Extreme function. Extreme Functions: Signs of existence, examples of solutions to find an extremum function if

07.09.2020

With this service you can find the greatest and smallest function One variable F (X) with the decoration of the solution in Word. If the function f (x, y) is specified, therefore, it is necessary to find the extremum function of two variables. You can also find intervals of increasing and descending function.

Rules for entering functions:

Required extremma condition of the function of one variable

Equation F "0 (x *) \u003d 0 is the required condition of the extremum function of one variable, i.e. at point x * The first derivative function should be in zero. It highlights stationary points x C, in which the function does not increase and does not decrease .

A sufficient condition of extremum functions of one variable

Let F 0 (x) be double-differentiable by x, belonging to the set d. If the condition is satisfied at the point x *:

F "0 (x *) \u003d 0
f "" 0 (x *)\u003e 0

That point x * is the local (global) minimum function.

If the condition is satisfied at the point x *:

F "0 (x *) \u003d 0
f "" 0 (x *)< 0

Then point x * - local (global) maximum.

Example number 1. Find the greatest and smallest values \u200b\u200bof the function: on the segment.
Decision.

The critical point is one x 1 \u003d 2 (f '(x) \u003d 0). This point belongs to the segment. (Point x \u003d 0 is not critical, as 0∉).
Calculate the values \u200b\u200bof the function at the ends of the segment and at a critical point.
f (1) \u003d 9, f (2) \u003d 5/2, f (3) \u003d 3 8/81
Answer: F min \u003d 5/2 at x \u003d 2; f max \u003d 9 at x \u003d 1

Example number 2. Using derivatives of higher orders to find the extremum function y \u003d x-2sin (x).
Decision.
Find a derivative function: y '\u003d 1-2cos (x). We will find critical points: 1-cos (x) \u003d 2, cos (x) \u003d ½, x \u003d ± π / 3 + 2πk, k∈Z. We find y '' \u003d 2sin (x), calculate, it means x \u003d π / 3 + 2πk, k∈Z - the minimum points of the function; So x \u003d - π / 3 + 2πk, k∈Z - the points of the maximum function.

Example number 3. Explore the extremum of FCCscation in the vicinity of the point x \u003d 0.
Decision. Here it is necessary to find extremmas functions. If the extremum X \u003d 0, then find out its type (minimum or maximum). If there are no x \u003d 0 among the points found, then calculate the value of the function f (x \u003d 0).
It should be noted that when the derivative on each side of this point does not change its mark, the possible situations are not exhausted even for differentiable functions: it may happen that for an arbitrarily small neighborhood according to one side of the point x 0 or on both sides, the derivative changes sign. At these points you have to use other methods for researching functions to extremum.

Example number 4. Spread the number 49 into two terms, the product of which will be the largest.
Decision. Denote x - the first term. Then (49-x) - the second term.
The work will be maximal: x · (49-x) → Max
or
49x - x 2

The greatest volume of the cylinder

Find the size of the cylinder of the greatest volume made from the billet in the form of a radius of R.
Decision:

The volume of the cylinder is: V \u003d πr 2 H
where H \u003d 2H,
Substitute these values \u200b\u200binto the target function.

V → Max
Find an extremum function. Since the volume V (H) function depends only on one variable, we will find a derivative using the service

2) find the first derivative;

3) find critical points;

2) find a derivative

5) Calculate the value of the function

2) find a derivative

5) Calculate the extremum function

2) Calculate the derivative

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The definition of the extremum function is given, an example is also given as using an online calculator to find an extremum function.

Example

There is a function (x ^ 3 -exp (x) + x) / (1 + x ^ 2).

We introduce it to the calculator by research features online:

We get the following result:

In order to find extremmas, you need to solve the equation $$ \\ FRAC (D) (DX) F (\\ Left (X \\ Right)) \u003d 0 $$ (the derivative is zero), and the roots of this equation will be extremes of this function: $ $ \\ FRAC (D) (DX) F (\\ Left (X \\ Right)) \u003d $$ The first derivative $$ - \\ FRAC (2 x) (\\ left (x ^ (2) + 1 \\ RIGHT) ^ (2 )) \\ left (x + x ^ (3) - E ^ (x) \\ Right) + \\ FRAC (3 x ^ (2) - E ^ (x) + 1) (x ^ (2) + 1) \u003d 0 $$ solve this equation
The roots of this URNA $$ x_ (1) \u003d 0 $$$$ x_ (2) \u003d 3.28103090528 $$$$ x_ (3) \u003d -0.373548376565 $$ zn. Extremes at points:
(0, -1)
(3.28103090528, 1.01984828342285)
(-0.373548376565, -0.977554081645009)
Intervals of increasing and descending function:
Find intervals where the function increases and decreases, as well as minima and maxima of the function, for this we look at how the function in extremums behave at the slightest deviation from extremum:
Minima function at points: $$ x_ (3) \u003d 0 $$ Maxons of function at points: $$ x_ (3) \u003d 3.28103090528 $$$$ x_ (3) \u003d -0.373548376565 $$ decreases
(-o, -0.373548376565] u u

Introduction of local maxima and minima does not do without differentiation and is necessary in the study of the function and building its schedule.

The point is called a local maximum (or minimum) function, if there is such a neighborhood of this point, belonging to the function of determining the function, and for all of this surrounding it is inequality (or).

The maximum and minimum points are called the points of the extremum function, and the values \u200b\u200bof the function in extreme points - its extreme values.

Required condition of local extremum:

If the function has a local extremum at a point, then either the derivative is zero or does not exist.

Points that satisfy the requirements discharged above are called critical points.

However, in each critical point, the function has an extremum.

The concept of extremum function

The answer to the question: there will be a critical point of the extremum point gives the following theorem.

A sufficient condition for the existence of extremum function

Theorem І. Suppose that the function is continuous in some interval containing a critical point and differentiated at all points of this interval (except for the point itself).

Then, for a point, the function has a maximum, if the condition is satisfied for the arguments that the derivative is larger than zero, and for the condition - the derivative is less than zero.

If the derivative is less than zero, and for more zero, then for a point, the function has a minimum.

