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yes....

it is the mean theorum but for a flat line.... really.....

ok thanks, how is it represented mathematically?

...that I dont know ;)

oh so in essence it's similiar to the mean value theorem?

yeah I wonder also I am gonna send you a question and u solve it n show me the steps

if I got the time, I gotta get to class in 30 minutes :)

ok u gonna be back anytime soon?

3 hours..if anything. post it up on the main side over there and see if anyone bites.

or let me see it and ill let you know if I got the time :)

ok

\[let f:[-1,1]rightarrowR be such that\]

let f:[−1,1]rightarrow R be such that

let f(-1) to f(1) be.....

\[f(x)= x^2 - 1/(1+x^2)\]

plug in -1 and solve: f(-1) = 0
plug in 1 and solve: f(1) = 0

yes and show that f satisfies all the conditions for the application of Rolle's theorem

if we can find the x = c such that f'(c) = 0 ..... then that shuold solve it.

Rolles theorum doesnt tell you HOW to find "c", just that itlll exist :)

yea it's continous

so it's saying find all c which satisfy the conclusion of Rolle's them

oh, so take the defferential of f(x) and equate it to zero

n solve for x?

yes..
(1+x^2)(2x) - (x^2 -1)(2x) = 0

4x = 0....when x = 0

f(0) = -1 :)

ok thanx, have a productive class

Ciao:)

\[_{?}\]