremum, emphasizing the fact that this concept is associated only with a sufficiently small neighborhood of the point x 0. On the same interval, the function can have several local maxima and minima, which do not necessarily coincide with global maximum or minimum(i.e., the largest or smallest value of the function over the entire interval).

A necessary condition for an extremum... In order for a function to have an extremum at a point, it is necessary that its derivative at this point is equal to zero or does not exist.

For differentiable functions, this condition follows from Fermat's theorem. In addition, it also provides for the case when the function has an extremum at the point at which it is not differentiable.

The points at which the necessary extremum condition is satisfied are called critical(or stationary for a differentiable function). These points must be within the scope of the function definition.

Thus, if at any point there is an extremum, then this point is critical (the necessity of the condition). Note that the converse is not true. The critical point is not necessarily the extreme point, i.e. the formulated condition is not sufficient.

The first sufficient condition for an extremum... If, when passing through a certain point, the derivative of the differentiable function changes its sign from plus to minus, then this is the maximum point of the function, and if from minus to plus, then the minimum point.

The proof of this condition follows from the sufficient condition for monotonicity (when the sign of the derivative changes, a transition occurs either from an increase in the function to a decrease, or from a decrease to an increase).

Second sufficient condition for an extremum... If the first derivative of a twice differentiable function at some point is equal to zero, and the second derivative at this point is positive, then this is the minimum point of the function; and if the second derivative is negative, then this is the maximum point.

The proof of this condition is also based on the sufficient monotonicity condition. Indeed, if the second derivative is positive, then the first derivative is an increasing function. Since at the point under consideration it is equal to zero, therefore, when passing through it, it changes sign from minus to plus, which returns us to the first sufficient condition for a local minimum. Similarly, if the second derivative is negative, then the first decreases and changes sign from plus to minus, which is a sufficient condition for a local maximum.

Examining a function for an extremum in accordance with the formulated theorems, it includes the following stages:

1. Find the first derivative of the function f` (x).

2. Check the fulfillment of the necessary extremum condition, i.e. find the critical points of the function f (x) at which the derivative f` (x) = 0 or does not exist.

3. Check the fulfillment of the sufficient condition for the extremum, i.e. either examine the sign of the derivative to the left and right of each critical point, or find the second derivative f,, (x) and determine its sign at each critical point. Make a conclusion about the presence of extrema of the function.

4. Find the extrema (extreme values) of the function.

Finding the global maximum and minimum of a function at a certain interval it is also of great practical importance. The solution to this problem on a segment is based on the Weierstrass theorem, according to which a continuous function takes its maximum and minimum values ​​on an interval. They can be reached both at the extremum points and at the ends of the segment. Therefore, the solution includes the following steps:

1. Find the derivative of the function f` (x).

2. Find the critical points of the function f (x) at which the derivative f` (x) = 0 or does not exist.

3. Find the values ​​of the function at critical points and at the ends of the segment and choose the largest and the smallest of them.

As you can see, this feature of the extremum of a function requires the existence of a derivative at least up to the second order at a point.

Example.

Find the extrema of the function.

Solution.

Let's start with the scope:

Let's differentiate the original function:

x = 1, that is, it is the point of a possible extremum. Find the second derivative of the function and calculate its value at x = 1:

Consequently, by the second sufficient condition for an extremum, x = 1 is the maximum point. Then - maximum function.

Graphic illustration.

Answer:

The third sufficient condition for the extremum of a function.

Let the function y = f (x) has derivatives up to n-th order in -neighborhood of a point and derivatives up to n + 1-th order at the point itself. Let and.

Example.

Find extremum points of a function .

Solution.

The original function is a whole rational, its domain of definition is the entire set of real numbers.

Let's differentiate the function:

The derivative vanishes at therefore, these are points of possible extremum. Let us use the third sufficient condition for an extremum.

We find the second derivative and calculate its value at the points of a possible extremum (we omit intermediate calculations):

Therefore, is a maximum point (for the third sufficient criterion for an extremum, we have n = 1 and ).

To find out the nature of the points find the third derivative and calculate its value at these points:

Therefore, is the inflection point of the function ( n = 2 and ).

It remains to deal with the point. Find the fourth derivative and calculate its value at this point:

Therefore, is the minimum point of the function.

Graphic illustration.

Answer:

The maximum point is the minimum point of the function.

10. Extrema of the function Determination of the extremum

The function y = f (x) is called increasing (diminishing) in some interval, if for x 1< x 2 выполняется неравенство (f(x 1) < f (x 2) (f(x 1) >f (x 2)).

If a differentiable function y = f (x) increases (decreases) on an interval, then its derivative on this interval f "(x)  0

(f "(x)  0).

Point x O called point of local maximum (minimum) of the function f (x) if there exists a neighborhood of the point x O, for all points of which the inequality f (x) ≤ f (x о) (f (x) ≥ f (x о)) is true.

The maximum and minimum points are called extremum points, and the values ​​of the function at these points are its extrema.

Extremum points

Necessary conditions for an extremum... If point x O is the extremum point of the function f (x), then either f "(x о) = 0, or f (x о) does not exist. Such points are called critical, moreover, the function itself is defined at the critical point. The extrema of a function should be sought among its critical points.

First sufficient condition. Let be x O- critical point. If f "(x) when passing through the point x O changes the plus sign to minus, then at the point x O the function has a maximum, otherwise it has a minimum. If the derivative does not change sign when passing through the critical point, then at the point x O there is no extreme.

Second sufficient condition. Let the function f (x) have a derivative f "(x) in a neighborhood of the point x O and the second derivative at the very point x O... If f "(x о) = 0,> 0 (<0), то точка x O is the point of the local minimum (maximum) of the function f (x). If = 0, then either use the first sufficient condition or involve higher derivatives.

On a segment, the function y = f (x) can reach the smallest or largest value either at critical points or at the ends of the segment.

Example 3.22. Find the extrema of the function f (x) = 2x 3 - 15x 2 + 36x - 14.

Solution. Since f "(x) = 6x 2 - 30x +36 = 6 (x -2) (x - 3), then the critical points of the function x 1 = 2 and x 2 = 3. Extrema can be only at these points. So as when passing through the point x 1 = 2 the derivative changes its sign plus to minus, then at this point the function has a maximum. When passing through the point x 2 = 3 the derivative changes sign from minus to plus, therefore at the point x 2 = 3 the function has a minimum. Having calculated the values ​​of the function at the points x 1 = 2 and x 2 = 3, we find the extrema of the function: the maximum f (2) = 14 and the minimum f (3) = 13.