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Math Help - Differentiable, continuous function... "show that"

  1. #1
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    Differentiable, continuous function... "show that"

    If f is continuous on [0,1] and differentiable at a point x \in [0,1], show that, for some pair m,n \in \mathbb{N},
    \left | \frac{f(t)-f(x)}{t-x}\right | \leq n whenever 0 \leq |t-x| \leq \frac{1}{m}

    Since it's differentiable at x I know \lim_{t \to x}\frac{f(t)-f(x)}{t-x} exists...
    And it's continuous so I know for that for any \epsilon > 0 there's a \delta > 0 so that |f(t)-f(x)| < \epsilon when |t-x| < \delta

    Not sure how to put it all together...
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  2. #2
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    Try using the mean value theorem instead.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Try using the mean value theorem instead.
    Wouldn't that require the function to be differentiable on (0,1) ? I only have differentiability at a single point in [0,1]...

    I don't follow :S
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  4. #4
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    Oops, right, I didn't see that! Sorry. I guess you were right to use the definition of "differentiable at a point".

    Let n be any integer strictly larger than f'(x)+1. Certainly there exist [itex]\delta> 0[/itex] such that \frac{|f(t)- f(x)||}{|t-x|}< f'(x)+ 1< n And, since \frac{1}{\delta} is a real number, there exist [tex]m> \frac{1}{\delta}[/itex]. If |t- x|< \frac{1}{m}< \delta then |\frac{f(t)- f(x)}{x-t}|< n.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Certainly there exist \delta> 0 such that \frac{|f(t)- f(x)|}{|t-x|}< f'(x)+ 1< n
    when |t-x| < delta?

    I don't get this bit

    Thanks for your help, I'm almost there
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