Life audit. Zero point NT The state of a person at point zero

The point x 0 is called the maximum point of the function f(x) if in some neighborhood of the point x 0 the inequality ()(0 xfxf) is satisfied

Point x 1 is called the minimum point of the function f(x), if in some neighborhood of the point x 1 the inequality ()(1 xfxf) The values ​​of the function at the points x 0 and x 1 are called the maximum and minimum of the function, respectively. The maximum and minimum of the function are called the extremum of the function .

On one interval, a function can have several extrema, and it may be that the minimum at one point is greater than the maximum at another. The maximum or minimum of a function on a certain interval is not, in general, the largest and smallest value of the function. If at some point xx 00 the differentiable function f(xf(x)) has an extremum, then in some neighborhood of this point Fermat’s theorem holds and the derivative of the function at this point is equal to zero: 0)(0 xf

However, a function may have an extremum at a point at which it is not differentiable. For example, the function xy has a minimum at the point 0 x but it is not differentiable at this point.

In order for the function y=f(x) to have an extremum at the point x 0, it is necessary that its derivative at this point be equal to zero or not exist.

The points at which the necessary extremum condition is satisfied are called critical or stationary. T. ob. , if there is an extremum at any point, then this point is critical. But the critical point is not necessarily the extremum point.

Let us apply the necessary extremum condition: xxy 2)(2 002 xprixy 0 0 y x - critical point

Let us apply the necessary extremum condition: 23 3)1(xxy 003 2 xprixy 1 0 y x - critical point

If, when passing through the point x 0, the derivative of the differentiable function y=f(x) changes sign from plus to minus, then x 0 is the maximum point, and if from minus to plus, then x 0 is the minimum point.

Let the derivative change sign from plus to minus, i.e., on a certain interval 0; xa 0)(xf and on some interval bx; 0 0)(xf Then the function y=f(x) will increase by 0; xa

and will decrease by bx; 0 By definition of an increasing function 00 ;)()(xaxallforxfxf For a decreasing function bxxallforxfxf;)()(00 0 x is the maximum point. It is proved similarly for the minimum.

1 Find the derivative of the function)(xfy 2 Find the critical points of the function at which the derivative is zero or does not exist.

3 Investigate the sign of the derivative to the left and to the right of each critical point. 4 Find the extremum of the function.

Let's apply the scheme for studying a function to an extremum: 1 Find the derivative of the function: 233)1(3)1())1((xxxxxy)14()1()31()1(22 xxxxx

3 We examine the sign of the derivative to the left and to the right of each critical point: x 4 1 1 y y There is no extremum at the point x=1 x=1.

If the first derivative of the differentiable function y=f(x) at the point x 0 is equal to zero, and the second derivative at this point is positive, then x 0 is the minimum point, and if the second derivative is negative, then x 0 is the maximum point.

Let 0)(0 xf therefore 0)(0 xf and in some neighborhood of the point x 00, i.e. 0)()(xfxf

functionba; will increase on)(xf containing the point x 00. But Ho 0)(0 xf on the interval 0; xa 0)(xf and on the interval bx; 0 0)(xf

Thus, when passing through the point x 00, the function changes sign from minus to plus, therefore this point is the minimum point.) (xf The case for the maximum of the function is proved in a similar way.

The scheme for studying the function at the extremum in this case is similar to the previous one, but the third point should be replaced with: 3 Find the second derivative and determine its sign at each critical point.

From the second sufficient condition it follows that if at a critical point the second derivative of the function is not equal to zero, then this point is an extremum point. The converse statement is not true: if at a critical point the second derivative of a function is equal to zero, then this point can also be an extremum point. In this case, to study the function it is necessary to use the first sufficient condition for the extremum.

I feel with my skin that earthly evolution has approached point ZERO today. Located at the zero point.

All social institutions are critically profaned.
Collapse now, this year could happen...

The mountains are moving apart
the oceans rise

The earth is shaking
The heavens open up -
The Great Cosmic Year is ending.
Only human stupidity
Remains standing still.
Human nature seems to be unchangeable in its immobility.

She imagines small things as big,
Big - insignificant.
There...

A person who wants to become a channel of universal power must learn to accept and rely on himself. He needs to let go of all internal conflicts and self-destructive tendencies. He is called to give his personality the love and nourishment it needs.

