# proof of Rolle’s theorem

Because $f$ is continuous on a compact (closed and bounded) interval $I=[a,b]$, it attains its maximum and minimum values. In case $f(a)=f(b)$ is both the maximum and the minimum, then there is nothing more to say, for then $f$ is a constant function and $f^{\prime}\equiv 0$ on the whole interval $I$. So suppose otherwise, and $f$ attains an extremum in the open interval $(a,b)$, and without loss of generality, let this extremum be a maximum, considering $-f$ in lieu of $f$ as necessary. We claim that at this extremum $f(c)$ we have $f^{\prime}(c)=0$, with $a.

To show this, note that $f(x)-f(c)\leq 0$ for all $x\in I$, because $f(c)$ is the maximum. By definition of the derivative, we have that

 $f^{\prime}(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}.$

Looking at the one-sided limits, we note that

 $R=\lim_{x\to c^{+}}\frac{f(x)-f(c)}{x-c}\leq 0$

because the numerator in the limit is nonpositive in the interval $I$, yet $x-c>0$, as $x$ approaches $c$ from the right. Similarly,

 $L=\lim_{x\to c^{-}}\frac{f(x)-f(c)}{x-c}\geq 0.$

Since $f$ is differentiable at $c$, the left and right limits must coincide, so $0\leq L=R\leq 0$, that is to say, $f^{\prime}(c)=0$.

Title proof of Rolle’s theorem ProofOfRollesTheorem 2013-03-22 12:40:19 2013-03-22 12:40:19 rmilson (146) rmilson (146) 5 rmilson (146) Proof msc 26A06