It is said that the function has inside
Region D. local maximum(minimum) if there is such a neighborhood reader
for each point
which is carried out inequality

If the function has at the point
local maximum or local minimum, then they say that it has at this point local extremum(or Just extremum).

Theorem (necessary condition for the existence of extremum). If the differentiated functions of the extremum at the point
, then each private derivative of the first order from the function at this point adds to zero.

Points in which all private derivatives of the first order appeal to zero are called stationary features of the function
. The coordinates of these points can be found by deciding the system from equations

.

The necessary condition for the existence of an extremum in the case of a differentiable function can be briefly formulated as follows:

There are cases when at regular points some private derivatives have endless values \u200b\u200bor do not exist (while the rest are zero). Such points are called critical points of function.These points also need to be considered as "suspicious" to extremum, as well as stationary.

In case of two variables prerequisite Extremum, namely equality zero of private derivatives (differential) at the point of extremum, has a geometric interpretation: tangent
at the point of extremum should be parallel to the plane
.

20. Sufficient conditions for the existence of extremum

Execution at some point of the required condition of existence of extremum does not guarantee the presence of extremum there. As an example, you can take a differential function differentiating
. Both of its private derivatives and the function itself appeal to zero at the point.
. However, in any neighborhood of this point there is both positive (large
) and negative (smaller
) The values \u200b\u200bof this function. Consequently, at this point, by definition, the extremum is not observed. Therefore, it is necessary to know sufficient conditions in which the point, suspicious to the extremum, is the extremum point of the function under study.

Consider the case of the function of two variables. Suppose that the function
it is determined continuous and has continuous private derivatives to second order inclusive in the vicinity of a certain point.
which is a fixed point function
that is, satisfies the conditions

,
.

We introduce notation:

Theorem (sufficient conditions for the existence of extremum). Let the function
satisfies the above conditions, namely: differentiating in some neighborhood of a stationary point
and twice differentiating at the very point
. Then, if


If
that function
at point
reaches

local maximumfor
and

local minimumfor
.

In general, for function
a sufficient condition for existence at the point
localminimum(maximum) is an positive(negative) Definition of the second differential.

In other words, the following statement is right.

Theorem . If at the point
for function

for anyone not equal at the same time zero
then at this point the function has minimum(similar to maximum, if a
).

Example 18.Find points local extremum functions

Decision. We find private derivatives and equating them to zero:

Solving this system, we find two points of possible extremum:

We will find private derivatives of the second order for this feature:

In the first stationary point, therefore, and
Therefore, this point requires an additional study. Meaning function
at this point is zero:
Further,

for

but

for

Consequently, in any neighborhood of the point
function
takes values \u200b\u200bas big
and smaller
, and, then, at the point
function
By definition, it does not have a local extremum.

In the second stationary point



therefore, so because
then at point
the function has a local maximum.

Definition: The point X0 is called a local maximum (or minimum) point if, in some neighborhood of the point x0, the function takes the greatest (or smallest) value, i.e. For all x from some neighborhood of the point x0, the condition f (x) f (x0) (or f (x) f (x0)) is performed.

Local maximum or minimum points are combined common title - points of local extremum function.

Note that at points of the local extremum, the function reaches its greatest or the least meaning Only in some local area. There may be cases when the value of weaxuin.

Required sign of the existence of local extremum function

Theorem . If the continuous function y \u003d f (x) has a local extremum at point x0, then at this point the first derivative is either zero or does not exist, i.e. Local extremum takes place at critical points of the form I.

At the local extremum points, either the tangential axis of the 0x axis or there are two tangents (see Figure). Note that the critical points are necessary, but the lack of local extremum. Local extremum takes place only at critical points of the type I, but the local extremum takes place at all critical points.

For example: Cubic parabola y \u003d x3, has a critical point x0 \u003d 0, in which the derivative y / (0) \u003d 0, but the critical point x0 \u003d 0 is not an extremum point, and there is a point of inflection (see below).

