Proof of Rolle's Theorem on a Closed Real Interval

By Basic Facts about Intervals of the Real Line (v), $[a,b]$ is an interval containing two distinct points, $a,b\in[a,b]$ and $(a,b)\subseteq[a,b]$. As $a\le b$ and $f$ is continuous on $[a,b]$, Extreme Value Theorem on a Closed Real Interval (ii) gives $p,q\in[a,b]$ with

$$f(p)\le f(t)\le f(q)\qquad\text{for every }t\in[a,b].$$

In each case below the point $\xi$ found satisfies $a<\xi<b$, so $f$ is differentiable at $\xi$ by hypothesis, and Vanishing of the Derivative at a Local Extremum Strictly Inside an Interval is applied with the interval $[a,b]$, the points $a,b\in[a,b]$ on either side of $\xi$, and $\delta=1$.

Case 1: $a<q<b$. Let $\xi=q$. Every $t\in[a,b]$ satisfies $f(t)\le f(\xi)$, in particular every $t\in[a,b]$ with $|t-\xi|<1$. So $f'(\xi)=0$ by Vanishing of the Derivative at a Local Extremum Strictly Inside an Interval (i).

Case 2: $a<p<b$. Let $\xi=p$. Every $t\in[a,b]$ satisfies $f(\xi)\le f(t)$, so $f'(\xi)=0$ by Vanishing of the Derivative at a Local Extremum Strictly Inside an Interval (ii).

Case 3: neither. Since $a\le p\le b$ and not $a<p<b$, $p=a$ or $p=b$; likewise $q=a$ or $q=b$. As $f(a)=f(b)$, this gives $f(p)=f(a)$ and $f(q)=f(a)$, so $f(a)\le f(t)\le f(a)$, that is $f(t)=f(a)$, for every $t\in[a,b]$. Let $\xi=(a+b)/2$; then $a<\xi<b$ by Rules of Arithmetic and Order in an Ordered Field (ix), so $\xi\in[a,b]$ and $f(\xi)=f(a)$. Hence every $t\in[a,b]$ satisfies $f(t)=f(a)=f(\xi)$, in particular $f(t)\le f(\xi)$, and $f'(\xi)=0$ by Vanishing of the Derivative at a Local Extremum Strictly Inside an Interval (i).

In every case $\xi\in(a,b)$ and $f'(\xi)=0$.