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Math Help - how to finish this question..

  1. #1
    MHF Contributor
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    Nov 2008
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    how to finish this question..

    f is differentiable continuously(which means that f(x) and f'(x) are continues and differentiable) on [a,b]
    suppose that |f'(x)|<1 for x in [a,b]

    prove that there is 0<=k<1 so there is x1,x2 in [a,b]
    |f(x1)-f(x2)|<=k|x1-x2|
    ??

    i started from the data that i was given and i know that f(x) is differentiable
    so
    <br />
f'(x)=\lim _{h->0}\frac{f(x+h)-f(h))}{h}<br />
    i use

    |f'(x)|<1
    so
    <br />
|f'(x)|=\lim _{h->0}\frac{|f(x+h)-f(h))|}{|h|}<1<br />

    so f(x) is continues and differentiable
    so i use mvt on x1 and x2
     <br />
f'(k)=\frac{f(x1)-f(x2)}{x1-x2}<br />
    and i combine that with
    |f'(x)|<1

     <br />
\frac{|f(x1)-f(x2)|}{|x1-x2|}<1<br />
    so
     <br />
   |f(x1)-f(x2)|  <|x1-x2|<br />

    what now?
    Last edited by transgalactic; March 13th 2009 at 07:12 AM.
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  2. #2
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    they sign L as the minimum of f'(x)
    and they sign l as the maximum of f'(x)

    then they say that k=max{|f(L)|,|f(l)|}

    but its not true
    i am given that 0=<k<1
    and i dont know if i get the same values

    why
    k=max{|f(L)|,|f(l)|} equals 0=<k<1
    ??
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