Theorem II. Let the function doublely differentiate in the neighborhood of the point and the derivative is zero. Then at the point the function has a local maximum if the second derivative is less than zero and a local minimum, if on the contrary.

If the second derivative is zero, the point may not be a point of extremum.

In the study of functions on extremes, both theorems are used. The first in practice is easier, because it does not require the second derivative.

Rules for finding Extremes (maxima and minima) with the help of the first derivative

1) find the definition area;

2) find the first derivative;

3) find critical points;

4) Explore the sign of the derivative at the intervals that received the definition area from the split points.

At the same time, the critical point is a point of a minimum, if during the transition through it from left to the right of the derivative changes the sign from a negative on a positive, otherwise a maximum point.

Instead of this rule, you can determine the second derivative and explore according to the second theorem.

5) Calculate the values \u200b\u200bof the function at extremum points.

We now consider the study of the function on extremes on specific examples.

Collection V.Yu. Klepko, V.L. Halts "Higher Mathematics in Examples and Tasks"

1) the definition area will be many valid numbers

2) find a derivative

3) Calculate critical points

They split the definition area for the following intervals

4) Explore the sign of the derivative at the found intervals method of substitution of values

Thus, the first point is a minimum point, and the second is the maximum point.

5) Calculate the value of the function

1) the field of definition will be a lot of valid numbers, so the root is always more united

and the Arcthangent function is determined on the entire valid axis.

2) find a derivative

3) With the conditions of equality, the derivative of zero find a critical point

It breaks the definition area for two intervals.

4) Determine the sign of the derivative in each of the regions

Thus, we find that at a critical point, the function takes the minimum value.

5) Calculate the extremum function

1) The function is defined when the denominator does not turn into zero

It follows from this that the definition area consists of three intervals

2) Calculate the derivative

3) equate the derivative to zero and find critical points.

4) Install the sign of the derivative in each of the areas, substituting the corresponding values.

Thus, the point is a local maximum point, and a local minimum. In the inflection of the function, but there will be more material about it in the following articles.

5) Find a value at critical points.

Despite the fact that the value of the function, the first point is the point of the local maximum, and the arc is a minimum. Do not be afraid if you have similar results, in identifying local extremums such situations are permissible.

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Literature

1. Bogomolov N.V. Practical classes in mathematics. - M.: Higher. Shk., 2009

2. P.T.Apanasov, M.I. Eornov. Collection of tasks in mathematics. - M.: Higher. Shk., 2009

Methodical instructions

Study of functions using a derivative. Finding the intervals of monotonality

Theorem1. If the function f (x) is defined and continuous on the interval (A; b) and f '(x) is positive everywhere (F' (x)\u003e 0), then the function increases at the interval (A; b).

Theorem2. If the function f (x) is defined and continuous on the interval (A; b) and f '(x) are negative (F' (x) throughout<0), тогда функция убывает на промежутке (а;b).

Example1. Explore the monotony of y \u003d.

Solution: U \u003d 2x-1

The numeric axis is divided into two intervals

It means that the function decreases in the interval (-; 5) and the function increases in the interval (5;).

Finding Extreme Functions

The function f (x) has a maximum (minimum) at point x0, if this point has a neighborhood in which F (x) f (x0)) for xx0.

Maximum and minimum are combined with the name of the extremum.

Theorem 1. (Prereremum required). If the point x0 is the point of extremum function y \u003d f (x) and at this point there is a derivative F '(x0), then it is zero: F' (x) \u003d 0.

The points where f '(x) \u003d 0 or are not called critical.

Theorem 2. (sufficient condition). Suppose that the function f (x) is continuous at the point x0 and in its neighborhood has a derivative, except perhaps the point x0. Then

a) if the derivative f '(x) during the transition through the point x0 changes the sign from the plus to minus, then the point x0 is the maximum point F (X);

b) if the derivative f '(x) during the transition through the point x0 changes the sign from a minus to plus, then the point x0 is the point of the minimum function f (x);

c) If there is a neighborhood (x0-; x0 +) of the point x0, in which the derivative F '(X) saves its mark, then at the point x0, this function f (x) does not have an extremum.

Example 2.Explore the function of the function y \u003d 3 -5x.

Solution: U '\u003d -5-2x

When switching through the point x \u003d - 2.5, the derivative in 'changes the sign from "+" to "-" \u003d\u003d\u003e x \u003d -2.5 point of maximum.

Sufficient conditions of extremum function.

xmax \u003d - 2.5; Umax \u003d 9.25.

Didn't find what you were looking for? Use the search:

Extreme function: signs of existence, examples of solutions

Then at the point the function has a local maximum if the second derivative is less than zero and a local minimum, if on the contrary.

If the second derivative is zero, the point may not be a point of extremum.

In the study of functions on extremes, both theorems are used. The first in practice is easier, because it does not require the second derivative.

Rules for finding Extremes (maxima and minima) with the help of the first derivative

1) find the definition area;

2) find the first derivative;

3) find critical points;

4) Explore the sign of the derivative at the intervals that received the definition area from the split points.

Instead of this rule, you can determine the second derivative and explore according to the second theorem.

5) Calculate the values \u200b\u200bof the function at extremum points.

We now consider the study of the function on extremes on specific examples.

Collection V.Yu. Klepko, V.L. Halts "Higher Mathematics in Examples and Tasks"

1) the definition area will be many valid numbers

2) find a derivative

3) Calculate critical points

They split the definition area for the following intervals

4) Explore the sign of the derivative at the found intervals method of substitution of values

Thus, the first point is a minimum point, and the second is the maximum point.

5) Calculate the value of the function

1) the field of definition will be a lot of valid numbers, so the root is always more united

and the Arcthangent function is determined on the entire valid axis.

2) find a derivative

3) With the conditions of equality, the derivative of zero find a critical point

It breaks the definition area for two intervals.

4) Determine the sign of the derivative in each of the regions

Thus, we find that at a critical point, the function takes the minimum value.

5) Calculate the extremum function

1) The function is defined when the denominator does not turn into zero

It follows from this that the definition area consists of three intervals

2) Calculate the derivative

3) equate the derivative to zero and find critical points.

4) Install the sign of the derivative in each of the areas, substituting the corresponding values.