Sufficient sign of the existence of local extremum function

Theorem . If during the transition of the argument through the critical point I of the genus left to the right of the first derivative in / (x)

changes the sign from "+" to "-", the continuous function of (x) in this critical point has a local maximum;

changes the sign from "-" on "+", then the continuous function of (x) has a local minimum in this critical point

does not change the sign, then in this critical point there is no local extremum, there is a point of inflection.

For a local maximum, the area of \u200b\u200bincrement of the function (y / 0) is replaced by the larger area of \u200b\u200bthe function (y / 0). For a local minimum, the decrease region of the function (y / 0) is replaced by the area of \u200b\u200bthe increase function (y / 0).

Example: Investigate the function y \u003d x3 + 9x2 + 15x - 9 on monotony, extremum and build a graph of a function.

We will find the critical points of I of the genus, determining the derivative (y /) and equating it with zero: at / \u003d 3x2 + 18x + 15 \u003d 3 (x2 + 6x + 5) \u003d 0

Spest square three decreases with the help of discriminant:

x2 + 6x + 5 \u003d 0 (a \u003d 1, B \u003d 6, C \u003d 5) d \u003d, x1k \u003d -5, x2k \u003d -1.

2) We break the numeric axis with critical points for 3 areas and we define the signs of the derivative (y /). According to these signs, we will find areas of monotony (increase and decrease) of functions, and by changing signs to determine the points of local extremum (maximum and minimum).

The results of the study will be submitted in the form of a table from which the following conclusions can be drawn:

  • 1. On the interval of / (- 10) 0, the function monotonically increases (the sign of the derivative y was estimated at the control point x \u003d -10 taken in this interval);
  • 2. On the interval (-5; -1) in / (- 2) 0, the function monotonically decreases (the sign of the derivative y was estimated at the control point x \u003d -2, taken in this interval);
  • 3. On the interval of / (0) 0, the function monotonously increases (the sign of the derivative y was estimated at the control point x \u003d 0 taken in this interval);
  • 4. When switching through the critical point x1k \u003d -5, the derivative changes the sign from "+" to "-", therefore this point is a local maximum point
  • (ymax (-5) \u003d (-5) 3 + 9 (-5) 2 +15 (-5) -9 \u003d -125 + 225 - 75 - 9 \u003d 16);
  • 5. When switching through the critical point X2K \u003d -1, the derivative changes the sign from "-" on "+", therefore, this point is a local minimum point
  • (ymin (-1) \u003d -1 + 9 - 15 - 9 \u003d - 16).

x -5 (-5; -1) -1

3) Constructing a graph to follow the results of the study with the attraction of additional calculations of the functions of the function at check points:

we build a rectangular coordinate system OhU;

we show the maximum point coordinates (-5; 16) and a minimum (-1; -16);

to clarify the graph, we calculate the value of the function at the control points, choosing them on the left and right on the maximum points and the minimum and inside the average interval, for example: y (-6) \u003d (- 6) 3 +9 (-6) 2 + 15 (-6 ) -9 \u003d 9; y (-3) \u003d (- 3) 3 + 9 (-3) 2 + 15 (-3) -9 \u003d 0;

(0) \u003d -9 (-6; 9); (-3; 0) and (0; -9) - estimated control points that are applied to build a schedule;

show the graph in the form of a curve by convexing up at the maximum point and convexing down at the point of the minimum and passing through the calculated control points.

\u003e\u003e Extremes

Extreme function

Determination of Extremum

Function y \u003d f (x) 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 differential function y \u003d f (x) on the segment increases (decreases), then its derivative on this segment F " (x)> 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 is faithful 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

Required conditions of extremum . If point x. about 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 switching through the point x. about changes the sign plus on minus, then at the point x O. The function has a maximum, otherwise, a 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 at the very point x O. . If f "(x O.) = 0, >0 ( <0), то точка x O. It is a point of local minimum (maximum) Function F (X). If \u003d 0, then you need to either use the first sufficient condition, or 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.