Believing in yourself does not imply isolation from the world. It is clear to any reasonable person that success and creative realization are possible only in conditions of active interaction with the world. However, we can approach the world and ourselves differently - in...

So, the human Light body changes under the influence of certain factors. First, a person can consciously work with the Lightbody through visualization. The purity of the Light body is gained so that your bodies connect around the Centering Point.

What is a Centering Point? This is a harmonious arrangement of the human Light body in a cell of the spatial grid. Each of you has only a few Points in the space-time continuum at which your multidimensional...

The path was long. Longer than usual. And although the place was already known, the road was difficult. The straps cut into the shoulders did not allow one to forget about the body, and only will and presence, and half a bag did not allow emotions to prevail.

And here is the starting point.
Camp, firewood, fire and the anxiety that arises from time to time from trying to understand the space around. The mystery is borderline, or rather turning into animal fear. The picture opening underfoot inspired respectful awe with its multidimensionality...

We want to draw your attention to a very wise, but at the same time completely obvious statement: you cannot run away from yourself. What did you think before and continue to think now? Here's the result! Your present life in all its glory.

But, as they say: what goes around comes around (or what goes around comes around)! Now we want to bring your closest attention to the vibrational relativity between where you are now and where you really want to be - because this is where...

There are two steps on the way up, and the first is to gain desire (hisaron)!

In our material world, we are already born with the desire to reveal it and dominate it, use it, contact it.

But in the spiritual world it’s not like that! We need to earn this desire ourselves.

And the light also helps us with this - we use it to build our desire in a reverse, egoistic form, which helps us receive the desire to bestow, the spiritual Kli (spiritual vessel is the desire to bestow).

The same question is repeated all the time: why couldn’t light from the very beginning create a finished creation already similar to the Creator?

But this is impossible! It is impossible to directly transfer the property of bestowal from the Creator to creation.

Therefore, we have to go through the splitting of the worlds (in the world of Nikudim) and the soul (the Fall from the Tree of Knowledge).

Otherwise, it is impossible to unite, to bring into each other the properties of bestowal and reception, Bina and Malchut, except through the breaking, the sin (breaking) of Adam or the breaking of the Temple.

“Point of return” is a temporary and energetic state, after overcoming which a person’s consciousness can no longer die - annihilate. It occurs when a person’s soul has accumulated a sufficient amount of energy, that is, it has acquired a certain energy intensity of consciousness that balances its karmic debts.

Karmic debt, in addition to moral, ethical and energetic connections between people, nature and other objects and subjects, is also always expressed in the account of time, which...

There is no light without a vessel, no filling without desire. The Upper Light is in absolute peace; it fills and surrounds the entire universe. Everything depends only on our desire, our vessels of perception.

If our desire strives precisely for that filling, then we will feel this filling.

And if the desire does not exactly correspond to the frequency or property of the filling, that is, there is no similarity of properties between the desire and the filling, then we do not feel the filling, as in many cases in our world.

Everything is turned upside down in this world! Partly out of naivety, partly to eliminate the illusion of the heavy burden of existence, the author tried to put in writing “The Quintessence of Illusoryness” for his friends. Nevertheless, the author still accepts well-deserved reproaches and complaints about the difficulty of understanding existence as a fact. Although, in his opinion, the “Point Zero” formula is stunning in its simplicity and leads to complete carefreeness.

This work is another attempt to express the inexpressible in words. The idea of ​​writing this text was suggested to the author in conversations about the imperfection of this world and the loss of modern man in the piles that

he builds it himself.

We know little about a person, but many of us are haunted by the idea that a person should achieve something. This is where the confusion begins. Confusion with achievement, gain, possession, loss and fears. A person constantly has to learn by looking around. Having looked around jealously, we want to receive conditional good, conditional happiness, conditional freedom. We long to have what we think others have. Outwardly imperceptible anxiety about the happiness that has bypassed us gives rise to unhappiness in us. Suddenly, a desire appears to change the world so that the realization of our desires finally comes. And if we manage to get something, we inevitably want to hold on to it and take credit for the achievement. It is not always obvious to us that we are playing cat and mouse with ourselves and with our plans for the future. And as you know, any attempts to change or create the “correct” future lead to thinking about the past, comparison, choice, anxiety and struggle in the present. Worries and fears have become the norm for humans, but this has little to do with the natural state of man. Anxiety will continue unless the false is seen as false in its creator.