Thus, the point is a local maximum point, and a local minimum. In the inflection of the function, but there will be more material about it in the following articles.

5) Find a value at critical points.

View materials:

Higher Mathematics »Functions of several variables» Extreme function of two variables

Extreme function of two variables. Examples of research of functions for extremum.

Let the function $ z \u003d f (x, y) $ is defined in some neighborhood of the point $ (x_0, y_0) $. It is said that $ (x_0, y_0) $ - point (local) maximum, if for all points $ (x, y) $ of some neighborhood of the point $ (x_0, y_0) $ is performed inequality $ f (x, y)< f(x_0,y_0)$. Если же для всех точек этой окрестности выполнено условие $f(x,y)> F (x_0, y_0) $, then the point $ (x_0, y_0) $ is called a point (local) minimum.

Maximum and minimum points are often called a common term - extremum points.

If $ (x_0, y_0) $ is a maximum point, then the value of the function $ F (x_0, y_0) $ at this point is called the maximum function $ z \u003d f (x, y) $. Accordingly, the value of the function at a minimum point is called a minimum of the function $ z \u003d f (x, y) $. Minima and maxima function are combined by a common term - extremes of the function.

Algorithm research function $ z \u003d f (x, y) $ for extremum

Find the private derivatives of $ \\ FRAC (\\ Partial z) (\\ Partial x) $ and $ \\ FRAC (\\ Partial Z) (\\ Partial y) $. Create and solve the system of equations $ \\ left \\ (\\ begin (aligned) \\ FRAC (\\ Partial z) (\\ partial x) \u003d 0; \\\\ \\ FRAC (\\ Partial z) (\\ partial y) \u003d 0. \\ Find $ \\ FRAC (\\ Partial ^ 2z) (\\ Partial x ^ 2) $, $ \\ FRAC (\\ Partial ^ 2z) (\\ Partial x \\ Partial y) $, $ \\ FRAC (\\ Partial ^ 2Z) (\\ Partial y ^ 2) $ and calculate the value of $ \\ delta \u003d \\ FRAC (\\ Partial ^ 2z) (\\ partial x ^ 2) \\ CDOT \\ FRAC (\\ Partial ^ 2z) (\\ Partial Y ^ 2) - \\ Left (\\ FRAC (\\ Partial ^ 2z) (\\ Partial X \\ Partial Y) \\ Right) ^ 2 $ in each stationary point. After that, use the following scheme:
If $ \\ Delta\u003e 0 $ and $ \\ FRAC (\\ Partial ^ 2z) (\\ Partial x ^ 2)\u003e 0 $ (or $ \\ FRAC (\\ Partial ^ 2z) (\\ Partial Y ^ 2)\u003e 0 $), That in the test point is a minimum point.

If $ \\ Delta\u003e 0 $ and $ \\ FRAC (\\ Partial ^ 2z) (\\ Partial x ^ 2)
If $ \\ Delta< 0$ (или $\frac{\partial^2z}{\partial y^2} < 0$), то в исследуемая точка есть точкой максимума.
If $ \\ delta \u003d 0 $, then nothing definite about the presence of extremum cannot be said; Additional research is required.< 0$, то в расматриваемой стационарной точке экстремума нет.
Note (desirable for a more complete text understanding): show / hide

If $ \\ delta\u003e 0 $, then $ \\ FRAC (\\ Partial ^ 2z) (\\ Partial x ^ 2) \\ Cdot \\ FRAC (\\ Partial ^ 2z) (\\ Partial Y ^ 2) - \\ left (\\ FRAC (\\ And hence it follows that $ \\ FRAC (\\ Partial ^ 2z) (\\ Partial x ^ 2) \\ CDOT \\ FRAC (\\ Partial ^ 2z) (\\ partial y ^ 2)\u003e \\ left (\\ FRAC (\\ Partial ^ 2z) (\\ Partial x \\ Partial y) \\ Right) ^ 2 ≥ 0 $. Those. $ \\ FRAC (\\ Partial ^ 2z) (\\ Partial x ^ 2) \\ CDOT \\ FRAC (\\ Partial ^ 2z) (\\ Partial y ^ 2)\u003e 0 $. If the product of certain values \u200b\u200bare more than zero, then these values \u200b\u200bof one sign. Those., For example, if $ \\ FRAC (\\ Partial ^ 2z) (\\ Partial x ^ 2)\u003e 0 $, then $ \\ FRAC (\\ Partial ^ 2z) (\\ Partial y ^ 2)\u003e 0 $. In short, if $ \\ Delta\u003e 0 $ is the signs of $ \\ FRAC (\\ Partial ^ 2z) (\\ Partial x ^ 2) $ and $ \\ FRAC (\\ Partial ^ 2z) (\\ Partial y ^ 2) $ coincide.

Example №1

Explore the Extreme function $ z \u003d 4x ^ 2-6xy-34x + 5y ^ 2 + 42y + $ 7.

$$ \\ FRAC (\\ Partial z) (\\ Partial x) \u003d 8x-6y-34; \\ FRAC (\\ Partial z) (\\ Partial y) \u003d - 6x + 10y + 42. $$.

$$ \\ left \\ (\\ begin (aligned) & 8x-6y-34 \u003d 0; \\\\ & -6x + 10y + 42 \u003d 0. \\ End (Aligned) \\ Right. $$

We will reduce each equation of this system by $ 2 $ and transferring the numbers to the right parts of the equations:

{!LANG-a705dddbcdfb2303a7a6424e6851a727!}

$$ \\ left \\ (\\ begin (aligned) & 4x-3y \u003d 17; \\\\ & -3x + 5y \u003d -21. \\ END (Aligned) \\ Right. $$

We received a system of linear algebraic equations. In this situation, it seems to me the most convenient use of the Cramer method to solve the system obtained.

$$ \\ Begin (Aligned) & \\ Delta \u003d \\ Left | \\ Begin (Array) (CC) 4 & -3 \\\\ -3 & 5 \\ END (Array) \\ Right | \u003d 4 \\ CDOT 5 - (- 3) \\ Cdot (-3) \u003d 20-9 \u003d 11; \\ \\ Begin (Array) (CC) 17 & -3 \\\\ -21 & 5 \\ END (Array) \\ Right | \u003d 17 \\ CDot 5 - (- 3) \\ Cdot (-21) \u003d 85-63 \u003d 22; \\ \\ Begin (Array) (CC) 4 & 17 \\\\ -3 & -21 \\ END (Array) \\ Right | \u003d 4 \\ Cdot (-21) -17 \\ Cdot (-3) \u003d - 84 + 51 \u003d -33 . \\ END (aligned) \\\\ x \u003d \\ FRAC (\\ Delta_ (x)) (\\ Delta) \u003d \\ FRAC (22) (11) \u003d 2; \\; Y \u003d \\ FRAC (\\ Delta_ (Y)) (\\ Delta) \u003d \\ FRAC (-33) (11) \u003d - 3. $$.

