The mean value theorem

[microsheva]

We already know the mean value theorem :"Let I=[a,b] ,K=(a,b) , f:I-->I f is continuous on I and differentiable on K . Then exists c in K s.t. f(b)-f(a)=f'(c)(b-a) ". If we replace I by open intevals such as (a,b) , [a,b),(a,b] , is the conclusion still true now ?

[Lelia]

It can't be I an open interval!
Because then it will not exist c so that f(b)-f(a)=f'(c)(b-a), because the function is not deffinate on a close interval I, so will not takes the a and b value!
It has to be an closed interval!

[Fibonacci]

Specific counterexample:
[itex]f(x)=0[/itex] if [itex]x=a[/itex] or [itex]b[/itex], and [itex]f(x)=x[/itex] if not.

[stevem]

Specific counterexample:
if or , and if not.

There was a change to the latex code since the above message was posted, making it incomprehensible so I have redone it

[fredbel6]

it can be pushed a little further

if f is continuous on [a,b] and differentiable on ]a,b[ it has a point f'(c)=(f(b)-f(a)/(b-a)

example : sqrt(1-x*x) is continuous on [-1,1] and differentiable on ]-1,1[ and
f'(0)=0

so this theorem has an obvious analogue for open intervals :

suppose f is continuous and differentiable on ]a,b[ (the latter implies the first) and has a one sided limit near both border points, there exists such point
(it is obvious because you can extend it to a function on [a,b])