Essentially innocent attempts to change the world around us already lead to anxiety, and most of us, starting to “professionally” improve our future, get lost in daydreams. We lose sight of the fact that the future is just a figment of the imagination and that the future can take care of itself. Attempts to influence the future are similar to magical rituals and turn into a desire to change and control a destiny that we don’t know anyway. Few people, in the process of searching for the best, notice that suddenly the categories Happiness, Suffering, Fate, etc. appeared from somewhere. But in the process of any search and action, something always remains unchanged and most often unnoticed. And that’s exactly what this piece is about.

Function f(x)±g(x) continuous at a point X 0 , if functions f(x\g(x) continuous at a point X 0 .

Function f(x)-g(x) continuous at a point X 0 , if functions f(x\g(x) continuous at a point X 0 .

Function - continuous at a point X 0 , if functions f(x), g(x) continuous at a point Xq And g(x 0 ) f 0.

Function f(g(x)) continuous at a point X 0 if function f(z) continuous at a point z 0 = g(x 0 ) , and the function g(x) continuous at a point X 0 .

Definition. The function is called continuous on the interval(A; b\ if it is continuous at every point of this interval. Function Dx ) called continuous on the segment , if it is continuous on the interval (a; b), and is also right continuous at the point A and is left continuous at the point Kommersant (i.e. lim fix)= f(a\ lim fix)= fib))

Definition. The function is called breaking pointX q, if at this point at least one of the conditions of the criterion for the continuity of a function at a point is violated. In this case, the point X 0 is called break point functions.

Classification of function discontinuity points

1) Point X 0 called removable break point, if at this point the limits on the right and left exist, are finite and are equal to each other, i.e.

ton /O) = ton /O). But at the same time, the value of the function at the point X 0 or not defined

divided or not equal to the specified one-sided limits.

2) Point X 0 is called discontinuity point of the 1st kind, if at this point the limits on the right and left exist, are finite and are not equal to each other, i.e.

ton /(jc)* ton /00

x^>x 0 +0 x^>x 0 -0

3) Point X 0 is called discontinuity point of the 2nd kind, if at this point at least one of the limits on the right and left does not exist or is infinite.

EXAMPLE. Explore functions/(x) , / 2 (jc) , / 3 (jc) for continuity, determine the points of discontinuity, if any, and determine the nature of the discontinuity. SOLUTION.

1) f1(x) = . Function not defined at point x = 0, so this point

ka is the breaking point. Let us determine the type of discontinuity. Using the first remarkable limit (see formula (1)), we obtain

sin x. sin x sin x Hm = Hm = lim = 1 therefore, x = 0 is the setting point

x^0 x x^0-0 x x^0+0 x

vulnerable breakup.

2) / 2 (jc) = 3 X . Function not defined at point x = 0, which means that the function breaks at this point. Let us show that this is a discontinuity of the 2nd kind. Let's find the limits on the right and left at the point x = 0. Let us recall the limiting properties of the exponential function d (a > 1), known from the school curriculum: Hm a 1 = +oo, Hmm A 1 = 0.

//->+yes f-»-oo

From here we get:

lim 3 X = lim3* =

x->0+0 *->0

at x-» 0, x> 0

lim 3 f = +oo,

£- » -00

lim 3 X =ShpZ*

I/7I X^O,X<0

Since the limit on the right is equal to infinity, then x = 0 is a discontinuity point of the 2nd kind. A schematic graph of the function under study is presented in Figure 1.

Rice. 1. Schematic graph of the function f 2 (x)

jc + 3 ,jc< 0

x 3 + 3, 0< X< 1 3-jv: , jc > 1

Function f 3 (x) is given by various analytical expressions on the intervals (-oo;0), " =f"-v"-u".

Table of derivatives of elementary functions

(x"Y = n-x"-\

(A X )" = a X \pa, (e X )" = e X .

3) (log jcV =^, (ln*)" = -.

A ) haaa ) X

1. (sin*)"= cos x

\+x 2

(ch x)" = sh x

(S hjc)" = chjc.