The values \u200b\u200bof $ x \u003d $ 2, $ y \u003d -3 $ are the coordinates of the stationary point $ (2; -3) $.

$$ \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x ^ 2) \u003d 8; \\ FRAC (\\ Partial ^ 2 z) (\\ Partial y ^ 2) \u003d 10; \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x \\ Partial y) \u003d - 6. $$.

Calculate the value of $ \\ delta $:

$$ \\ Delta \u003d \\ FRAC (\\ Partial ^ 2z) (\\ Partial x ^ 2) \\ CDOT \\ FRAC (\\ Partial ^ 2z) (\\ Partial Y ^ 2) - \\ Left (\\ FRAC (\\ Partial ^ 2Z) ( \\ Partial X \\ Partial y) \\ RIGHT) ^ 2 \u003d 8 \\ CDOT 10 - (- 6) ^ 2 \u003d 80-36 \u003d 44. $$.

Since $ \\ delta\u003e 0 $ and $ \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x ^ 2)\u003e 0 $, according to the algorithm, the point $ (2; -3) $ is a point of a minimum function $ z $. The minimum function is $ z $ will find, substituting the coordinate of the point $ (2; -3) $:

$$ z_ (min) \u003d z (2; -3) \u003d 4 \\ CDOT 2 ^ 2-6 \\ CDOT 2 \\ CDOT (-3) -34 \\ CDOT 2 + 5 \\ CDOT (-3) ^ 2 + 42 \\ $$.

Answer: $ (2; -3) $ - minimum point; $ z_ (min) \u003d - 90 $.

Example number 2.

Examine on the extremum function $ z \u003d x ^ 3 + 3XY ^ 2-15x-12y + $ 1.

We will follow the algorithm specified above. To begin to find private derivatives of the first order:

$$ \\ FRAC (\\ Partial z) (\\ Partial x) \u003d 3x ^ 2 + 3Y ^ 2-15; \\ FRAC (\\ Partial z) (\\ Partial y) \u003d 6XY-12. $$.

We will make a system of equations $ \\ left \\ (\\ begin (aligned) \\ FRAC (\\ Partial z) (\\ Partial x) \u003d 0; \\\\ & \\ FRAC (\\ Partial z) (\\ Partial y) \u003d 0. \\ END ( Aligned) \\ Right. $:

$$ \\ left \\ (\\ begin (aligned) & 3x ^ 2 + 3Y ^ 2-15 \u003d 0; \\\\ & 6XY-12 \u003d 0. \\ END (ALIGNED) \\ RIGHT. $$

Sperate the first equation by 3, and the second - by 6.

$$ \\ left \\ (\\ begin (aligned) & x ^ 2 + y ^ 2-5 \u003d 0; \\\\ & xy-2 \u003d 0. \\ END (ALIGNED) \\ RIGHT. $$

If $ x \u003d 0 $, then the second equation will lead us to a contradiction: $ 0 \\ cdot y-2 \u003d 0 $, $ -2 \u003d 0 $. Hence the output: $ X \\ NEQ 0 $. Then, from the second equation, we have: $ xy \u003d 2 $, $ y \u003d \\ FRAC (2) (x) $. Substituting $ Y \u003d \\ FRAC (2) (x) $ in the first equation, we will have:

$$ x ^ 2 + \\ left (\\ FRAC (2) (X) \\ RIGHT) ^ 2-5 \u003d 0; \\\\ x ^ 2 + \\ FRAC (4) (x ^ 2) -5 \u003d 0; \\\\ $$.

Received a biquette equation. We make a replacement $ t \u003d x ^ 2 $ (while we mean that $ T\u003e 0 $):

$$ T ^ 2-5T + 4 \u003d 0; \\\\ \\ begin (aligned) & d \u003d (- 5) ^ 2-4 \\ Cdot 1 \\ Cdot 4 \u003d 9; \\\\ & T_1 \u003d \\ FRAC (- (- 5) - \\ SQRT (9)) (2) \u003d \\ FRAC (5-3) (2) \u003d 1; \\\\ & T_2 \u003d \\ FRAC (- (- 5) + \\ sqrt (9)) (2) \u003d \\ FRAC (5 + 3) (2) \u003d 4. \\ End (Aligned) $$

If $ t \u003d 1 $, then $ x ^ 2 \u003d 1 $. From here we have two values \u200b\u200bof $ x $: $ x_1 \u003d 1 $, $ x_2 \u003d -1 $. If $ T \u003d $ 4, then $ x ^ 2 \u003d $ 4, i.e. $ x_3 \u003d $ 2, $ x_4 \u003d -2 $. Remembering that $ Y \u003d \\ FRAC (2) (X) $, we get:

\\ begin (aligned) & y_1 \u003d \\ FRAC (2) (x_1) \u003d \\ FRAC (2) (1) \u003d 2; \\\\ & y_2 \u003d \\ FRAC (2) (x_2) \u003d \\ FRAC (2) (- 1 ) \u003d - 2; \\\\ & y_3 \u003d \\ FRAC (2) (x_3) \u003d \\ FRAC (2) (2) \u003d 1; \\\\ & y_4 \u003d \\ FRAC (2) (x_4) \u003d \\ FRAC (2) ( -2) \u003d - 1. \\ End (Aligned)

So, we have four stationary points: $ M_1 (1; 2) $, $ M_2 (-1; -2) $, $ M_3 (2; 1) $, $ M_4 (-2; -1) $. This is the first step of the algorithm over.

