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...
Answer
Neighborhood of a point $p$ is a set $N_r(p)$ consisting of all points such that $d(p,q)
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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