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Find the derivative and critical numbers of a cubed root function

  1. Apr 11, 2013 #1
    1. Find the intervals of increase and decrease



    2. [itex] C(x)=x^{1/3}(x+4) [/itex]



    3. [itex] C(x)=x^{4/3}+4x^{1/3}; C'(x)=\frac{4}{3}x^{1/3}+\frac{4}{3}x^{-2/3}=\frac{4x^{1/3}}{3}+\frac{4}{3x^{2/3}}=\frac{x^{2/3}}{x^{2/3}}*\frac{4x^{1/3}}{3}+\frac{4}{3x^{2/3}}=\frac{4x+4}{3x^{2/3}}[/itex] I am wondering why the only critical number is -1, when 0 should also be considered.
     
  2. jcsd
  3. Apr 11, 2013 #2

    LCKurtz

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    Yes, 0 should be considered. Whether it gives a relative extremum is another matter.
     
  4. Apr 11, 2013 #3

    Mark44

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    A slightly better way to write the derivative is:
    C'(x) = (4/3)x-2/3(1 + x)
     
  5. Apr 11, 2013 #4
    Unless 0 isn't considered, because it would make C(x) 0.
     
  6. Apr 11, 2013 #5

    Mark44

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    Unless there are some constraints you haven't shown, it wouldn't matter that C(x) = 0.
     
  7. Apr 11, 2013 #6
    Unless, because it is still increasing from -1 to 0 and from 0 to ∞.
     
  8. Apr 11, 2013 #7

    LCKurtz

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    I think most calculus texts define a critical point as a point on the graph where either f'(x) = 0 or where the derivative doesn't exist. You have to include them when examining for relative extrema or changes in concavity. There's no "unless".
     
  9. Apr 11, 2013 #8
    I think the answer at the back of the book was referring to the intervals it was increasing or decreasing on. That would explain it.
     
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