Now proceed to the second step of the algorithm. Find the second-order private derivatives:

$$ \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x ^ 2) \u003d 6x; \\ FRAC (\\ Partial ^ 2 z) (\\ partial y ^ 2) \u003d 6x; \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x \\ Partial y) \u003d 6Y. $$.

Find $ \\ Delta $:

$$ \\ Delta \u003d \\ FRAC (\\ Partial ^ 2z) (\\ Partial x ^ 2) \\ CDOT \\ FRAC (\\ Partial ^ 2z) (\\ Partial Y ^ 2) - \\ Left (\\ FRAC (\\ Partial ^ 2Z) ( \\ Partial x \\ Partial y) \\ Right) ^ 2 \u003d 6x \\ Cdot 6x- (6Y) ^ 2 \u003d 36x ^ 2-36Y ^ 2 \u003d 36 (x ^ 2-y ^ 2). $$.

Now we will calculate the value of $ \\ delta $ in each of the previously found stationary points. Let's start from the point $ M_1 (1; 2) $. At this point we have: $ \\ delta (m_1) \u003d 36 (1 ^ 2-2 ^ 2) \u003d - $ 108. Since $ \\ Delta (M_1)< 0$, то согласно алгоритму в точке $M_1$ экстремума нет.

We explore the point $ M_2 (-1; -2) $. At this point we have: $ \\ delta (m_2) \u003d 36 ((- 1) ^ 2 - (- 2) ^ 2) \u003d - $ 108. Since $ \\ Delta (M_2)< 0$, то согласно алгоритму в точке $M_2$ экстремума нет.

We explore the point $ M_3 (2; 1) $. At this point we get:

$$ \\ DELTA (M_3) \u003d 36 (2 ^ 2-1 ^ 2) \u003d 108; \\; \\; \\ left. \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x ^ 2) \\ Right | _ (M_3) \u003d 6 \\ Cdot 2 \u003d 12. $$.

Since $ \\ Delta (M_3)\u003e 0 $ and $ \\ left. \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x ^ 2) \\ Right | _ (M_3)\u003e 0 $, then according to the $ M_3 algorithm (2 ; 1) $ There is a point of a minimum function $ z $. At least the function of $ z $ will find, substituting the coordinate point of $ M_3 in the specified function:

$$ z_ (min) \u003d z (2; 1) \u003d 2 ^ 3 + 3 \\ CDOT 2 \\ CDOT 1 ^ 2-15 \\ CDOT 2-12 \\ CDOT 1 + 1 \u003d -27. $$.

It remains to explore the $ M_4 (-2; -1) $ point. At this point we get:

$$ \\ DELTA (M_4) \u003d 36 ((- 2) ^ 2 - (- 1) ^ 2) \u003d 108; \\; \\; \\ left. \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x ^ 2) \\ Right | _ (m_4) \u003d 6 \\ Cdot (-2) \u003d - 12. $$.

Since $ \\ Delta (M_4)\u003e 0 $ and $ \\ left. \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x ^ 2) \\ Right | _ (M_4)< 0$, то согласно алгоритму $M_4(-2;-1)$ есть точкой максимума функции $z$. Максимум функции $z$ найдём, подставив в заданную функцию координаты точки $M_4$:

$$ z_ (max) \u003d z (-2; -1) \u003d (- 2) ^ 3 + 3 \\ Cdot (-2) \\ Cdot (-1) ^ 2-15 \\ CDOT (-2) -12 \\ CDOT (-1) + 1 \u003d 29. $$.

Research on the extremum is completed. It remains only to record the answer.

$ (2; 1) $ - a minimum point, $ z_ (min) \u003d - 27 $;
$ (- 2; -1) $ - Maximum point, $ z_ (max) \u003d $ 29.

Note

Calculate the value of $ \\ Delta $ in the general case is not necessary, because we are only interested in the sign, and not the specific value of this parameter. For example, for the example discussed above, No. 2 at point $ M_3 (2; 1) $ we have $ \\ delta \u003d 36 \\ Cdot (2 ^ 2-1 ^ 2) $. It is obvious here that $ \\ delta\u003e 0 $ (as both factors $ 36 $ and $ (2 ^ 2-1 ^ 2) $ are positive) and you can not find a specific value of $ \\ Delta $. True, for typical calculations, this remark is useless - they require the calculation to the number 🙂

Example number 3.

Investigate on the extremum function $ z \u003d x ^ 4 + y ^ 4-2x ^ 2 + 4xy-2y ^ 2 + 3 $.

We will follow the algorithm. To begin to find private derivatives of the first order:

$$ \\ FRAC (\\ Partial Z) (\\ Partial x) \u003d 4x ^ 3-4x + 4y; \\ FRAC (\\ Partial z) (\\ Partial y) \u003d 4Y ^ 3 + 4x-4y. $$.

$$ \\ left \\ (\\ begin (aligned) & 4x ^ 3-4x + 4y \u003d 0; \\\\ & 4Y ^ 3 + 4x-4y \u003d 0. \\ End (Aligned) \\ Right. $$

We will reduce both equations for $ 4 $:

$$ \\ left \\ (\\ begin (aligned) & x ^ 3-x + y \u003d 0; \\\\ & y ^ 3 + x-y \u003d 0. \\ End (Aligned) \\ Right. $$

Add to the second equation first and express $ y $ than $ x $:

$$ y ^ 3 + x-y + (x ^ 3-x + y) \u003d 0; \\\\ y ^ 3 + x ^ 3 \u003d 0; y ^ 3 \u003d -x ^ 3; y \u003d -x. $$.

Substituting $ Y \u003d -x $ in the first equation of the system, we will have:

$$ x ^ 3-x-x \u003d 0; \\\\ x ^ 3-2x \u003d 0; \\\\ x (x ^ 2-2) \u003d 0. $$.

