Re: mean value theorem

In article <Pine.BSI.4.58.0612040130500.26543@xxxxxxxxxxxxxxxxx>, William Elliot writes:
On Mon, 4 Dec 2006, Kira Yamato wrote:
On 2006-12-04 00:00:57 -0500, William Elliot <marsh@xxxxxxxxxxxxxxxxxx> said:

Assume that f has a derivative (finite or infinte) at each point of an
open interval (a,b), and assume also that f is continuous at both
endpoints a, b.

Let f(x) = 1/x. Then f'(0) = -oo. Restrict f to (-1,1).
f fulfills the premise with a = -1, b = 1. f(a) = -1, f(b) = 1

f(b) - f(a) = 2 = b - a. Thus f'(c) = 1.
However you have f'(x) = -1/x^2 < 0. Tilt!

However, 1/x is not continuous at x=0.

It does not require continuity except at the end points.

1/x doesn't have a derivative at x=0, either, and that's specifically
required above.

--
Michael F. Stemper
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