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Find domain of f(x)=x^(5/3)-5x^(2/3)

  1. Feb 7, 2009 #1
    f(x)= x^(5/3) - 5x^(2/3)

    Need to find the domain of the above function?
    The graphing calculator shows x cannot be a negative number, and I don't see why not? Please help!
     
  2. jcsd
  3. Feb 7, 2009 #2

    d_b

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    Think of this function as (x^2 - x^3) and so if you have x to the power of anything will there be a value of x that function f(x) does not exists? ----if there is no value of x that f(x) does not exists then your domain is everything, but if there is a value(s) of x that f(x) does not exists(undefined) then your domain is everything except that value,

    does this help??
     
  4. Feb 8, 2009 #3
    My guess: Since the exponent is not an integer, the graphing calculator is using the formula xa = ea ln x, which is undefined for negative x. Really, its domain is the whole real number line (assuming you're working with real numbers), where we can let (-x)1/3 = -(x1/3).

    Are you using a TI-83 or similar? The TI-83 (and probably any related calculator) is peculiar in that if you calculate, say, -11/2, it will give back -1, as if it's doing (-x)a = -(xa), for any a. Well, it's pretty wrong about that, but for some reason it does it differently (more correctly) when graphing.
     
    Last edited: Feb 8, 2009
  5. Feb 8, 2009 #4
    If you factorise f(x) as x^2/3(x^5/2 -5), the x term in the bracket cannot be negative (being the square root of x^5).
     
  6. Feb 8, 2009 #5
    That's wrong. If you factor it correctly you get x2/3(x - 5).
     
  7. Feb 8, 2009 #6
    Of course - sorry about that! I think we can just say the domain is all real x and the calculator is wrong (and so is my factorising...)
     
  8. Jul 29, 2011 #7
    The domain of F(x) = (2/5)^x is all negative numbers, right?
     
  9. Jul 29, 2011 #8

    eumyang

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    Homework Helper

    No, it's not just "all negative numbers." You may want to look at its graph and see. Remember that the domain is the set of all INPUT values (usually x-values) of a function.

    Also, may I ask why did you reply to a 2-year old thread with a new question?
     
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