From the resulting equation, we have: $ x \u003d 0 $ or $ x ^ 2-2 \u003d 0 $. From the equation $ x ^ 2-2 \u003d 0 $ it follows that $ x \u003d - \\ sqrt (2) $ or $ x \u003d \\ sqrt (2) $. So, there are three values \u200b\u200bof $ x $, namely: $ x_1 \u003d 0 $, $ x_2 \u003d - \\ sqrt (2) $, $ x_3 \u003d \\ sqrt (2) $. Since $ y \u003d -x $, then $ y_1 \u003d -x_1 \u003d 0 $, $ y_2 \u003d -x_2 \u003d \\ sqrt (2) $, $ y_3 \u003d -x_3 \u003d - \\ sqrt (2) $.

The first step of the solution is over.

How to find extremum (minimum and maximum points) functions

We got three stationary points: $ M_1 (0; 0) $, $ m_2 (- \\ sqrt (2), \\ sqrt (2)) $, $ M_3 (\\ SQRT (2), - \\ SQRT (2)) $ .

Now proceed to the second step of the algorithm. Find the second-order private derivatives:

$$ \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x ^ 2) \u003d 12x ^ 2-4; \\ FRAC (\\ Partial ^ 2 z) (\\ Partial y ^ 2) \u003d 12y ^ 2-4; \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x \\ Partial y) \u003d 4. $$.

Find $ \\ Delta $:

$$ \\ Delta \u003d \\ FRAC (\\ Partial ^ 2z) (\\ Partial x ^ 2) \\ CDOT \\ FRAC (\\ Partial ^ 2z) (\\ Partial Y ^ 2) - \\ Left (\\ FRAC (\\ Partial ^ 2Z) ( \\ Partial x \\ Partial y) \\ Right) ^ 2 \u003d (12x ^ 2-4) (12Y ^ 2-4) -4 ^ 2 \u003d \\\\ \u003d 4 (3x ^ 2-1) \\ CDOT 4 (3Y ^ 2 -1) -16 \u003d 16 (3x ^ 2-1) (3Y ^ 2-1) -16 \u003d 16 \\ CDOT ((3x ^ 2-1) (3Y ^ 2-1) -1). $$.

Now we will calculate the value of $ \\ delta $ in each of the previously found stationary points. Let's start from the point $ M_1 (0; 0) $. At this point we have: $ \\ delta (M_1) \u003d 16 \\ CDOT ((3 \\ Cdot 0 ^ 2-1) (3 \\ Cdot 0 ^ 2-1) -1) \u003d 16 \\ Cdot 0 \u003d 0 $. Since $ \\ Delta (M_1) \u003d 0 $, according to the algorithm, an additional study is required, because it is impossible to say anything definite about the presence of extremum in the point under consideration. Let us drop this point alone and move in other points.

We investigate the point $ M_2 (- \\ SQRT (2), \\ SQRT (2)) $. At this point we get:

\\ Begin (Aligned) \\ Delta (M_2) \u003d 16 \\ CDOT ((3 \\ CDOT (- \\ SQRT (2)) ^ 2-1) (3 \\ CDOT (\\ SQRT (2)) ^ 2-1) - 1) \u003d 16 \\ CDOT 24 \u003d 384; \\\\ \\ LEFT. \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x ^ 2) \\ Right | _ (M_2) \u003d 12 \\ CDOT (- \\ SQRT (2) ) ^ 2-4 \u003d 24-4 \u003d 20. \\ End (Aligned)

Since $ \\ Delta (M_2)\u003e 0 $ and $ \\ left. \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x ^ 2) \\ Right | _ (M_2)\u003e 0 $, according to the $ M_2 algorithm (- \\ SQRT (2), \\ SQRT (2)) $ There is a point of a minimum function $ z $. At least the function of $ z $ will find, substituting in the specified function of the coordinate point $ M_2 $:

$$ z_ (min) \u003d z (- \\ sqrt (2), \\ sqrt (2)) \u003d (- \\ sqrt (2)) ^ 4 + (\\ SQRT (2)) ^ 4-2 (- \\ sqrt ( 2)) ^ 2 + 4 \\ CDOT (- \\ SQRT (2)) \\ SQRT (2) -2 (\\ sqrt (2)) ^ 2 + 3 \u003d -5. $$.

Similar to the previous paragraph, we investigate the point $ M_3 (\\ SQRT (2), - \\ SQRT (2)) $. At this point we get:

\\ begin (Aligned) \\ Delta (M_3) \u003d 16 \\ CDOT ((3 \\ CDOT (\\ SQRT (2)) ^ 2-1) (3 \\ CDOT (- \\ SQRT (2)) ^ 2-1) - 1) \u003d 16 \\ CDOT 24 \u003d 384; \\\\ \\ LEFT. \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x ^ 2) \\ Right | _ (M_3) \u003d 12 \\ CDOT (\\ SQRT (2)) ^ 2-4 \u003d 24-4 \u003d 20. \\ End (Aligned)

Since $ \\ Delta (M_3)\u003e 0 $ and $ \\ left. \\ FRAC (\\ Partial ^ 2 z) (\\ Partial x ^ 2) \\ Right | _ (M_3)\u003e 0 $, according to the $ M_3 algorithm (\\ At least the function of $ z $ will find, substituting the coordinate point of $ M_3 in the specified function:

$$ z_ (min) \u003d z (\\ sqrt (2), - \\ sqrt (2)) \u003d (\\ sqrt (2)) ^ 4 + (- \\ sqrt (2)) ^ 4-2 (\\ SQRT (2 )) ^ 2 + 4 \\ Cdot \\ SQRT (2) (- \\ SQRT (2)) - 2 (- \\ sqrt (2)) ^ 2 + 3 \u003d -5. $$.

It has arrived back to the point $ M_1 (0; 0) $, in which $ \\ Delta (M_1) \u003d 0 $. According to the algorithm, an additional study is required. Under this evasive phrase is meant "Do what you want" :). There is no general way to resolve such situations - and this is understandable. If such a way was, he would long come to all textbooks. And Pesima has to look for a special approach to each point in which $ \\ delta \u003d 0 $. Well, let's share the behavior of the function in the neighborhood of the point $ M_1 (0; 0) $. Immediately note that $ z (m_1) \u003d z (0; 0) \u003d $ 3. Suppose $ M_1 (0; 0) $ is a minimum point. Then for any point $ m $ from some neighborhood of the point $ M_1 (0; 0) $ we obtain $ z (m)\u003e z (m_1) $, i.e. $ z (m)\u003e $ 3. And what if any neighborhood contains the points in which $ z (M)< 3$? Тогда в точке $M_1$ уж точно не будет минимума.

