determine if Rolle's Theorem can be applied. If the theorem can be applied, find all c values.

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

I've done a few of these, but I am very confused about this one.

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- Nov 22nd 2009, 11:43 AMthatonegurlrolle's theorem help
determine if Rolle's Theorem can be applied. If the theorem can be applied, find all c values.

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

I've done a few of these, but I am very confused about this one. - Nov 22nd 2009, 11:47 AMArturo_026
First you want to make sure that f(1) = f(-1), Then make sure that the function is continuous on [-1,1], and differentiable on (-1,1)

Then if those conditions hold, you can apply Rolle's theorem.

Then find f'(x), and find for what x value or values does f'(x)=0. - Nov 22nd 2009, 11:52 AMRockHard
Does the rolle theorem state, it can be applied if the function is continuous and is differentiable at x = 0, then you can apply the theorem?

This function is undefined at x = 0, I am not too familiar with it so perhaps you can shed some light for me - Nov 22nd 2009, 11:57 AMArturo_026
- Nov 22nd 2009, 11:59 AMRockHard
Ah, so both condition's must hold true in order to apply the theorem? Thanks for the confirmation arturo.

- Nov 22nd 2009, 12:01 PMthatonegurl
well, the problem's directions say exactly this:

determine wether Rolle's Theorem can be applied to f on the losed interval [a,b]. If Rolle's Theorem can be applied, find all values of c in the open interval (a,b) such that f'(c)=0.

I just don't understand calculus really at all, if you want the honest truth. About 80% of my calc. AP class doesnt get whats going on because of the teacher. He says we need to do research and figure out our own ways of doing this, yet still grades us on wether or not we get it right. I'm only taking this class because it looks great on a college app. I'm stressing so bad over here. - Nov 22nd 2009, 12:07 PMRockHard
It says determine if Rolle's Theorem can be applied, it cannot be here, thanks to Arturo's confirmation. So you simply state Rolle's Theorem cannot be applied here because the function is discontinuous at x = 0 on the interval [-1,1]