From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Help with a math problem???
Date: 8 Aug 2000 08:10:56 -0400
Newsgroups: sci.math
Summary: [missing]

In article <8moqp3$pnl$1@nnrp1.deja.com>,  <mathpath1@my-deja.com> wrote:
: I got this math problem at an analytical math
:course and I can't solve it? So is there anyone
:who would know the answer or know where I could
:search for it? It is probably easy for you but a
:really hard one for me:
:
: f:[a,b]-> R is derivetive at [a,b] and
: f'(a) < y(0) < f'(b).
: Show that there is x(0) that belongs to ]a,b[
: so that f'(x(0))= y(0)
:
 This is a standard theorem (attributed to Darboux):

Suppose f is differentiable on [a, b] - with the stipulation that the
derivative at b is taken from the left and the derivative at a is taken
from  the right (so no extension of f beyond [a, b] is needed).

Suppose, as you did, that  f'(a) < y(0) < f'(b), and look for x(0) such
that  f'(x(0)) = y(0).

Motivation: We recall Fermat's theorem - that a function f which has an
extreme value at an interior point c of [a, b] and which also has a
derivative at c must have f'(c)=0. So, we can try reduce the given problem
of prescribed derivative y(0) to the problem of prescribed derivative 0 od
a related function.

Now the technicalities - you finish some of the steps on your own, I
trust: We invent a function

 g(x) = f(x) - x * y(0)

observe that

 g'(x) = f'(x) - y(0)

hence  g'(a) < 0  and  g'(b) > 0 .

The function g is continuous on [a, b], so by the extreme value property
it must have a minimum at some x(0) in [a, b].

Now pick it up: Can x(0) = a or x(0) = b? (Can f(a) or f(b) be minimal?)

You can find out that x(0) must be strictly between a and b, and the
conclusion follows from Fermat's Theorem.

Good luck, ZVK(Slavek).
==============================================================================

From: spamless@Nil.nil
Subject: Re: Help with a math problem???
Date: 8 Aug 2000 08:50:18 -0400
Newsgroups: sci.math

mathpath1@my-deja.com wrote:
>  I got this math problem at an analytical math
> course and I can't solve it? So is there anyone
> who would know the answer or know where I could
> search for it? It is probably easy for you but a
> really hard one for me:

>  f:[a,b]-> R is derivetive at [a,b] and
                  ^^^^^^^^^^
                  should be "differentiable"

>  f'(a) < y(0) < f'(b).
>  Show that there is x(0) that belongs to ]a,b[
>  so that f'(x(0))= y(0)


>  Sorry for my bad English skills and that I
> couldn't find all the right symbols from my
> computer but I think you get the idea!

It looks fine - don't worry about it. Also, don't use special symbols in
USENET, for it is an ascii forum. You did a good job.

Of course, if f' is continuous, and f'(a)<y_0<f'(b), then there is some
point x_0 between a and b with f'(x_0)=y_0. The point is that a function
may be differentiable on [a,b] (so the function is continuous on
[a,b]) but the derivative NOT continuous in [a,b]. What you are trying to
prove here is that even in that case, the discontinuity in the derivative
can't be a "jump" discontinuity (that is, it can't jump over and miss an
intermediate, like y_0, value).

There isn't much one can use to show "there exists an x_0 with
f'(x_0)=..." except the mean value theorem, that is, if we can find a u,v
(say a<=u,v<=b) with (f(v)-f(u))/(u-v) = y_0, then we can use the mean
value theorem to say there is some x_0 (in ]u,v[ and so a<=u<x_0<v<=b so x
is in ]a,b[) with f'(x_0)=(f(v)-f(u))/(v-u)=y_0 (since f is differentiable
on ]u,v[). The trick is to do that and then use the mean value theorem.
-----------

Let's consider the obvious thing to try: u=a, v=b. IF y_0 just HAPPENS to
be equal to (f(b)-f(a))/(b-a), then you can use the mean value theorem on
]a,b[ to say there is an x_0 with f'(x_0)=(f(b)-f(a))/(b-a).
------------

If not ... well, there are two possibilities:
y_0<(f(b)-f(a))/(b-a) or y_0>(f(b)-f(a))/(b-a)
-------------

If y_0<(f(b)-f(a))/(b-a).

We know f'(a)<y_0
Consider the function (f(v)-f(a))/(v-a)=g(v)
g is a *continuous*(+) function of v on [a,b] (if we give it the value of
f'(a) at v=a, since the limit exists).
We have: f'(a)=g(a)<y_0<g(b)
AHA! That means there exists some V between a and b where g(V)=y_0!

(since g is continuous, IT (g) can't jump over that value - the
intermediate value theorem applied to "g").

But then we have some a<V<b where g(V)=y_0 (or (f(V)-f(a))/(b-a)=y_0) and
by the mean value theorem, some x_0 between a and V (and, so, between a
and b as well) with f'(x_0)=(f(V)-f(a))/(V-a)=y_0

(+): Since f is differentiable, f is continuous. The only problem
     with g is at v=a, where the denominator vanishes, but this is
     a removable discontinuity since g has the limit
     lim(g(v):v decreases to x)=f'(a)
------------

On the other hand, if y_0>(f(b)-f(a))/(b-a) then consider
g(u)=(f(b)-f(u))/(b-u)....