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What is the domain of f(x)=x^(2/3)?

  1. Sep 7, 2012 #1
    What is the domain of f(x)=x^(2/3)?

    If I take a negative number and square it, the result is positive. For example, (-4)^2 is 16. The cube root is 16 is also a positive number. So I conclude that the domain of this function is "all real numbers". But in Wolfram the answer is "all non-negative real numbers". Where is my thinking going awry?
     
  2. jcsd
  3. Sep 7, 2012 #2
    Re: Domain

    I don't think that [itex]a^{b/c} = \sqrt[c]{a^b}[/itex] can be used when a is negative.
     
  4. Sep 7, 2012 #3
    Re: Domain

    acabus: I am still having difficulty understanding. If "b" is an even power doesn't the negative "a" under the square root sign become positive?
     
  5. Sep 7, 2012 #4

    Mark44

    Staff: Mentor

    Re: Domain

    You are correct - the domain of your function is all real numbers. The reason, I believe, that WolframAlpha gives a different domain is that it is converting x2/3 to a different form, using e as the base rather than x.

    y = x2/3
    => ln(y) = (2/3) ln(x)

    => eln(y) = e(2/3)ln(x)
    => y = e(2/3)ln(x)

    Now, x must be greater than 0. This is almost the same as what WA shows, except for zero.
     
  6. Sep 7, 2012 #5
    Re: Domain

    Mark 44 Thank you so much.
     
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