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Relative extrema of|f(x)|

  1. Jan 24, 2010 #1
    Use any method to find the relative extrema of [tex]|1+\sqrt[3]{x}|[/tex].
    I get the following [tex] |1+\sqrt[3]{x}| = \left\{\begin{array}{cc} 1+\sqrt[3]{x}, & \mbox{if} x >-1 \\ -1-\sqrt[3]{x}, & \mbox{if} x<-1\end{array}\right[/tex]
    Using the first derivative test, I get [tex] f'(x) = \frac{1}{3x^{2/3}} \mbox{and} -\frac{1}{3x^{2/3}}[/tex]. These suggest that there is a relative extremum at x=0 but not at x=-1, actually it is an inflection point at x=0. What is next? Maybe I am confused about absolute values! I think I have the function piece wise ok.Please help. Thanks.
     
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  3. Jan 24, 2010 #2

    mathman

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    At x=0, the derivative is infinite. I am confused about what you are confused about.
     
  4. Jan 25, 2010 #3
    Should not f'(x) =0 show a relative minimum at x=-1, i.e., if f(x) is differentiable at x=-1. According to my graphing calculator there is a relative minimum at x=-1. But I do not know to find the relative minimum at x=-1 by calculus? Thanks.
     
  5. Jan 25, 2010 #4

    arildno

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    f(x) is not differentiable at x=-1.

    That can be seen by the different values the right-hand derivative and left-hand derivative gets there.

    Thus, x=-1 is a sharp corner, and it is, indeed, a relative extremum.
     
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