anonymous
  • anonymous
could someone explain Rolle theorem?
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
yes....
amistre64
  • amistre64
it is the mean theorum but for a flat line.... really.....
amistre64
  • 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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amistre64
  • 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
  • anonymous
ok thanks, how is it represented mathematically?
amistre64
  • amistre64
...that I dont know ;)
amistre64
  • 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
  • anonymous
oh so in essence it's similiar to the mean value theorem?
amistre64
  • 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
  • anonymous
yeah I wonder also I am gonna send you a question and u solve it n show me the steps
amistre64
  • amistre64
if I got the time, I gotta get to class in 30 minutes :)
anonymous
  • anonymous
ok u gonna be back anytime soon?
amistre64
  • amistre64
3 hours..if anything. post it up on the main side over there and see if anyone bites.
amistre64
  • amistre64
or let me see it and ill let you know if I got the time :)
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[let f:[-1,1]rightarrowR be such that\]
anonymous
  • anonymous
let f:[−1,1]rightarrow R be such that
amistre64
  • amistre64
let f(-1) to f(1) be.....
anonymous
  • anonymous
\[f(x)= x^2 - 1/(1+x^2)\]
amistre64
  • amistre64
plug in -1 and solve: f(-1) = 0 plug in 1 and solve: f(1) = 0
anonymous
  • anonymous
yes and show that f satisfies all the conditions for the application of Rolle's theorem
amistre64
  • amistre64
since the bottom of the function is never 0, we have no restrictions there and the function is continuous thruout the interval...
amistre64
  • amistre64
if we can find the x = c such that f'(c) = 0 ..... then that shuold solve it.
amistre64
  • amistre64
Rolles theorum doesnt tell you HOW to find "c", just that itlll exist :)
amistre64
  • 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
  • anonymous
yea it's continous
amistre64
  • 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
  • anonymous
so it's saying find all c which satisfy the conclusion of Rolle's them
amistre64
  • 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
  • anonymous
oh, so take the defferential of f(x) and equate it to zero
anonymous
  • anonymous
n solve for x?
amistre64
  • 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
  • amistre64
yes.. (1+x^2)(2x) - (x^2 -1)(2x) = 0
amistre64
  • amistre64
4x = 0....when x = 0
amistre64
  • amistre64
f(0) = -1 :)
anonymous
  • anonymous
ok thanx, have a productive class
amistre64
  • amistre64
Ciao:)
anonymous
  • anonymous
\[_{?}\]

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