Re: f continuous on [a,b] and differentiable on (a,b)


Ronald Bruck wrote:
> In article <1136496599.486523.85940@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> <deniz.bahar@xxxxxxxxx> wrote:
>
> > Why do many theorems in calculus/analysis have the hypotheses, f
> > continuous on [a,b] and differentiable on (a,b)?
> >
> > My question is really why the endpoints are not included for the
> > differentiable hypothesis. Aren't functions that are continous on
> > [a,b] and differentiable on (a,b) also differentiable from one side at
> > the endpoints?
> >
> Seriously, my advice is to think mathematically. You suspect there's a
> reason WHY the theorem is stated as it is, and the reason MUST be that
> functions which are continuous on [a,b] and differentiable on (a,b)
> aren't necessarily differentiable at a and b (in the one-sided
> sense--although, if they're differentiable in the one-sided sense, you
> can extend them outside the interval just by continuing the graph to be
> a certain straight line segment).
>
> Well, how can a function not be differentiable? Consider [0,1]. Can
> we find a function which is continuous at 0 but not differentiable
> there? How can the derivative fail to exist at 0? Well, the
> difference quotients (f(h)-f(0))/h must fail to converge as h --> 0+.
> How can that happen? Well, in lots of ways, but two stand out: the
> limit might be infinite, or the quotient might oscillate wildly.
>
> With this little bit of thought, we now build examples of both
> phenomena. One I already gave: f(x) = sqrt(1-x^2). The graph is a
> semicircle, and it's obvious the difference quotients go to +infinity
> on geometric grounds at the left-hand endpoint -1 (and not hard to show
> analytically).
>
> For another example, use f(x) = x sin(1/x) for x > 0, f(0) = 0. Then f
> is continuous at 0 by the squeeze theorem (it's trapped between x and
> -x), but the difference quotients oscillate wildly between -1 and 1.
>
> Really, it's called mathematical ANALYSIS. And analyzing a problem is
> the major skill we try to teach.
>

Thanks to all. Makes sense now.

Do any of you have a counterexample to the claim that if a function is
differentiable on (a,b) then it must be continuous on [a,b] - I am
pretty sure there are no counterexamples. If there can be no
counterexample, my second question is why the continuous hypothesis (to
MVT and RT) is even given?

Wouldn't it be cleaner to state the theorems as "If f is differentiable
on (a,b), then ... " ?

.

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