Consider the points whose $ y \u003d 0 $, i.e. Points of type $ (x, 0) $. At these points, the $ z $ function will take such values:

$$ z (x, 0) \u003d x ^ 4 + 0 ^ 4-2x ^ 2 + 4x \\ Cdot 0-2 \\ Cdot 0 ^ 2 + 3 \u003d x ^ 4-2x ^ 2 + 3 \u003d x ^ 2 (x ^ 2-2) +3. $$.

In all sufficiently small surroundings $ M_1 (0; 0) $ we have $ x ^ 2-2< 0$, посему $x^2(x^2-2) < 0$, откуда следует $x^2(x^2-2)+3 < 3$. Вывод: любая окрестность точки $M_1(0;0)$ содержит точки, в которых $z < 3$, посему точка $M_1(0;0)$ не может быть точкой минимума.

But maybe the point $ M_1 (0; 0) $ is the maximum point? If so, then for any point $ m $ from some neighborhood of the point $ M_1 (0; 0) $ we get $ z (M)< z(M_1) $, т.е. $z(M) < 3$. А вдруг любая окрестность содержит точки, в которых $z(M) > $ 3? Then at point $ M_1 $ will definitely not be a maximum.

Consider the points whose $ y \u003d x $, i.e. The points of the form $ (x, x) $. At these points, the $ z $ function will take such values:

$$ z (x, x) \u003d x ^ 4 + x ^ 4-2x ^ 2 + 4x \\ cdot x-2 \\ cdot x ^ 2 + 3 \u003d 2x ^ 4 + 3. $$.

Since in any neighborhood of the point $ M_1 (0; 0) $ we have $ 2x ^ 4\u003e 0 $, then $ 2x ^ 4 + 3\u003e $ 3. Conclusion: Any neighborhood of a point $ M_1 (0; 0) $ contains the points in which $ z\u003e $ 3, therefore the point $ M_1 (0; 0) $ cannot be a maximum point.

Point $ M_1 (0; 0) $ is not a point of maximum, no point of minimum. Conclusion: $ M_1 $ is generally not an extremum point.

Answer: $ (- \\ SQRT (2), \\ SQRT (2)) $, $ (\\ sqrt (2), - \\ sqrt (2)) $ - minimum points $ z $. In both points $ z_ (min) \u003d - $ 5.

Online classes in higher mathematics

Function \u003d F (x) is called increasing (descending) in some interval, if at x 1< x 2 выполняется неравенство(f(x 1) < f (x 2) (f(x 1) >f (x 2)).

If the differentiable function \u003d f (x) on the segment increases (decreases), then its derivative on this segment f "(x)\u003e 0, (f" (x)< 0).

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

Maximum and minimum points are called points of extremum, and the values \u200b\u200bof the function at these points - it extremes.

Points of extremum

The necessary conditions Extremum. If point x. about It is an extremum point F (X), then either f "(x o) \u003d 0, or F (x o) does not exist. Such points are called critical Moreover, the function itself is defined at the critical point. Extreme function should be sought among its critical points.

The first sufficient condition. Let be x. about - critical point. If f "(x) when moving through a point x. about changes the sign plus on minus, then at the point x O. The function has a maximum in otherwise - Minimum. If during the transition through the critical point the derivative does not change the sign, then at the point x. about Extremum is not.

The second sufficient condition. Let the function f (x) be f "(x) in the neighborhood of the point X. about and second derivative f "" (x 0) at the point itself x O.. If f "(x o) \u003d 0, f" "(x 0)\u003e 0, (f" "(x 0)<0), то точка X O. It is a point of local minimum (maximum) Function F (X). If F "" (x 0) \u003d 0, then you need to either use the first sufficient condition, or to attract the highest.

On the segment, the function y \u003d f (x) can reach the smallest or greatest value or at critical points, or at the ends of the segment.

Example 3.22.Find extremmas function f (x) \u003d 2x 3 - 15x 2 + 36x - 14.

Decision.Since f "(x) \u003d 6x 2 - 30x +36 \u003d 6 (x -2) (x - 3), then the critical points of the function x 1 \u003d 2 and x 2 \u003d 3. Extremes can only be at these points. So As in the transition through the point x 1 \u003d 2, the derivative changes the sign plus per minus, then at this point the function has a maximum. When switching through the point x 2 \u003d 3, the derivative changes the minus sign on plus, therefore at point x 2 \u003d 3 in the function at least. Calculate the function value at the points x 1 \u003d 2 and x 2 \u003d 3, we find the extremums of the function: the maximum f (2) \u003d 14 and the minimum F (3) \u003d 13.

Tasks for finding extremum functions

Example 3.23.a.

Decision. x. and y.. The area area is equal to S \u003d XY. Let be y. - This is the length of the side adjacent to the wall. Then, by the condition, the equality 2x + y \u003d a should be performed. Therefore, y \u003d a - 2x and s \u003d x (a - 2x), where 0 ≤x ≤a / 2 (the length and width of the site cannot be negative). S "\u003d a - 4x, a - 4x \u003d 0 with x \u003d a / 4, from where y \u003d a - 2 × a / 4 \u003d a / 2. Since X \u003d A / 4 is the only critical point, check whether the sign changes derivative when switching through this point. at x< a/4, S " > 0, and with x\u003e a / 4, s "< 0, значит, в точке x = a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед). Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет наибольшим значением функции. Таким образом, наиболее выгодным соотношением сторон площадки при данных условиях задачи является y = 2x.

Example 3.24.

Decision.
R \u003d 2, N \u003d 16/4 \u003d 4.

Example 3.22.Find extremmas functionF (x) \u003d 2x 3 - 15x 2 + 36x - 14.

