Re: Help w/calculus problem


Russell wrote:

matt271829-news@xxxxxxxxxxx wrote:
Dean_Travers wrote:
Hello--

I'm currently studying arc length and have reached a problem at the end
of the chapter's exercises which has me stumped; I've tried a couple of
tutors and haven't had any luck. The problem is as follows:

Let y = f(x) be a smooth curve and suppose f'(x) is greater than or
equal to 0 on the closed interval [a, b]. Prove: there are numbers m
and M such that m is less than or equal to f'(x) is less than or equal
to M for all x in [a, b].

I am lousy at these sort of questions, so you'd be better off ignoring
me and listening to someone else. However... m = 0 obviously satisfies
the lower bound. Then, if such M doesn't exist then f'(x) would have to
be somewhere larger than any number, and hence undefined, for some x.

But if it's undefined, there isn't an "it" to talk about in
the first place. In other words, you have to phrase
your argument a little differently.

Besides, the function

g = 1/x for x > 0
= 0 for x <=0

is unbounded on the closed interval [0,1], yet it
*is* defined everywhere on that interval. You
need to show that f' cannot be a function of this
kind, and unfortunately you haven't yet.

Thanks, I think I see this. ("Yet" might be a bit optimistic though!!)


Kind of like y = x^(1/3) over the interval [-1, 1], say (though I'm not
sure that's an entirely legitimate example).

It's an OK example as far as it goes, but you
would have to show that functions like this
are the only ones that have unbounded
derivatives.

.

Relevant Pages

  • Re: Help w/calculus problem
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  • Re: Help w/calculus problem
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  • Re: Help w/calculus problem
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  • Help w/calculus problem
    ... I'm currently studying arc length and have reached a problem at the end ... of the chapter's exercises which has me stumped; ... tutors and haven't had any luck. ... If anyone here would be kind enough to offer their insights, advice, ...
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