Proof of Rolle's Theorem on a Closed Real Interval

Throughout, $|\cdot|$ is the absolute value on $\mathbb{R}$, and local extrema of $f$ are understood in the sense of Local Extremum at a Point, for the function $f$ on the interval $[a,b]$.

Since $a<b$ we have $a\le b$, so Extreme Value Theorem on a Closed Real Interval applies to $f$ and yields $x_{\min},x_{\max}\in[a,b]$ with

We distinguish three cases; they are exhaustive, because if neither the condition of Case 1 nor that of Case 2 holds, then $f(x_{\max})=f(a)=f(x_{\min})$.

Case 1: $f(x_{\max})\ne f(a)$. Since $f(a)=f(b)$, the point $x_{\max}$ is neither $a$ nor $b$; being in $[a,b]$, it therefore satisfies $a<x_{\max}<b$, so $x_{\max}\in(a,b)$ is an interior point of $[a,b]$. The function $f$ has a local maximum, hence a local extremum, at $x_{\max}$: with $\delta=1$, every $x\in[a,b]$ with $|x-x_{\max}|<\delta$ satisfies $f(x)\le f(x_{\max})$, since every $x\in[a,b]$ does. By hypothesis $f$ is differentiable at $x_{\max}$. By Fermat Stationary Point Criterion, $f'(x_{\max})=0$, and we take $c=x_{\max}$.

Case 2: $f(x_{\min})\ne f(a)$. Symmetrically, $x_{\min}\in(a,b)$, and $f$ has a local minimum, hence a local extremum, at $x_{\min}$: with $\delta=1$, every $x\in[a,b]$ with $|x-x_{\min}|<\delta$ satisfies $f(x)\ge f(x_{\min})$. By hypothesis $f$ is differentiable at $x_{\min}$, so Fermat Stationary Point Criterion gives $f'(x_{\min})=0$, and we take $c=x_{\min}$.

Case 3: $f(x_{\min})=f(a)=f(x_{\max})$. Then for every $x\in[a,b]$,

so $f(x)=f(a)$ by antisymmetry of $\le$. Let $c=a+(b-a)/2$; since $b-a>0$, claim 8 of Elementary Order Arithmetic in an Ordered Field gives $0<(b-a)/2<b-a$, hence $a<c<b$, so $c\in(a,b)$ is an interior point of $[a,b]$. With $\delta=1$, every $x\in[a,b]$ with $|x-c|<\delta$ satisfies $f(x)=f(a)=f(c)\le f(c)$, so $f$ has a local extremum at $c$, and $f$ is differentiable at $c$ by hypothesis. By Fermat Stationary Point Criterion, $f'(c)=0$. $\blacksquare$