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Show that if f is differentiable on a neighborhood of $[a,b]$ and

$f'(a) < m < f'(b)$

then there exists $a$ in $(a,b)$ such that $f'(c) = m.$

First off, what is a neighborhood of an interval? Looking at my books definition of a neighborhood, it defines a neighborhood of a point, like so:

A neighborhood of a point $x \in \mathbb{R}^n$ is a subset $X \subset \mathbb{R}^n$ such that there exists $e>0$ with $B_e(x) \subset X.$

Also, the original statement looks an awful lot like the MVT...

terrible at math
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1 Answers1

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Neighborhood of a point $p$ is a set $N_r(p)$ consisting of all points such that $d(p,q)<r$. The number $r$ is called the radius of $N_r(p)$. Here $d$ is the distance function.

It may look like intermediate value theorem but there are things to be noted.

  • For Intermediate value theorem to be true, a function should be continuous in the interval
  • A function is differentiable does not mean that the derivative is continuous( Unless specified that the function is continuously differentiable). For a counter-example see here

When the function is continuously differentiable as noted earlier intermediate value theorem would prove the result

But surprisingly, the theorem is true without the continuity constraint of derivative.

For a proof of that part see Darboux theorem.

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hrkrshnn
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  • I thought that if a function was differentible on $[a,b]$ then it is automatically true that the function is continuous on $[a,b]$ – terrible at math Feb 25 '14 at 05:51
  • If $f$ is differentiable, then $f$ has to be continuous but it is not necessary that $f'$ should be continuous. – hrkrshnn Feb 25 '14 at 05:53
  • That's a good point. Thank you. Now that i'm thinking about the problem, i think we are given that f is differentiable $[a,b]$ just so that we have it is differentiable on the interval we're working with.. I don't think the fact that it is a neighborhood is significant.

    Also, that wikipedia proof uses a lot of things we don't have in class:fermat's theorem, for one. Also first line says "if y equals f'(a) and f'(b)" but by the statement of the problem we can't have that, we have strict inequality.

     – terrible at math Feb 25 '14 at 06:02
  • @terribleatmath. I think you must be knowing the Fermat's theorem. – hrkrshnn Feb 25 '14 at 06:06
  • Oh, I do know that, but i've never heard it called that. That's weird that they never gave it a name. – terrible at math Feb 25 '14 at 06:08
  • that still doesn't clear up the wikipedia confusion about "$y equals f'(a) and f'(b)$ though. – terrible at math Feb 25 '14 at 06:11
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    Wikipedia has a slightly different version of the theorem. It says if $f'(x)\le m \le f'(b)$ and the conditions then there is some $c$ in $[a,b]$ such that $f'(c)=m$. And they split it into two cases either $f'(a)$ or $f'(b)$ equals $m$ or $f'(a)<m<f'(b)$. For the first case the proof is obvious and they go on to prove the second case. – hrkrshnn Feb 25 '14 at 06:14