proof of Rolle’s theorem

Because f is continuousMathworldPlanetmathPlanetmath 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 f0 on the whole interval I. So suppose otherwise, and f attains an extremumMathworldPlanetmath 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(c)=0, with a<c<b.

To show this, note that f(x)-f(c)0 for all xI, because f(c) is the maximum. By definition of the derivativePlanetmathPlanetmath, we have that


Looking at the one-sided limits, we note that


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


Since f is differentiableMathworldPlanetmathPlanetmath at c, the left and right limits must coincide, so 0L=R0, that is to say, f(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