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Math Help - Can someone help me with this problem on continuity? :(

  1. #1
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    Can someone help me with this problem on continuity? :(

    I asked so many people and they don't understand it either. If you anybody could help it be much appreciated

    Prove that f(x)=x^1/3 is continous at x=c for c>0
    [Note: a-b=(a-b)(a+ab+b)]
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  2. #2
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    All you have to do is to show that the function is differentiable in the domain x>0 because a function must be continuous at a point in order for it to be differentiable at that point. Hope that helps!
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  3. #3
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    Quote Originally Posted by Linnus View Post
    All you have to do is to show that the function is differentiable in the domain x>0 because a function must be continuous at a point in order for it to be differentiable at that point. Hope that helps!
    It looks like he needs to do it by definition, ie. an \epsilon - \delta proof.

    @swtbellrose: Assume |x-c| < \delta and use the hint: \displaystyle |x-c| = \left| \left( x^{\frac{1}{3}} \right) ^3 - \left( c^{\frac{1}{3}} \right) ^3 \right| = \left| \left( x^{\frac{1}{3}} - c^{\frac{1}{3}} \right) \left(x^{\frac{2}{3}} + x^{\frac{1}{3}}c^{\frac{1}{3}} + c^{\frac{2}{3}} \right) \right|

    Now try to find a good \delta so that \left| x^{\frac{1}{3}} - c^{\frac{1}{3}} \right| < \epsilon
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  4. #4
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    But of course, since swtbellrose did not show any work at all, we have no idea what the problem really is. My first thought was just to note that \lim_{x\to 0}x^{1/3}= 0 which is the value of 0^{1/3} so the function is continuous at x= 0.
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