At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
it is the mean theorum but for a flat line.... really.....
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?
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......
ok thanks, how is it represented mathematically?
...that I dont know ;)
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 :)
oh so in essence it's similiar to the mean value theorem?
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....
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 :)
\[let f:[-1,1]rightarrowR be such that\]
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
since the bottom of the function is never 0, we have no restrictions there and the function is continuous thruout the interval...
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 :)
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?
yea it's continous
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
so it's saying find all c which satisfy the conclusion of Rolle's them
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?
oh, so take the defferential of f(x) and equate it to zero
n solve for x?
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
yes.. (1+x^2)(2x) - (x^2 -1)(2x) = 0
4x = 0....when x = 0
f(0) = -1 :)
ok thanx, have a productive class