Decision.As f. " (

Tasks for finding extremum functions

Example 3.23. a.

Decision. x. and y. y.
0 ≤ x.≤
\u003e 0, and when x\u003e A / 4 S " < 0, значит, в точке x=a /4 функция S имеет максимум. Значение Functions kv.. elf).

Example 3.24.p ≈

Decision.p P.
S "
R \u003d 2, N \u003d 16/4 \u003d 4.

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

Decision.As 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. Since when switching through the point x 1 \u003d 2, the derivative changes the sign plus on minus, then at this point the function has a maximum. When switching through the point x 2 \u003d 3, the derivative changes the sign minus on plus, therefore at point x 2 \u003d 3 in the function at least. Calculate function values \u200b\u200bat points
x 1 \u003d 2 and x 2 \u003d 3, we find the extremums of the function: the maximum f (2) \u003d 14 and at least f (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 area?

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 - 2 × A / 4 \u003d A / 2. Insofar as x \u003d A / 4 is the only critical point, check whether the sign is changing during the transition through this point. With x a / 4 s "\u003e 0, and when x\u003e A / 4 S " < 0, значит, в точке x=a /4 функция S имеет максимум. Значение Functions S (A / 4) \u003d A / 4 (A - A / 2) \u003d A 2/8 (kv.. elf). Since S is continuous on and its values \u200b\u200bat the ends of S (0) and S (A / 2) are zero, the value found will be the greatest function value. Thus, the most advantageous aspect ratio of the site under these conditions of the problem is y \u003d 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 total surface of the cylinder is S \u003d 2p. R (R + H). We know the volume of the cylinder V \u003dp R 2 H þ H \u003d V / P R 2 \u003d 16 p / p R 2 \u003d 16 / R 2. So, S (R) \u003d 2p. (R 2 + 16 / R). Find a derivative of this feature:
S "(R) \u003d 2 P (2R- 16 / R 2) \u003d 4 P (R- 8 / R 2). S " (R) \u003d 0 at R 3 \u003d 8, therefore,
R \u003d 2, N \u003d 16/4 \u003d 4.

$ E \\ Subset \\ Mathbb (R) ^ (n) $. It is said that $ F $ has local maximum At point $ x_ (0) \\ in E $, if there is such a neighborhood of $ U $ points $ x_ (0) $, that for all $ X \\ in U $, an inequality is $ f \\ left (x \\ right) \\ leqslant f \\ left (x_ (0) \\ Right) $.

Local maximum called strict If the neighborhood of $ u $ can be chosen so that for all $ x \\ in u $, different from $ x_ (0) $, there was $ F \\ Left (X \\ Right)< f\left(x_{0}\right)$.

Definition
Let $ F $ be the actual function on an open set $ E \\ Subset \\ Mathbb (R) ^ (n) $. It is said that $ F $ has local minimum At point $ x_ (0) \\ in E $, if there is such a neighborhood of $ U $ points $ x_ (0) $, that for all $ x \\ in U $, an inequality of $ F \\ Left (X \\ Right) \\ GEQSLANT F \\ left (x_ (0) \\ Right) $.

The local minimum is called strict if the neighborhood of $ U $ can be chosen so that for all $ x \\ in u $, different from $ x_ (0) $, there was $ F \\ Left (X \\ Right)\u003e F \\ Left (X_ ( 0) \\ Right) $.

Local extremum combines the concepts of a local minimum and a local maximum.

