FAQFAQ   SearchSearch   MemberlistMemberlist   UsergroupsUsergroups 
 ProfileProfile   PreferencesPreferences   Log in to check your private messagesLog in to check your private messages   Log inLog in 
Forum index » Science and Technology » Math
Another nice mean value like theorem
Post new topic   Reply to topic Page 1 of 1 [4 Posts] View previous topic :: View next topic
Author Message
eugene
science forum Guru


Joined: 24 Nov 2005
Posts: 331

PostPosted: Wed Jul 12, 2006 3:09 pm    Post subject: Another nice mean value like theorem Reply with quote
Let f be differentiable on [a,b] such that f'(a) = f'(b). Prove that
there exist a c in (a,b) such that

f'(c) = (f(c) - f (a) ) / (x - a).

Thanks.
Back to top
eugene
science forum Guru


Joined: 24 Nov 2005
Posts: 331

PostPosted: Wed Jul 12, 2006 3:31 pm    Post subject: Re: Another nice mean value like theorem Reply with quote
eugene wrote:
Quote:
Let f be differentiable on [a,b] such that f'(a) = f'(b). Prove that
there exist a c in (a,b) such that

f'(c) = (f(c) - f (a) ) / (x - a).

Thanks.

Edited: f'(c) = (f(c) - f (a) ) / (c - a).
Back to top
The World Wide Wade
science forum Guru


Joined: 24 Mar 2005
Posts: 790

PostPosted: Wed Jul 12, 2006 7:28 pm    Post subject: Re: Another nice mean value like theorem Reply with quote
In article
<1152718319.656278.239750@i42g2000cwa.googlegroups.com>,
"eugene" <jane1806@rambler.ru> wrote:

Quote:
eugene wrote:
Let f be differentiable on [a,b] such that f'(a) = f'(b). Prove that
there exist a c in (a,b) such that

f'(c) = (f(c) - f (a) ) / (x - a).

Thanks.

Edited: f'(c) = (f(c) - f (a) ) / (c - a).

Let g(x) = (f(x) - f(a))/(x - a), for x in (a,b], g(a) = f'(a).
Then g is continuous on [a,b]. Note g(b) = (f(b) - f(a))/(b - a).
Compare g(a) to g(b) using the hypotheses to see that an extreme
value of g must occur at some c, a < c < b. At this c we have
g'(c) = 0, and that's the c you want.
Back to top
Robert B. Israel
science forum Guru


Joined: 24 Mar 2005
Posts: 2151

PostPosted: Wed Jul 12, 2006 7:57 pm    Post subject: Re: Another nice mean value like theorem Reply with quote
In article <1152718319.656278.239750@i42g2000cwa.googlegroups.com>,
eugene <jane1806@rambler.ru> wrote:
Quote:

eugene wrote:
Let f be differentiable on [a,b] such that f'(a) = f'(b). Prove that
there exist a c in (a,b) such that

f'(c) = (f(c) - f (a) ) / (x - a).

Thanks.

Edited: f'(c) = (f(c) - f (a) ) / (c - a).


Hint:
Show that WLOG you can take a = f(a) = 0, f'(a) = f'(b) = 0.
Let g(x) = f(x)/x for 0 < x <= b,
g(0) = 0.
If g is not constant, show it attains a max or min at
some point of [0,b] other than 0 and b.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Back to top
Google
Back to top
Display posts from previous:   
Post new topic   Reply to topic Page 1 of 1 [4 Posts] View previous topic :: View next topic
The time now is Mon Jun 20, 2011 4:57 am | All times are GMT
Forum index » Science and Technology » Math
Jump to:  

Similar Topics
Topic Author Forum Replies Last Post
No new posts Tarski fixed-point theorem William Elliot Math 14 Tue Jul 18, 2006 10:24 am
No new posts Quasi chinese remainder theorem cliomseerg@kriocoucke.mai Math 2 Mon Jul 17, 2006 1:22 pm
No new posts *unique* prime factorizations; the fundamental theorem of... DGoncz@aol.com Math 5 Sun Jul 16, 2006 9:53 am
No new posts The Fundamental Theorem of Calculus Maury Barbato Math 19 Tue Jul 11, 2006 10:06 pm
No new posts Irrationality and the Fundamental Theorem of Arithmetic J. B. Kennedy Math 30 Mon Jul 10, 2006 12:02 pm

Copyright © 2004-2005 DeniX Solutions SRL
Other DeniX Solutions sites: Electronics forum |  Medicine forum |  Unix/Linux blog |  Unix/Linux documentation |  Unix/Linux forums  |  send newsletters
 


Powered by phpBB © 2001, 2005 phpBB Group

[ Time: 0.0532s ][ Queries: 16 (0.0260s) ][ GZIP on - Debug on ]