October 5th, 2012, 09:25 PM  #1 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Rolle's Theorem 2
y = f(x) = (x^(2)  1) / (x) on the closed interval [1,1] Can Rolle's Theorem be applied here? if so apply it now to the open interval (1,1) such that f ' (c) = 0 Thanks to any and to all for the help. We should have the solution handout in a few days but I can't wait lol 
October 5th, 2012, 10:01 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs  Re: Rolle's Theorem 2
Is the given function continuous on the given interval?

October 6th, 2012, 03:15 PM  #3 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: Rolle's Theorem 2
Mark, I graphed it and see it is not continuous. It has quite a jump. Algebraically, I am getting x = root 1 and x = root 1 throw out root 1 and root 1 is just 1. 1 is an endpoint so I beleive u said we throw out so I have learned from visual inspection of the graph it is not continuous on the closed interval. so right there I believe I can throw out but algebraically I verified that there is no point c between a and b on closed interval so we end right here. correct? 
October 6th, 2012, 07:22 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs  Re: Rolle's Theorem 2
Once you determine the function is not continuous on the given interval, then Rolle's theorem cannot be applied.

October 6th, 2012, 07:35 PM  #5 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: Rolle's Theorem 2
Ok


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