Finding the critical numbers

  1. Oct 28, 2006 #1
    I have a problem that involves trying to find the critical numbers of a function.

    g(x) = x^1/3 - x^-2/3

    I would assume in this instance that Dg = [0, +infinity)
    g'(x) = 1/3x^-2/3 - (-2/(3(x^-5/3))
    g'x = 1/ 3(x^2/3) + 2/ 3(x^5/3)

    Now could you say that g'(0) = undefined, so 0E[0, +infinity) and g'(0) d.n.e therefore, 0 is a critical number of g. However, are there are any other critical numbers in this instance? And if so how do you solve for that?
     
  2. jcsd
  3. Oct 29, 2006 #2

    HallsofIvy

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    Yes, certainly, since g' is not defined at x= 0, that is, by definition, a critical number. That is the only value of x where the derivative is not defined but there might be other critical numbers where the derivative is equal to 0. How do you solve for that? Solve the equation 1/(3x^(2/3))+ 2/(3x^(5/3))= 0, of course. Looks pretty straight forward to me.
     
  4. Oct 29, 2006 #3
    Ahhhh I follow you now. So the other critical number after solving 1/(3x^(2/3))+ 2/(3x^(5/3))= 0, I got was x= -2 in this instance. I'm assuming this is another critical number in this question. Therefore since this question is only asking for the "critical numbers", they are then just 0 and -2 I would assume.
     
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