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 .

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

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

Looking at the one-sided limits, we note that

$$R=\underset{x\to c^{+}}{lim}\frac{f(x)-f(c)}{x-c}\le 0$$

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

$$L=\underset{x\to c^{-}}{lim}\frac{f(x)-f(c)}{x-c}\ge 0.$$

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

Title	proof of Rolle’s theorem
Canonical name	ProofOfRollesTheorem
Date of creation	2013-03-22 12:40:19
Last modified on	2013-03-22 12:40:19
Owner	rmilson (146)
Last modified by	rmilson (146)
Numerical id	5
Author	rmilson (146)
Entry type	Proof
Classification	msc 26A06