Image of Interval by Derivative

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

Let $I \subseteq \operatorname{Dom}(f)$ be an interval.

Then $f'(I)$ is an interval.

Proof
Let $x_1, x_2 \in f'(I) : x_1 < x_2$.

Let $\xi \in (x_1..x_2)$.

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

Let $a,b \in I : f'(a) = x_1 \land f'(b) = x_2$.

Assume $a < b$. The case $b < a$ is handled similarly.

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

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

By Differentiable Function is Continuous, $g$ is continuous.

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

From Corollary 3 to Continuous Image of a Compact Space is Compact, $g\restriction_{[a..b]}$ attains a minimum.

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

Let $N^*_\epsilon(x)$ denote the deleted $\epsilon$-neighborhood of $x$.

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

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

Hence $g(a)$ and $g(b)$ are not minima of $g\restriction_{[a..b]}$, 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 $f'(m) = \xi$.

The result follows.