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Math Help - Rolle's theorem question...

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    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?
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    Eater of Worlds
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    Check it out. Rolles theorem is not applicable. Can you see why?.

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

    Now, your function is f:[-8,8]\to\mathbf{R},f(x)=\sqrt[3]{x}-1.
    Which hypothesis is not verified?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by galactus View Post
    Check it out. Rolles theorem is not applicable. Can you see why?.

    Hint: Complex number.
    What about complex numbers??

    -Dan
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    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.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by driver327 View Post
    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?
    Quote Originally Posted by galactus View Post
    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
    f(x) = \sqrt[3]{x} - 1
    not
    f(x) = \sqrt{x} - 1

    -Dan
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    Quote Originally Posted by galactus View Post
    Check it out. Rolles theorem is not applicable. Can you see why?.

    Hint: Complex number.
    No complex numbers involved since (-8)^(1/3)=-2
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    Quote Originally Posted by driver327 View Post
    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.
    Attached Thumbnails Attached Thumbnails Rolle's theorem question...-july35.gif  
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    Eater of Worlds
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    Quote Originally Posted by curvature View Post
    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.
    Last edited by galactus; July 28th 2007 at 11:31 AM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by curvature View Post
    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
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    Quote Originally Posted by topsquark View Post
    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.
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    Quote Originally Posted by topsquark View Post
    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.
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  14. #14
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    The derivative does not need to be continous for Rolle's theorem to work.
    Au contrare!

    -Dan
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    Quote Originally Posted by topsquark View Post
    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".
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