could someone explain Rolle theorem?

- anonymous

could someone explain Rolle theorem?

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- amistre64

yes....

- amistre64

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

- amistre64

the mean theorum states that any slope between two points can be moved to a spot on the curve where it only touches at one point right?

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## More answers

- amistre64

the Rolle theorum states that if the slope between two poits is a flat horizontal line, then there is a point on the curve that the flat line will only touch in one spot......

- anonymous

ok thanks, how is it represented mathematically?

- amistre64

...that I dont know ;)

- amistre64

if an interval I exists and is continuous between it end points and the interval I [a,b] has a slope of zero between f(a) and f(b) then there exists a point in the interval [a,b] called "c" such that the f'(c) = 0....matbe like that :)

- anonymous

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

- amistre64

in essense...it IS the mean value theorum ;) only with a flat line. Why we need it to have its own name is a mystery to me....

- anonymous

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

- amistre64

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

- anonymous

ok u gonna be back anytime soon?

- amistre64

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

- amistre64

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

- anonymous

ok

- anonymous

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

- anonymous

let f:[−1,1]rightarrow R be such that

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

since the bottom of the function is never 0, we have no restrictions there and the function is continuous thruout the interval...

- amistre64

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

- amistre64

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

- amistre64

are there any values of x in the interval [-1,1] that we should avoid using? I dont see any, so it is continuous right?

- anonymous

yea it's continous

- amistre64

what value of x will make the denominator go to zero?
1+x^2 = 0 .....none, nade, zip, ziltch.... its consinuous :) so there must be a c value in the interval [-1,1] that satisfies Rolles theorum

- anonymous

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

- amistre64

if you want to find it, yo could look for the hgihest or lowest value of f(x) and that would be it; or take the derivative and make it equal to 0 right?

- anonymous

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

- anonymous

n solve for x?

- amistre64

yep, at f'(x) = 0 we have what they call "critical points" is is when the slope of the curve becomes "0"...a horizontal line

- amistre64

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

- amistre64

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

- amistre64

f(0) = -1 :)

- anonymous

ok thanx, have a productive class

- amistre64

Ciao:)

- anonymous

\[_{?}\]

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