Theorem (the required condition of the extremum differentiable function)
Let $ F $ be the actual function on an open set $ E \\ Subset \\ Mathbb (R) ^ (n) $. If at point $ x_ (0) \\ in E $, the $ F function has a local extremum and at this point, then $$ \\ Text (D) F \\ Left (x_ (0) \\ Right) \u003d 0. $$ Equality zero The differential is equivalent to the fact that all are zero, i.e. $$ \\ DISPLAYSTYLE \\ FRAC (\\ Partial F) (\\ Partial x_ (i)) \\ left (x_ (0) \\ right) \u003d 0. $$

In the one-dimensional case it is. Denote by $ \\ phi \\ left (t \\ right) \u003d f \\ left (x_ (0) + th \\ right) $, where $ h $ is an arbitrary vector. The function $ \\ phi $ is defined with sufficiently small modulo values \u200b\u200bof $ t $. In addition, according to, it is differentiable, and $ (\\ phi) '\\ left (T \\ Right) \u003d \\ Text (D) F \\ Left (x_ (0) + Th \\ Right) H $.
Let $ F $ have a local maximum at $ 0 $ point. It means that the function $ \\ Phi $ with $ T \u003d 0 $ has a local maximum and, according to the farm theorem, $ (\\ phi) '\\ left (0 \\ right) \u003d 0 $.
So, we got that $ df \\ left (x_ (0) \\ right) \u003d 0 $, i.e. Functions $ F $ At point $ x_ (0) $ is zero on any vector $ H $.

Definition
Points in which the differential is zero, i.e. Such in which all private derivatives are zero, are called stationary. Critical points The functions of $ F $ are called such points in which $ F $ is not differentiated or equal to zero. If the point is stationary, then it does not yet follow that at this point the function has an extremum.

Example 1.
Let $ f \\ left (x, y \\ right) \u003d x ^ (3) + y ^ (3) $. Then $ \\ displaystyle \\ FRAC (\\ Partial F) (\\ Partial x) \u003d 3 \\ CDOT X ^ (2) $, $ \\ displaystyle \\ FRAC (\\ Partial F) (\\ Partial y) \u003d 3 \\ Cdot y ^ (2 ) $, so $ \\ left (0.0 \\ Right) $ is a stationary point, but at this point the function does not have an extremum. Indeed, $ F \\ Left (0.0 \\ Right) \u003d 0 $, but it is easy to see that in any neighborhood of the $ \\ left point (0.0 \\ RIGHT) $ function takes both positive and negative values.

Example 2.
The function is $ F \\ left (x, y \\ right) \u003d x ^ (2) - y ^ (2) $ start - stationary point, but it is clear that there is no extremum at this point.

Theorem (sufficient extremum condition).
Let the function $ F $ double-continuously differentiable on an open set $ E \\ Subset \\ MathBB (R) ^ (N) $. Let $ x_ (0) \\ in E $ - a stationary point and $$ \\ displaystyle q_ (x_ (0)) \\ left (h \\ right) \\ Equiv \\ Sum_ (i \u003d 1) ^ n \\ Sum_ (J \u003d 1) ^ n \\ FRAC (\\ Partial ^ (2) f) (\\ Partial x_ (i) \\ Partial x_ (j)) \\ left (x_ (0) \\ right) h ^ (i) H ^ (J). $$ Then

  1. if $ q_ (x_ (0)) $ -, then the function $ F $ at $ x_ (0) $ has a local extremum, namely, at a minimum, if the form is positively defined, and the maximum, if the form is negatively defined;
  2. if the quadratic form $ q_ (x_ (0)) $ is indefinite, then the function $ F $ at $ x_ (0) $ does not have an extremum.

