1. ## rolle'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]

2. 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.

3. 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

4. Originally Posted by RockHard
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
You are right!. I didn't notice the x in the denominator.
Yes, since Rolle's theorem asks for the function to be continuos on [a,b] and differentiable on (a.b), then it cannot be applied in this function.

5. Ah, so both condition's must hold true in order to apply the theorem? Thanks for the confirmation arturo.

6. 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.

7. 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]