Re: Darboux intermediate value theorem



speedy wrote:
> This is about an exercise in the textbook 'Introduction to Mathematical
> Analysis' by SA Douglass.
>
> I quote the text litterally:
> " Prove Darboux's Intermediate Value Theorem for derivatives: If f is
> differentiable on [a,b] and if d is some number between f'(a) and
> f'(b), then there exists a c in (a,b) such that f'(c) = d. (Hint: For x
> in [a,b], let g(x) = d(x-a) - f(x). Show that g must have a critical
> point in (a,b).)
>
> Now my question is: can anybody see why g(x) is defined here as d(x-a)
> - f(x), and not as dx - f(x) (as it is for example in the proof found
> at planetmath.org/?op=getobj&from=objects&id=3056. Of course, the a
> does not matter further, because it vanishes in the derivative, but why
> bother then?

Only his hairdresser knows.I found this theorem as an exercise in
the 1968 Advanced Calculus by Alan Devinatz with the same hint(pg 158
exercise11) and also found the same proof written out clearly in the
1967 book :An Introduction to Analysis by Wilson Zaring (pg 245)
In both cases the function g(x)=f(x)-d(x-a) (note this differs
from your g by a minus sign).My guess is that d(x-a) is used instead of
just dx is to facilitate slightly drawing a picture (d(x-a) has value 0
at a) .Draw the graph y=f(x) and y=d(x-a) and you are looking for a
point on f where the tangent line is parallel to that of :y=d(x-a)
similar to the picture for the mean value thorem for f.
Well thats a guess (I see that you could draw y=dx just as
well),logically it does not matter,dx is fine

There is a proof in Tom Apostal ,Mathematical Analysis which just uses
the mean value theorem for f .
Consider G(x)=(f(x)-f(a)/x-a for x>a (=f'(c) ,a<c<x by MVT for f)
and define G(a)=f'(a) (the limit as x ->a ) so G is continuous on [a,b]
so by the intermediate value theorem G takes on all values d between
G(a)=f'(a) and K=G(b)=f(b)-f(a) /b-a. The same argument applied to
the function
G_(x) =f(b)-f(x)/b-x (G_(b)=f'(b)) gets that all values d between K
and f'(b) are of the form f'(c) also .Regards,Stuart M Newberger

.

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