(thjc)" l

hyperbolic cosine ch x hyperbolic sine sh x

e X +e~ X

shjc e X -e~ X

chjc e x +e~ x chjc e X +e~ X

hyperbolic cotangent cth* =

If we combine property 7 for finding the derivative of a complex function with a table of derivatives of elementary functions, we obtain the following formulas, which are convenient to use when calculating derivatives of complex functions if and = and(X):

(And n )" = p-i n - 1 -And".

(A And )" = a And \pa-i\ (e And )" = e And -And".

3) (log i)" =, (1rm)" = -.

A iLpa And

(cosu)" =-smu-u".

(sin And)"=cos and -and".

And And"

^/G^ 2 ~

And"

(arcsinw)" =

And"

And"

(arctgw)" =

\+ and 2

(chw)" = shww".

(shw)" = chww". And"

ch 2 w -And"

EXAMPLE. Find derivatives of given functions: 1)y= 4 X\ 2)y= ^5,

3)_y = 7 x2 "8x,

4)^ = 1n(x 4 -2x 3 + 6),

5)>> = cos 3x.

1) / = 4-(jc 2)" = 4-2jc = 8jc,

P 1 ^-i 1 - 2 1

(X- 5)

3) U = (7 l2 - 8l)" = 7 x2 - 8x -1p7-(x 2 -8xU = 7 l2 - 8l -1p7-(2x-8),

, n 4 z™ (x 4 -2x 3 +6)" 4x 3 -6x 2 +0 2x 2 (2x-3)

4) v = (\n(x - 2x + 6U) = = =

U U ) } x 4-2x 3+6 x 4-2x 3+6 x 4-2x 3+6"

5) y" =(cos 3jc)" = - sin 3x■(3 jc)" = - sin 3x■ 3 = -3 sin 3 jc.

20 EXAMPLE. Find derivatives of functions

1) y= x-mctgx, 2) y= arccos^,3) at = log^(3 + 5" x).

X

1 + x 2

Y = O arctg jc)" = x" ■ arctg x+ x ■ (arctg jc)" = arctg x+

y= (arccos^y = - . X -Ш =- . * G

Vi-(V^) 2 v ; Vi-(^ ) 2 2v*

3) y= (log 3 2 (3 + 5- x))" = 31og 2 2 (3 + 5- x)-(log 2 (3 + 5- x))" =

= 31og 2 (3 + 5-x)- " = 31og 2 (3 + 5-x) l j

(3 + 5" x)ln2

- 3 log 2 (3 + 5"* )

EXAMPLE. Find derivatives of functions1) ^=-l/ 1- 4l: 2 , 2) at 2 =\ь

1) Calculate the derivative y[ , using properties 6 and 7, as well as tabular derivative 1:

(- 8x)- x-Vl- 4x 2 -l

2-Vl-4 x 2

Wow

X

4x 2 -(1-4x 2)

2 x 2 -1-4x 2

x 2 -1-4x 2 2) Using the properties of logarithms and the 2nd property of derivatives, we obtain:

U 2

( l l + Vl-4 x 2 l I n

2 X

(ln(l +Vl- 4:c 2)) -(1p 2)"-(1shs)"

-(1W1-4 x 2) -0 - =

(1 + 1-4x 2) v " *

1.(-8jc)-- 4 x

(1 + 1-4x 2) 2-1-4x 2 * (1 + 1-4x 2)-1-4x 2 *

4x2-(1 + 1G4 2)-l1G47 -4 x 2-^1Г47-(^1::47)2

zz

4J 2 + ^1G47 + 1-4J 2 1 + l1G47 -1

EXAMPLE. Write the equation of the normal to the curve y = 3(3Jc - 2<У*) в точ­ке с абсциссой X 0 = 1.

SOLUTION. In order to create the normal equation, we find at 0 = y(x 0 ) And / (jc 0):

at 0 = 3(1 - 2) = - 3;

/"(jc) = (3-(3*-2V*)) = 3-(- -jc 3 -2 - jc 2)

32 3 [x2 J*"

/"(jc 0) = /"(1) = 1-3 = -2. Let's substitute x 0 =1 , y 0 = - 3, / / ​​(x 0) = - 2 into the normal equation (see formula (4)):

j/-(-3) = (l-1).

