1. ## Rolle's theorem question...

I would like to know, when could Rolle's theorem be applied? The book focused on it briefly(not enough), Well, here is a problem: f(x)=(x^(1/3))-1 on[-8,8]. (oddly enough), it only ask if Rolle's theorem can be applied in this case?

2. Check it out. Rolles theorem is not applicable. Can you see why?.

Hint: Complex number.

3. Is it because there is a possibility a negative number can be replaced for x thus making it undefined(for lack of better word)? Again, I don't know much about Rolle's theorem.

4. Rolle's Theorem
Let $\displaystyle f:[a,b]\to\mathbf{R}$ a function such that:
1) $\displaystyle f$ is continuos on $\displaystyle [a,b]$;
2) $\displaystyle f$ is differentiable on $\displaystyle (a,b)$;
3) $\displaystyle f(a)=f(b)$.
Then, exists at least one point $\displaystyle c\in(a,b)$ such that $\displaystyle f'(c)=0$.

Now, your function is $\displaystyle f:[-8,8]\to\mathbf{R},f(x)=\sqrt[3]{x}-1$.
Which hypothesis is not verified?

5. Originally Posted by galactus
Check it out. Rolles theorem is not applicable. Can you see why?.

Hint: Complex number.

-Dan

6. $\displaystyle f(8)=1, \;\ f(-8)=\sqrt{3}i$

Did I look at it wrong?. Rolle's theorem is not applicable because of the complex value which arises.

7. Originally Posted by driver327
I would like to know, when could Rolle's theorem be applied? The book focused on it briefly(not enough), Well, here is a problem: f(x)=(x^(1/3))-1 on[-8,8]. (oddly enough), it only ask if Rolle's theorem can be applied in this case?
Originally Posted by galactus
$\displaystyle f(8)=1, \;\ f(-8)=\sqrt{3}i$

Did I look at it wrong?. Rolle's theorem is not applicable because of the complex value which arises.
Ahhhh... I see the problem. The function is
$\displaystyle f(x) = \sqrt[3]{x} - 1$
not
$\displaystyle f(x) = \sqrt{x} - 1$

-Dan

8. Originally Posted by galactus
Check it out. Rolles theorem is not applicable. Can you see why?.

Hint: Complex number.
No complex numbers involved since (-8)^(1/3)=-2

9. Originally Posted by driver327
I would like to know, when could Rolle's theorem be applied? The book focused on it briefly(not enough), Well, here is a problem: f(x)=(x^(1/3))-1 on[-8,8]. (oddly enough), it only ask if Rolle's theorem can be applied in this case?
Rolle's theorem is not applicable since f(-8)=-3 and f(8)=1 are different.

10. Originally Posted by curvature
No complex numbers involved since (-8)^(1/3)=-2
Yes, I'm an idiot. I done this quick with my calculator and had it on rectangular instead of real. Plus, the pint of Tangeuray didn't help. The new Tanguray Rangpur.

I shouldn't have needed a calculator to see that. Just a major brainfart. Sorry.

11. Originally Posted by curvature
Rolle's theorem is not applicable since f(-8)=-3 and f(8)=1 are different.
Not to mention the discontinuity in the first derivative at x = 0...

-Dan

12. Originally Posted by topsquark
Not to mention the discontinuity in the first derivative at x = 0...

-Dan
The derivative does not need to be continous for Rolle's theorem to work.

13. Originally Posted by topsquark
Not to mention the discontinuity in the first derivative at x = 0...

-Dan
Yes, you are right. The function is not differentiable at x=0.

14. Originally Posted by ThePerfectHacker
The derivative does not need to be continous for Rolle's theorem to work.
Au contrare!

-Dan

15. Originally Posted by topsquark
I am familar with Rolle's theorem. It is just that you said "derivative is not continous". Which is a strange way (and probably wrong way) of saying "not differenciable".