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

• May 19th 2009, 05:40 AM
MMath09
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...
• May 19th 2009, 06:29 AM
HallsofIvy
Try using the mean value theorem instead.
• May 19th 2009, 07:21 PM
MMath09
Quote:

Originally Posted by HallsofIvy
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]...

• May 20th 2009, 12:55 PM
HallsofIvy
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 $\delta> 0$ 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.
• May 21st 2009, 03:01 AM
MMath09
Quote:

Originally Posted by HallsofIvy
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 :)