[SPAM] Re: [obm-l] função contínua

[Artur Costa Steiner <artur_steiner@xxxxxxxxx>]

SPAM: -------------------- Start SpamAssassin results ----------------------
SPAM: This mail is probably spam.  The original message has been altered
SPAM: so you can recognise or block similar unwanted mail in future.
SPAM: See http://spamassassin.org/tag/ for more details.
SPAM: 
SPAM: Content analysis details:   (6.20 hits, 5 required)
SPAM: IN_REP_TO          (-0.8 points) Found a In-Reply-To header
SPAM: X_MAILING_LIST     (-0.3 points) Found a X-Mailing-List header
SPAM: SPAM_PHRASE_00_01  (0.8 points)  BODY: Spam phrases score is 00 to 01 (low)
SPAM:                    [score: 0]
SPAM: FORGED_YAHOO_RCVD  (1.4 points)  'From' yahoo.com does not match 'Received' headers
SPAM: RCVD_IN_ORBS       (2.2 points)  RBL: Received via a relay in orbs.dorkslayers.com
SPAM:                    [RBL check: found 58.85.191.209.orbs.dorkslayers.com., type: 68.178.232.99]
SPAM: RCVD_IN_OSIRUSOFT_COM (0.4 points)  RBL: Received via a relay in relays.osirusoft.com
SPAM:                    [RBL check: found 58.85.191.209.relays.osirusoft.com.]
SPAM: X_OSIRU_OPEN_RELAY (2.7 points)  RBL: DNSBL: sender is Confirmed Open Relay
SPAM: AWL                (-0.2 points) AWL: Auto-whitelist adjustment
SPAM: 
SPAM: -------------------- End of SpamAssassin results ---------------------

Como f é continua, existe c em (a , b) tal que f(c) =
(a+b)/2.

Aplicando o TVM a [a , c], obtemos x1 em (a , c) tal
que f'(x1) = (f(c) -f(a))/(c -a) =(b - a)/(2(c - a).
Aplicando o TVM agora a [c , b], obtemos x2 em (c , b)
tal que f'(x2) = (f(b) -f(c))/(b - c) =(b - a)/(2(b -
c). 

Temos, entao, que a < x1 < x2 < b e que

1/f'(x1) + 
1/f'(x2) = (2(c - a))/(b - a) + (2(b - c))/(b - a =
(2(b - a))/(b - a) = 2 , provando a afirmacao.

Artur

Ps. O merito desta prova nao e meu, um amigo sugeriu o
ponto chave c e eu so dei os arremates finais com o
TVM.

>     From: Carlos Gomes 
>     To: obm-l@xxxxxxxxxxxxxx 
>     Sent: Saturday, February 09, 2008 7:45 AM
>     Subject: função contínua
> 
> 
>     Olá amigos...será que alguém pode me ajudar com
> essa?
> 
>     Seja f uma função contínua em [a,b] e
> diferenciável em (a,b) tal que f(a)=a e f(b)=b.
> Mostre que existem x_1 e x_2 tais que a< x_1 < x_2 <
> b tais que 1/f ' (x_1)  +  1/f ' (x_2) = 2.
> 
> 
>     Valew, Cgomes



      ____________________________________________________________________________________
Be a better friend, newshound, and 
know-it-all with Yahoo! Mobile.  Try it now.  http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ 

=========================================================================
Instruções para entrar na lista, sair da lista e usar a lista em
http://www.mat.puc-rio.br/~obmlistas/obm-l.html
=========================================================================