After the transformation, we obtain the required normal equation:

x - 2y - 7 = 0. ANSWER: x - 2y - 7= 0 - normal equation.

22 Derivative of a power-exponential function y = (F)U (X) found using the logarithmic differentiation method. This method consists of first taking the original function logarithmically; then convert to a product using the properties of logarithms, and find the derivative of the left and right sides of the equation that contains the given function; Finally, the desired derivative is expressed from the resulting equation. Let us show the derivation of the formula for the derivative of a power-exponential function using the logarithmic differentiation method:

y= u v , 1p>" = 1p(m at), \ny= v-\nu,(\ny)"=v"-\nu + v(\nu)\

£ = v"-1nu + v- - ,

U And

V 1J y" = y. ( V "-\ nU + ),

And

And Thus, we obtained a formula for calculating the derivative of a power-exponential function:

VU

(u v)" = u v-(v"-\mi +). (5)

EXAMPLE. Find the derivative of a function y =(sin x)cos x . SOLUTION. In our case and = sin x, v = cosx, hence, and" = cos x, v" = - sin x Therefore, from formula (5) it follows

((sin*)00 ")" = (smx)""* .(-smx.\nsmx + C0SX - C ° SX \ v 7 sin x

y" =(sin x)cos x (ctg x■ cos x-sin x■ In sin x). ANSWER: y" = (sin x)cos x (ctg x■ cos x- sin x■ I n sin x).

Statement.Derivative y" (X ) by variable X from the function given

(x = X(t),

V parametric form i , is determined by the formula [ y = Y(t)

X:

Definition.Derivative of nth order f (n) (x) from function j/= f(x) is called the first derivative of (and - 1)th derivative, i.e.

/ (And) (*) = (/ (And_ 1) (* ))"

From the definition it follows that the second derivative of a given function is the first derivative of the first derivative of this function, and the calculation of derivatives of order greater than the first is reduced to the calculation of the first derivative of new functions.

Let's get the formula for calculating the second derivative y" according to change-

\x = X(t\ Noah X from a function specified in parametric form j= y(t) :

at " =(at" U

U XX \UGH

It is necessary to calculate the first derivative of the new function, specified in parametric form, by applying formula (6) to the function

\x = X(t),

\>Yt

V-

Vх x\

Hence,

U

( at U

X[

24 EXAMPLE. Find the first and second derivatives of a function given in parametric form

f x= lncos?, [.у = In sin?.

SOLUTION. To find the first derivative at, let's calculate the first derivatives with respect to the variable? from X And y:

x t -

cos t

Then, according to formula (6) we have

. COS?COS? 2

Y\ = -, =-ctg 2 ?. sin? sin?

We find the second derivative using formula (7):

(y"J t (- ctg 2 p; g K S m 2 ?

V = =

U XX

sint

Uhh sm4?

25 I.4. DIFFERENTIAL OF A FUNCTION OF ONE VARIABLE

Definition.Differential functions at = (*) is the product of the derivative of a function and the differential of a variable and is denoted dy= f(x)-dx. The differential of an independent variable is the increment of this variable: dx= Ax.

From the definition of a differential follows another representation of the first derivative through differentials:

S„ y

1 (X)= -

dx.Geometric meaning of differential : the differential of a function at a certain point is equal to the increment of the ordinate of the tangent drawn at this point: D at cas = dy(see Figure 3). This follows directly from the equation of the tangent at a point (see formula (3)). If we designate y~y 0 =Ay To ac. , X- X 0 = A X= dx, then the tangent equation can be written as: Ay cas = f"(x 0 ) ■ Ah , or Du cas = dy

Rice. 3. Geometric meaning of the differential

26 From the geometric meaning of the differential follows the formula for approximate calculation of the function through the differential. With small increments of the variable, the increment of the function can be replaced by the increment of the tangent (see Fig. 3), i.e. at small AX approximate equality

Au~Au ka With = f"(x 0 )-Ax.

Because Ay= f(x 0 + Oh ) - f(x 0 ), then we get formula for approximate calculation of the function at some point close to X 0 :

f(x 0 + Oh ) « Dh 0) + f(x 0 ) Oh. (8)