We use the decomposition by the Taylor formula (12.7 p. 292). Considering that the individual derivatives of the first order at point $ x_ (0) $ are zero, we get $$ \\ displaystyle f \\ left (x_ (0) + h \\ right) -f \\ left (x_ (0) \\ right) \u003d \\ \\ left (x_ (0) + \\ theta h \\ right) h ^ (i) H ^ (J), $$ where $ 0<\theta<1$. Обозначим $\displaystyle a_{ij}=\frac{\partial^{2} f}{\partial x_{i} \partial x_{j}} \left(x_{0}\right)$. В силу теоремы Шварца (12.6 стр. 289-290) , $a_{ij}=a_{ji}$. Обозначим $$\displaystyle \alpha_{ij} \left(h\right)=\frac{\partial^{2} f}{\partial x_{i} \partial x_{j}} \left(x_{0}+\theta h\right)−\frac{\partial^{2} f}{\partial x_{i} \partial x_{j}} \left(x_{0}\right).$$ По предположению, все непрерывны и поэтому $$\lim_{h \rightarrow 0} \alpha_{ij} \left(h\right)=0. \left(1\right)$$ Получаем $$\displaystyle f \left(x_{0}+h\right)−f \left(x_{0}\right)=\frac{1}{2}\left.$$ Обозначим $$\displaystyle \epsilon \left(h\right)=\frac{1}{|h|^{2}}\sum_{i=1}^n \sum_{j=1}^n \alpha_{ij} \left(h\right)h_{i}h_{j}.$$ Тогда $$|\epsilon \left(h\right)| \leq \sum_{i=1}^n \sum_{j=1}^n |\alpha_{ij} \left(h\right)|$$ и, в силу соотношения $\left(1\right)$, имеем $\epsilon \left(h\right) \rightarrow 0$ при $h \rightarrow 0$. Окончательно получаем $$\displaystyle f \left(x_{0}+h\right)−f \left(x_{0}\right)=\frac{1}{2}\left. \left(2\right)$$ Предположим, что $Q_{x_{0}}$ – положительноопределенная форма. Согласно лемме о положительноопределённой квадратичной форме (12.8.1 стр. 295, Лемма 1) , существует такое положительное число $\lambda$, что $Q_{x_{0}} \left(h\right) \geqslant \lambda|h|^{2}$ при любом $h$. Поэтому $$\displaystyle f \left(x_{0}+h\right)−f \left(x_{0}\right) \geq \frac{1}{2}|h|^{2} \left(λ+\epsilon \left(h\right)\right).$$ Так как $\lambda>0 $, and $ \\ epsilon \\ left (H \\ Right) \\ RightarRow 0 $ with $ H \\ Rightarrow 0 $, then the right side will be positive with any vector $ H $ sufficiently small length.
So, we came to the fact that in some neighborhood of the point $ x_ (0) $ it was the inequality of $ f \\ left (x \\ right)\u003e F \\ left (x_ (0) \\ right) $, if only $ x \\ neq x_ (0) $ (we put $ x \u003d x_ (0) + h $ \\ right). This means that at point $ x_ (0) $ function has a strict local minimum, and thus proved the first part of our theorem.
Suppose now that $ q_ (x_ (0)) $ is an indefinite form. Then there are vectors $ H_ (1) $, $ H_ (2) $, such as $ Q_ (x_ (0)) \\ left (H_ (1) \\ Right) \u003d \\ lambda_ (1)\u003e 0 $, $ Q_ (x_ (0)) \\ left (H_ (2) \\ RIGHT) \u003d \\ Lambda_ (2)<0$. В соотношении $\left(2\right)$ $h=th_{1}$ $t>0 $. Then we get $$ f \\ left (x_ (0) + th_ (1) \\ right) -f \\ left (x_ (0) \\ right) \u003d \\ FRAC (1) (2) \\ left [t ^ (2) \\ left [\\ lambda_ (1) + | h_ (1) | ^ (2) \\ Epsilon \\ left (TH_ (1) \\ RIGHT) \\ Right]. $$ with sufficiently small $ T\u003e 0 $ The right-hand part is positive. This means that in any neighborhood of the point $ x_ (0) $, the $ F $ function takes the values \u200b\u200bof $ F \\ left (x \\ right) $, large than $ F \\ left (x_ (0) \\ Right) $.
Similarly, we obtain that in any neighborhood of the point $ x_ (0) $ function $ F $ takes the values \u200b\u200bless than $ F \\ left (x_ (0) \\ right) $. This, along with the previous one, means that at point $ x_ (0) $ function $ F $ does not have an extremum.