Decision.Since f "(x) \u003d 6x 2 - 30x +36 \u003d 6 (x -2) (x - 3), then the critical points of the function x 1 \u003d 2 and x 2 \u003d 3. Extremes can only be at these points. So As in the transition through the point x 1 \u003d 2, the derivative changes the sign plus per minus, then at this point the function has a maximum. When switching through the point x 2 \u003d 3, the derivative changes the minus sign on plus, therefore at point x 2 \u003d 3 in the function at least. Calculate the values \u200b\u200bof the function at points x 1 \u003d 2 and x 2 \u003d 3, we find the extremums of the function: maximum (2) \u003d 14 and minimaf (3) \u003d 13.

Example 3.23.It is necessary to build a rectangular platform near the stone wall so that it is drilled off with a wire mesh from three sides, and adjoined the wall to the wall. For this is available a. Running mesh patterns. With what aspect ratio will have the highest square?

Decision.Denote the side of the site through x. and y.. The area area is equal to S \u003d XY. Let be y. - This is the length of the side adjacent to the wall. Then, by the condition, the equality 2x + y \u003d a should be performed. Therefore y \u003d a - 2x and s \u003d x (a - 2x), where
0 ≤x ≤a / 2 (the length and width of the site cannot be negative). S "\u003d a - 4x, a - 4x \u003d 0 at x \u003d a / 4, from where
Y \u003d A - 2A / 4 \u003d A / 2. Since X \u003d A / 4 is the only critical point, check whether the sign is changing during the transition through this point. Atx< a/4, S " > 0, and with x\u003e a / 4 s "< 0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед). Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет наибольшим значением функции. Таким образом, наиболее выгодным соотношением сторон площадки при данных условиях задачи является y = 2x.

Example 3.24.It is required to make a closed cylindrical tank with a capacity V \u003d 16p ≈ 50 m 3. What should be the sizes of the tank (R radius and height H) so that the least amount of material goes on its manufacture?

Decision.The area of \u200b\u200bthe full surface of the cylinder is S \u003d 2PR (R + H). We know the volume of the cylinder V \u003d PR 2 H þ H \u003d V / PR 2 \u003d 16P / PR 2 \u003d 16 / R 2. So, S (R) \u003d 2p (R 2 + 16 / R). Find a derivative of this feature:
S "(R) \u003d 2p (2R- 16 / R 2) \u003d 4P (R- 8 / R 2). S" (R) \u003d 0 at R 3 \u003d 8, therefore,
R \u003d 2, N \u003d 16/4 \u003d 4.

Consider two teeth well all the famous saw profile. We will send the axis along the smooth side of the saw, and the axis is perpendicular to it. We obtain a graph of some function shown in Fig. one.

It is obvious that at the point, and at the point, the values \u200b\u200bof the function turn out to be the highest in comparison with the values \u200b\u200bin the adjacent points on the right and left, and at the point - the smallest in comparison with adjacent points. The points are called the extremum points of the function (from Latin Extremum - "extreme"), points and - maximum points, and the point is a point of a minimum (from Latin Maximum and Minimum - "the largest" and "smallest").

We clarify the definition of extremum.

It is said that the function at the point has a maximum if there is an interval containing the point and belonging the function of determining the function, such that for all points of this interval it turns out. Accordingly, the function at the point has a minimum if the condition is performed for all points of some interval.

In fig. 2 and 3 are graphs of functions having an extremum at point.

We draw attention to the fact that by definition the point of extremum should lie inside the set of function setting, and not at its end. Therefore, for the function shown in Fig. 1, it is impossible to assume that at the point it has a minimum.

If in this definition of the maximum (minimum) of the functions to replace strict inequality on the non-stroke , I get the definition of a non-stroke maximum (incredible minimum). Consider for example the top profile of the mountain (Fig. 4). Each point of a flat platform - a segment is a point of a non-strict maximum.

In differential calculus, the study of the function on extremes is very effective and fairly simply carried out using a derivative. One of the main theorems of differential calculus, which establishes the necessary extremum condition of the differentiable function, is the farm theorem (see the farm farm). Let the function at the point have an extremum. If at this point there is a derivative, then it is zero.

On the geometric language, the farm theorem means that at the extremum point tangent to the graphics of the horizontal function (Fig. 5). The opposite statement, of course, is incorrect, which shows, for example, a graph in Fig. 6.

The theorem is named after the French mathematics P. Farm, who one of the first solved a number of tasks to extremum. He has not yet aroused the concept of a derivative, but applied a method in the study, the essence of which is expressed in the approval of the theorem.

A sufficient condition for the extremum of the differentiable function is to change the sign of the derivative. If at the point the derivative changes the sign from a minus to plus, i.e. Its decrease is replaced by an increase, the point will be a point of minimum. On the contrary, the point will be a maximum point if the derivative changes the sign from the plus to minus, i.e. Goes from increasing to descending.

The point where the derived function is zero, is called stationary. If the differential function is examined for an extremum, then all its stationary points should be found and consider the signs of the derivative left and right from them.

Explore the extremum function.

We will find its derivative: .

A simple algorithm for finding extremes ..

Find a derivative function
Equate this derivative to zero
We find the values \u200b\u200bof the variable of the resulting expression (the values \u200b\u200bof the variable in which the derivative is converted to zero)
We divide these values \u200b\u200bto the coordinate straight on the intervals (you do not need to forget about the gap points, which should also be applied to direct), all these points are called "suspicious" points for extremum
Calculate, on which of these gaps the derivative will be positive, and on what is negative. To do this, substitute the value from the gap in the derivative.

From points suspicious to extremum, you need to find exactly. For this, we look at our gaps on the coordinate direct. If when passing through some point, the derivative sign changes from the plus to minus, then this point will be maximum, and if with a minus on plus, then minimum.

To find the largest and smallest function value, you need to calculate the value of the function at the sections of the segment and at the extremum points. Then choose the largest and smallest value.

Consider example
We find a derivative and equate it to zero:

The obtained values \u200b\u200bof the variables are applied to the coordinate direct and calculate the sign of the derivative on each of the gaps. Well, for example, for the first we take-2 , then the derivative will be equal-0,24 for the second we take0 , then the derivative will be2 , and for the third, take2 , then the derivative will be-0.24. I put the corresponding signs.

We see that when passing through the point -1, the derivative changes a sign from a minus to plus, that is, it will be a minimum point, and when passing through 1 - from the plus to minus, this is a maximum point, respectively.

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