User:Torarin/Sandbox

Theorem
Let $I$ be a real interval.

Let $f : I \rightarrow \R$ be everywhere differentiable.

Then $f'(I)$ is an interval. That is, $f'$ has the Intermediate Value Property.

Proof
Let $a,b \in I, \xi \in \R : a < b \land f'(a) < \xi < f'(b)$. (The case $f'(b) < \xi < f'(a)$ is handled similarly).

We need to show that $\xi \in f'(I)$, or equivalently, that $\exists x \in I : f'(x) = \xi$.

Let $g(x) = f(x) - \xi x$.

Then $g'(x) = f'(x) - \xi$.

By Differentiable Function is Continuous, $g$ is continuous on $[a..b]$.

By Restriction of Continuous Mapping, $g\restriction_{[a..b]}$ is continuous.

From the Continuity Property, $g\restriction_{[a..b]}$ attains a minimum.

By hypothesis, $g'(a) < 0$ and $g'(b) > 0$.

Then from Behaviour of Function Near Limit:
 * $\displaystyle\exists N^*_\epsilon(a) : \forall x \in N^*_\epsilon(a) : \frac{g(x) - g(a)}{x-a} < 0$

and
 * $\displaystyle\exists N^*_\delta(b) : \forall x \in N^*_\delta(b) : \frac{g(x) - g(b)}{x-b} > 0$

Thus:
 * $\forall x \in N^*_\epsilon(a) \cap [a..b]: g(x) < g(a)$

and
 * $\forall x \in N^*_\delta(b) \cap [a..b]: g(x) < g(b)$

Hence $g(a)$ and $g(b)$ are not minima in $g\restriction_{[a..b]}$, and so $g\restriction_{[a..b]}$ must attain its minimum at some $m \in (a..b)$.

By Derivative at Maximum or Minimum, $g'(m) = 0$.

Hence $f'(m) = \xi$.

$\blacksquare$