Consider private case This theorem for the function $ f \\ left (x, y \\ right) $ two variables defined in some neighborhood of the point $ \\ left (x_ (0), y_ (0) \\ Right) $ and having continuous private derivatives in this neighborhood and second orders. Suppose that $ \\ left (x_ (0), y_ (0) \\ right) $ is a stationary point, and denote $$ \\ displayStyle A_ (11) \u003d \\ FRAC (\\ Partial ^ (2) f) (\\ Partial x ^ (2)) \\ left (x_ (0), y_ (0) \\ Right), A_ (12) \u003d \\ FRAC (\\ Partial ^ (2) f) (\\ Partial x \\ Partial y) \\ left (x_ ( 0), y_ (0) \\ RIGHT), A_ (22) \u003d \\ FRAC (\\ Partial ^ (2) f) (\\ Partial y ^ (2)) \\ left (x_ (0), y_ (0) \\ RIGHT ) $$ Then the previous theorem will take the following form.

Theorem
Let $ \\ delta \u003d a_ (11) \\ CDOT A_ (22) - A_ (12) ^ $ 2. Then:

  1. if $ \\ Delta\u003e 0 $, then the $ F function has a $ \\ left (x_ (0), y_ (0) \\ right) $ local extremum, namely, at least if $ A_ (11)\u003e 0 $ , and maximum if $ A_ (11)<0$;
  2. if $ \\ Delta<0$, то экстремума в точке $\left(x_{0},y_{0}\right)$ нет. Как и в одномерном случае, при $\Delta=0$ экстремум может быть, а может и не быть.

Examples of solving problems

Algorithm for finding extremum functions of many variables:

  1. We find stationary points;
  2. We find a 2nd order differential in all stationary points
  3. Using a sufficient condition for extremum functions of many variables, we consider the 2nd order differential in each stationary point
  1. Explore the function on the extremum $ F \\ left (x, y \\ right) \u003d x ^ (3) + 8 \\ Cdot y ^ (3) + 18 \\ Cdot x - 30 \\ Cdot y $.
    Decision

