Re: f continuous on [a,b] and differentiable on (a,b)
- From: Ronald Bruck <bruck@xxxxxxxxxxxxxxxxx>
- Date: Thu, 05 Jan 2006 21:28:26 -0800
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.
--Ron Bruck
.
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