Image of Interval by Derivative

Theorem

Let $f$ be a real function that is everywhere differentiable.

Let $I \subseteq \Dom f$ be a real interval.

Then:

$f' \sqbrk I$ is a real interval

where $f'$ denotes the derivative of $f$

Let $x_1, x_2 \in f' \sqbrk I: x_1 < x_2$.

Let $\xi \in \openint {x_1} {x_2}$.

We need to show that $\xi \in f' \sqbrk I$.

Let $a, b \in I : \map {f'} a = x_1 \land \map {f'} b = x_2$.

Without loss of generality, assume $a < b$. The case $b < a$ is handled similarly.

Let $\map g x = \map f x - \xi x$.

Then:

$\map {g'} x = \map {f'} x - \xi$

$g \restriction_{\closedint a b}$ attains a minimum.

By hypothesis, $\map {g'} a < 0$ and $\map {g'} b > 0$.

Let $\map {N^*_\epsilon} x$ denote the deleted $\epsilon$-neighborhood of $x$.

$\exists \map {N^*_\epsilon} a: \forall x \in \map {N^*_\epsilon} a: \dfrac {\map g x - \map g a} {x - a} < 0$

and:

$\exists \map {N^*_\delta} b: \forall x \in \map {N^*_\delta} b: \dfrac {\map g x - \map g b} {x - b} > 0$

Thus:

$\exists x \in \map {N^*_\epsilon} a \cap \closedint a b: \map g x < \map g a$

and

$\exists x \in \map {N^*_\delta} b \cap \closedint a b: \map g x < \map g b$

Hence $\map g a$ and $\map g b$ are not minima of $g \restriction_{\closedint a b}$.

So $g \restriction_{\closedint a b}$ must attain its minimum at some $m \in \openint a b$.

$\map {g'} m = 0$

Hence:

$\map {f'} m = \xi$

The result follows.

$\blacksquare$