    We will find private derivatives of the 1st order: $$ \\ DisplayStyle \\ Frac (\\ Partial f) (\\ Partial x) \u003d 3 \\ CDOT X ^ (2) - 6 \\ CDOT Y; $$$$ \\ DISPLAYSTYLE \\ FRAC (\\ Partial f) (\\ partial y) \u003d 24 \\ Cdot y ^ (2) - 6 \\ CDOT x. $$ Make and resolve system: $$ \\ displayStyle \\ Begin (Cases) \\ FRAC (\\ Partial F) (\\ Partial x) \u003d 0 \\\\\\ FRAC (\\ Partial F) (\\ Partial Y) \u003d 0 \\ End (Cases) \\ Rightarrow \\ Begin (Cases) 3 \\ CDOT X ^ (2) - 6 \\ Cdot Y \u003d 0 \\\\ 24 \\ CDOT Y ^ (2) - 6 \\ CDOT X \u003d 0 \\ END (Cases) \\ Rightarrow \\ Begin (Cases) x ^ (2) - 2 \\ Cdot y \u003d 0 \\\\ 4 \\ Cdot y ^ (2) - x \u003d 0 \\ end (Cases) $$ from the 2nd equation Express $ x \u003d 4 \\ cdot y ^ (2) $ - we substitute in the 1st equation: $$ \\ displayStyle \\ left (4 \\ Cdot y ^ (2) \\ Right ) ^ (2) -2 \\ cdot y \u003d 0 $$$$ 16 \\ Cdot y ^ (4) - 2 \\ Cdot y \u003d 0 $$$$ 8 \\ Cdot y ^ (4) - y \u003d 0 $$$$ y \\ left (8 \\ Cdot y ^ (3) -1 \\ Right) \u003d 0 $$ As a result, 2 stationary points were obtained:
    1) $ y \u003d 0 \\ rightarrow x \u003d 0, m_ (1) \u003d \\ left (0, 0 \\ right) $;
    2) $ \\ displaystyle 8 \\ cdot y ^ (3) -1 \u003d 0 \\ rightarrow y ^ (3) \u003d \\ FRAC (1) (8) \\ rightarrow y \u003d \\ FRAC (1) (2) \\ rightarrow x \u003d 1 , M_ (2) \u003d \\ left (\\ FRAC (1) (2), 1 \\ RIGHT) $
    Check the implementation of sufficient conditions of extremum:
    $$ \\ DISPLAYSTYLE \\ FRAC (\\ Partial ^ (2) f) (\\ Partial x ^ (2)) \u003d 6 \\ CDOT x; \\ FRAC (\\ Partial ^ (2) f) (\\ partial x \\ partial y) \u003d - 6; \\ FRAC (\\ Partial ^ (2) f) (\\ partial y ^ (2)) \u003d 48 \\ Cdot y $$
    1) for the point $ M_ (1) \u003d \\ left (0.0 \\ RIGHT) $:
    $$ \\ displayStyle A_ (1) \u003d \\ FRAC (\\ Partial ^ (2) f) (\\ Partial x ^ (2)) \\ left (0.0 \\ RIGHT) \u003d 0; B_ (1) \u003d \\ FRAC (\\ Partial ^ (2) f) (\\ Partial x \\ Partial y) \\ left (0.0 \\ Right) \u003d - 6; C_ (1) \u003d \\ FRAC (\\ Partial ^ (2) f) (\\ partial y ^ (2)) \\ left (0.0 \\ right) \u003d 0; $$
    $ A_ (1) \\ Cdot B_ (1) - C_ (1) ^ (2) \u003d -36<0$ , значит, в точке $M_{1}$ нет экстремума.
    2) for the $ M_ (2) point $:
    $$ \\ displayStyle A_ (2) \u003d \\ FRAC (\\ Partial ^ (2) f) (\\ Partial x ^ (2)) \\ left (1, \\ FRAC (1) (2) \\ RIGHT) \u003d 6; B_ (2) \u003d \\ FRAC (\\ Partial ^ (2) f) (\\ partial x \\ partial y) \\ left (1, \\ FRAC (1) (2) \\ Right) \u003d - 6; C_ (2) \u003d \\ FRAC (\\ Partial ^ (2) f) (\\ Partial y ^ (2)) \\ left (1, \\ FRAC (1) (2) \\ Right) \u003d 24; $$
    $ A_ (2) \\ CDOT B_ (2) - C_ (2) ^ (2) \u003d 108\u003e 0 $, it means, at point $ M_ (2) $ there is an extremum, and since $ A_ (2)\u003e 0 $, That is the minimum.
    Answer: Point $ \\ DisplayStyle M_ (2) \\ Left (1, \\ FRAC (1) (2) \\ RIGHT) $ is a point of minimum function $ F $.

  2. Explore the function on the extremum $ f \u003d y ^ (2) + 2 \\ CDOT X \\ CDOT Y - 4 \\ CDOT X - 2 \\ CDOT Y - $ 3.
    Decision

