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

1. ## 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...

2. Try using the mean value theorem instead.

3. 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]...

4. 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 $\frac{|f(t)- f(x)||}{|t-x|}< f'(x)+ 1< n$ 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.

5. Originally Posted by HallsofIvy
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