Basic theorems of differential calculus and their application
Yarmetova Mukaddas Ruzimovna-UrSU teacher
Saidova Aziza Yuldash qizi-UrSU student
Annotation. This article gives you some easy ways to solve common. Basic theorems of differential calculus and their application
Key words: vector, inequality, angle, identity
We can often use theorems on derivative functions to solve some problems. These theorems play an important role in checking functions.
Theorem 1 (Farm theorem).)(xf
functionRX given in the package.Xx 0 for the circumference of the point)0(),()( 000 +−=
XxxxU The following conditions must be met:
1))( 0xUx
da)),()(()()( 00 xfxfxfxf
2))( 0xf
be available and limited. Then0)( 0 = xf is being..
Let's say,)( 0xUx
in)()( 0xfxf let it beObviously, in this case0)()( 0 − xfxf
will be.Conditionally)(xf function0x limited in point)( 0xf yield.Then0
lim
lim
lim)( 000 xx
xfxf
xx
xfxf
xx
xfxf
xf xxxxxx −
= −→+→→ Will be.At the moment ,0xx will be,0)('
lim0
)()( 0
=
+→ xf
xx
xfxf
xx
xfxf
xx0xx
will be0)('
lim0
)()( 0
=
−→ xf
xx
xfxf
xx
xfxf
xx
from0)(' 0 =xf It
turns out that.
Theorem 2 (Roll theorem).Suppose,)(xf function],[ ba to meet the following conditions:
1)],,[)( baCxf
2)),( bax in)(xf available and limited,
3))()( bfaf = let it be.),(0 bax 0)( 0 = xf
Conditionally],[)( baCxf . According to Weierstrass's second theorem)(xf function],[ ba at its maximum and minimum values,21 , cс points]),[,( 21 bacc found,]},,[|)(max{)( 1 baxxfcf =]},[|)(min{)( 2 baxxfcf = Has been.
If)()( 21 cfcf = been, Then],[ ba inconstxf =)( is being,),(0 bax at0)(' 0 =xf
. If)()( 21 cfcf be, thats)()( bfaf = because)(xf function)( 1cf and)( 2cf to at least one of the values],[ ba the interior of the segment)( 00 bxax .reach the point According to the farm theorem0)( 0 = xf will be. ►
3-theorem (Lagranj by theorem).Suppose,)(xf function],[ ba at will be,fulfill the following conditions:
1)],[)( baCxf ,
2)),( bax at)(xf the product is available and limitedIn that case it is so),( baс
poind found ,))(()()( abcfafbf −=− will be.
This)(
)()()( ax
ab
afxfxF −
−−= (1) Let's look at the function. This function satisfies all the conditions of the Roll theorem. At the same time, its a productab
xfxF −
−= )()( will be.According to the roll theorem, so)),(( baсc the point is found,0)(' =cF
(2) Will be.
(1) and (2) from equations0
)( =
− ab
,that is))(()()( abcfafbf −=−
1-result. Let's say,)(xf function),( ba at)(xf ,having a product),( bax at0)( = xf
being to. Then),( bax atconstxf =)( will be.),(, 0 baxx
take, edgesx and0x in the segment)(xf using Lagrange's theorem on the functionconstxfxf == )()( 0 being found. ►
2-result.)(xf and)(xg function),( ba at)(xf ,)(xg ,products),( bax in)()( xgxf =
been. Then),( bax inconstxgxf += )()( will be.
This is proof of the result)()( xgxf − by applying result 1 to the function.Theorem 4 (Cauchy Theorem). Let, and let the functions fulfill the following conditions.
1)],[)( baCxf ,],[)( baCxg ,
2)),( bax da)(xf va)(xg crops are available and limited;
3)),( bax da0)(' xg will be. Then),( baс ,the point is found)( Willbe. First of all)()( agbg We emphasize that because)()( agbg = if so, then according to Roll's theorem),( baс the point would be found0)( = cg would be This is contrary to condition 3) The following]),[()]()([
)()()( baxagxg
afxfxФ −
−−=),( baс
found point ,
7150)( = cФ
will be (3) Obviouly,)(
)()( xg
xfxФ
−=
(4) (3) and (4) relationship0)(
)( =
− cg
that is)(
Example 1.Rxx '',' for|'''||''sin'sin| xxxx −− prove the inequality.
Let's say,''' xx will be.xxf sin)( = in]'','[ xx We apply Lagrange's theorem.That's it)'','( xxc the point is that,)'''(|cos||''sin'sin| xxcxx −=− willbe.IfRt at1|cos| t Given that, then from the above relationship,)'','(|'''||''sin'sin| Rxxxxxx −−
Being.
References
1.Mirzaahmedov.M , Sotiboldiyev T “Matematikadanolimpiadamasalalari” Toshkent-2003 y
2.Гелфанд.F “методы доказательства неравенства” Москва-1972г
3.Сивяшинский.В “доказательство тригонометрических неравенств” Москва-1985г
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