Squeeze Theorem for derivatives

[benf.stokes]

Homework Statement

Show, with appropriate examples, that the conditions g(x) < f(x) < h(x) and derivative(g(x0))=derivative(h(x0)) = m does not imply derivative(f(x0)) = m or even exists. And with some additional condition.

Homework Equations

derivative g(x) = lim(h tends to zero) (g(x+h)-g(x))/h

The Attempt at a Solution

How do I even get started?
Thanks

 

[jbunniii]

Start by identifying the simplest function you can think of which has no derivative at, say, x = 0.

By the way, the problem seems almost too simple as stated. But what do you mean "and some additional condition"? Are there some more constraints on g, f, and/or h? E.g., do their values have to be close to each other at $x_0$?

 

[benf.stokes]

Thanks for the reply. The problem does require a formal proof although I know it may not appear because of my sloppy translation.
And the aditional conditional means that something other than g < f < h is required for deriv g < deriv f = deriv h = m