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eugene
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Joined: 24 Nov 2005
Posts: 331

Posted: Wed Jul 12, 2006 3:09 pm    Post subject: Another nice mean value like theorem

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.
eugene
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Joined: 24 Nov 2005
Posts: 331

Posted: Wed Jul 12, 2006 3:31 pm    Post subject: Re: Another nice mean value like theorem

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).
science forum Guru

Joined: 24 Mar 2005
Posts: 790

Posted: Wed Jul 12, 2006 7:28 pm    Post subject: Re: Another nice mean value like theorem

In article
"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.
Robert B. Israel
science forum Guru

Joined: 24 Mar 2005
Posts: 2151

Posted: Wed Jul 12, 2006 7:57 pm    Post subject: Re: Another nice mean value like theorem

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

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