Extension: Darboux's Theorem from Mean and Intermediate Value Theorems

In this section, we use the Mean Value Theorem and the Intermediate Value Theorem to prove Darboux's Theorem.

First, let's recall what Darboux's Theorem says:

Suppose $f(x)$ is differentiable on an open interval $I$. Then for all $a,b$ in $I$ with $a\le b$, and for all $r$ between $f'(a)$ and $f'(b)$, there exists $d\in I$ such that $f'(d)=r$.

So, we'll start out by assuming that $r$ lies strictly between $f'(a)$ and $f'(b)$. We then define two continuous functions $f_a,f_b: I\to \mathbb{R}$ by $$f_a(x) = \left\{ \begin{array}{lr} \frac{f(a)-f(t)}{a-t} & : t\ne a\\ f'(a) & : t=a \end{array} \right.$$ and $$f_b(x) = \left\{ \begin{array}{lr} \frac{f(t)-f(b)}{t-b} & : t\ne b\\ f'(b) & : t=b \end{array} \right.$$ We first remark that since $$\lim_{t\to a}\frac{f(a)-f(t)}{a-t}=f'(a)\text{ and } \lim_{t\to b}\frac{f(t)-f(b)}{t-b}=f'(b),$$ the functions are indeed continuous everywhere.

Now, by definition, we have $f_a(a)=f'(a), f_a(b)=f_b(a),$ and $f_b(b)=f'(b)$. Therefore, we must have that either $r$ lies between $f_a(a)$ and $f_a(b)$, or it lies between $f_b(a)$ and $f_b(b)$.

So without loss of generality, suppose $r$ is between $f_a(a)$ and $f_a(b)$. Since $f_a(x)$ is continuous, by the Intermediate Value Theorem there exists $D$ between $a$ and $b$ such that $$r=f_a(D)=\frac{f(D)-f(a)}{D-a}.$$ Finally, by the Mean Value Theorem on $f(x)$, it follows that there exists $d$ between $a$ and $b$ for which $$r=\frac{f(D)-f(a)}{D-a}=f'(d).$$ This completes the proof.