1. ## uniform continuity

I need to show f(x)=x^(1/3) is uniformly continuous on (-1,1).

I can't find a delta st |x-y|<delta => |f(x)-f(y)|<epsilon..

any ideas?

2. Originally Posted by CarmineCortez
I need to show f(x)=x^(1/3) is uniformly continuous on (-1,1).
Do you understand that $f(x) = \sqrt[3]{x}$ is uniformly continuous on $[-1,1]$?
Do you understand that any continuous function is uniformly continuous on any closed interval?
If so why re-invent the wheel?

3. is there some simple argument to show its uniformly continuous on the closed set?

4. Originally Posted by CarmineCortez
is there some simple argument to show its uniformly continuous on the closed set?
Well that depends upon what one knows.
Do you know that any compact set has a finite cover of arbitrary length?

5. ## Use the extension theorem

You can try and find a delta but I am quite sure it's going to be hard, (been there done that), two aproaches that are easier are:
1) Show that it is a Lipschitz function (I'm not saying that it is, I don't know, haven't tried it)
2) Use extension theorem: a function is uniform continuous in (a,b) if and only if it can be defined in the end points a, b such that the extended function is continuous.
** Clearly the case of this function. The theorem is in the book Introduction to Real Analysis by Bartle.

6. Originally Posted by Diego
You can try and find a delta but I am quite sure it's going to be hard, (been there done that), two aproaches that are easier are:
1) Show that it is a Lipschitz function (I'm not saying that it is, I don't know, haven't tried it)
It isn't Lipschitz since if $f$ is differentiable and Lipschitz then $f'$ is bounded. $f'(x)=\frac{1}{3\sqrt[3]{x^2}}$ which is unbounded as $x\to 0$

2) Use extension theorem: a function is uniform continuous in (a,b) if and only if it can be defined in the end points a, b such that the extended function is continuous.
** Clearly the case of this function. The theorem is in the book Introduction to Real Analysis by Bartle.
This only works for the exact reason that Plato said namely a continuous function on a compact metric space (in this case simply $[a,b]$) is automatically u.c.