On-line Math 21

On-line Math 21

4.1  The Mean Value Theorem

Theorem 5 If f is continuous on a closed interval [a,b] , and differentiable on the open interval (a,b) , and if f(a) = f(b) , then there is some point c Î (a,b) so that f¢(c) = 0 .

Proof. If, for some x Î (a,b) , f(x) > f(a) , then there is a maximum point by the theorem we proved a while back about the existence of a maximum of a continuous function. That maximum value has to be greater than f(a) , so it isn't going to happen at x = b , since f(a) = f(b) . So, it's at some c Î (a,b) . Our assumption was that the function was differentiable on (a,b) , so f¢(c) exists. By the previous result, then, f¢(c) = 0 .

The same argument works if, for some x , f(x) < f(a) (all the inequalities are reversed). The only other possibility is that f(x) is never bigger or smaller than f(a) , which says that f(x) is constant, so has 0 derivative at all x Î (a,b) .

Copyright (c) 2000 by David L. Johnson.


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On 30 Nov 2000, 23:55.