Results 1 to 5 of 5

Thread: Differentiable, continuous function... "show that"

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    9

    Differentiable, continuous function... "show that"

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

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

    Not sure how to put it all together...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,795
    Thanks
    3035
    Try using the mean value theorem instead.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2009
    Posts
    9
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,795
    Thanks
    3035
    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 $\displaystyle \frac{|f(t)- f(x)||}{|t-x|}< f'(x)+ 1< n$ And, since $\displaystyle \frac{1}{\delta}$ is a real number, there exist [tex]m> \frac{1}{\delta}[/itex]. If $\displaystyle |t- x|< \frac{1}{m}< \delta$ then |\frac{f(t)- f(x)}{x-t}|< n.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2009
    Posts
    9
    Quote Originally Posted by HallsofIvy View Post
    Certainly there exist $\displaystyle \delta> 0$ such that $\displaystyle \frac{|f(t)- f(x)|}{|t-x|}< f'(x)+ 1< n$
    when |t-x| < delta?

    I don't get this bit

    Thanks for your help, I'm almost there
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Sep 23rd 2011, 05:43 AM
  2. Replies: 2
    Last Post: Apr 24th 2011, 07:01 AM
  3. Replies: 1
    Last Post: Oct 25th 2010, 04:45 AM
  4. Replies: 0
    Last Post: Sep 12th 2010, 09:43 PM
  5. Replies: 1
    Last Post: Nov 19th 2008, 12:08 AM

Search Tags


/mathhelpforum @mathhelpforum