    Find stationary points: $$ \\ displayStyle \\ FRAC (\\ Partial f) (\\ Partial x) \u003d 2 \\ Cdot y - 4; $$$$ \\ DISPLAYSTYLE \\ FRAC (\\ Partial f) (\\ Partial y) \u003d 2 \\ CDOT Y + 2 \\ CDOT X - 2. $$
    We will also solve the system: $$ \\ displayStyle \\ Begin (Cases) \\ FRAC (\\ Partial F) (\\ Partial x) \u003d 0 \\\\\\ FRAC (\\ Partial F) (\\ Partial Y) \u003d 0 \\ End (Cases) \\ 1 \\ End (Cases) \\ Rightarrow X \u003d -1 $$
    $ M_ (0) \\ Left (-1, 2 \\ Right) $ - stationary point.
    Check the execution of sufficient extremum condition: $$ \\ displayStyle A \u003d \\ FRAC (\\ Partial ^ (2) f) (\\ Partial x ^ (2)) \\ left (-1.2 \\ right) \u003d 0; B \u003d \\ FRAC (\\ Partial ^ (2) f) (\\ Partial x \\ Partial y) \\ left (-1.2 \\ right) \u003d 2; C \u003d \\ FRAC (\\ Partial ^ (2) f) (\\ partial y ^ (2)) \\ left (-1.2 \\ right) \u003d 2; $$
    $ A \\ Cdot B - C ^ (2) \u003d -4<0$ , значит, в точке $M_{0}$ нет экстремума.
    Answer: Extremes are absent.

Time limit: 0

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    Task 1 of 4

    1 .
    Number of points: 1

    Explore the function $ F $ for extremes: $ f \u003d e ^ (x + y) (x ^ (2) -2 \\ Cdot y ^ (2)) $

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  1. Task 2 of 4

    2 .
    Number of points: 1

    Is there an extremum in the function $ f \u003d 4 + \\ sqrt (((x ^ (2) + y ^ (2)) ^ (2)) $

Changing the function at a specific point and is defined as the limit of increasing the function to the increment of the argument, which tends to zero. To find it use the Table of Derivatives. For example, the derivative function y \u003d x3 will be equal to y '\u003d x2.

Eclay this derivative to zero (in this case x2 \u003d 0).

Find the value of the variable of this. These will be the values, with this derivative will be equal to 0. To do this, substitute arbitrary digits instead of x, in which all expression becomes zero. For example:

2-2X2 \u003d 0.
(1-x) (1 + x) \u003d 0
x1 \u003d 1, x2 \u003d -1

Apply the obtained values \u200b\u200bto the coordinate direct and calculate the sign of the derivative for each of the obtained. The coordinate direct points are noted that are accepted for the beginning of the reference. To calculate the value at intervals, substitute arbitrary values \u200b\u200bsuitable for criteria. For example, for the previous function to the interval -1, you can select the value -2. From -1 to 1, you can choose 0, and for more than 1 values, select 2. Substitute the numbers in the derivative and find out the derivative sign. In this case, the derivative with x \u003d -2 will be -0.24, i.e. Negative and at this interval there will be a minus sign. If x \u003d 0, then the value will be 2, and a sign is installed at this gap. If x \u003d 1, the derivative will also be -0.24 and is minus.

If, when passing through the point on the coordinate direct, the derivative changes its sign from a minus to plus, then this is a minimum point, and if from a plus for minus, then this is a maximum point.

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To find a derivative there are online services that count the desired values \u200b\u200band output the result. On such sites you can find a derivative of up to 5 orders.

Sources:

  • One of the service calculation services
  • point of maximum function

The maximum points of the function along with the points of the minimum are called extremum points. At these points, the function changes the nature of behavior. Extremes are determined in limited numerical intervals and are always local.

Instruction

The process of finding local extremums is called function and is performed by analyzing the first and second derivative function. Before starting the study, make sure that the specified argument value interval belongs to valid values. For example, for the function f \u003d 1 / x, the value of the argument x \u003d 0 is unacceptable. Or for the Y \u003d TG (X) function, the argument cannot have a value x \u003d 90 °.

Ensure that the function y is differentiable on the all specified segment. Find the first derivative Y. "It is obvious that until the point of the local maximum is reached, the function increases, and when moving, the function becomes decreasing. The first derivative in its physical meaning characterizes the speed of change of function. While the function increases, the speed of this process is positive. When switching Through a local maximum, the function starts to decrease, and the speed of the change process of the function becomes negative. The transition of the speed of change of function through zero occurs at the local maximum point.