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Math Help - rolle's theorem

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    rolle's theorem

    rolle's theorem
    f must be continuous on a closed interval [a,b]
    differential on (a,b) with f(a)=f(b)
    Then, there is atleast one point C in (a,b) such that f'(c)=0

    I am suppose to determine if rolle's theorem applies to the fuction and solve
    f(x)=1-x^{2/3} in the inverval [-1,1]

    Mathab says the assumptions for the Theorem have not been satisfied. I am under the impression they are satisfied. I was wonderifing if someone could provide and explanation as to why the theorem has not been satisfied.
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    Re: rolle's theorem

    Quote Originally Posted by delgeezee View Post
    rolle's theorem
    f must be continuous on a closed interval [a,b]
    differential on (a,b) with f(a)=f(b)
    Then, there is atleast one point C in (a,b) such that f'(c)=0

    I am suppose to determine if rolle's theorem applies to the fuction and solve
    f(x)=1-x^{2/3} in the inverval [-1,1]

    Mathab says the assumptions for the Theorem have not been satisfied. I am under the impression they are satisfied. I was wonderifing if someone could provide and explanation as to why the theorem has not been satisfied.
    f(-1) = 0 and f(1) = 0, so Rolle's Theorem is satisfied.
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    Re: rolle's theorem

    is f(x) differentiable on (-1,1)? let's see:

    f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}

    hmm....looks like we have a problem at x = 0.....
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    Re: rolle's theorem

    Is f(-1)= f(1) ?
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    Re: rolle's theorem

    Quote Originally Posted by pickslides View Post
    Is f(-1)= f(1) ?
    It certainly is...

    f(-1) = 1 - (-1)^(2/3) = 1 - [(-1)^2]^(1/3) = 1 - 1^(1/3) = 1 - 1 = 0

    and

    f(1) = 1 - 1^(2/3) = 1 - [1^2]^(1/3) = 1 - 1^(1/3) = 1 - 1 = 0.

    However, Deveno is right about the function not being differentiable in this entire region. So I was wrong about Rolle's Theorem being satisfied
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