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Homework Help: Find the domain of this function

  1. Jan 22, 2013 #1
    What is the domain of sqrt(2x-x^3)

    I thought this would be pretty straight forward but I am completely stumped. So obviously square root is not defined for (-inf, 0), therefore I said:

    2x-x^3 >= 0

    Working this out you get x <= sqrt(2) as the domain.

    That is partially correct but I can't for the life of me get the rest of the answer.
    The final answer is (-inf,-sqrt(2)] and [0, sqrt(2)]

    HOW do you find out that (-sqrt(2), 0) is not defined?

    Thanks for the read and help :D
  2. jcsd
  3. Jan 22, 2013 #2


    Staff: Mentor

    You oversimplified things. I think this is what you did:
    2x - x3 ≥ 0

    x(2 - x2) ≥ 0 (so far, so good)

    Next, it looks like you divided both sides by x - not good.

    You need to factor the quadratic part, and you can't throw away the x factor as you did. Note that x = -2 is in the domain, which your work doesn't show.
  4. Jan 23, 2013 #3
    Actually I did something worse i think, but I didn't know it was not correct? I started with

    2x - x3 ≥ 0

    but then I did this: 2x ≥ x3 and then canceled the Xs....
    so then I was left with

    2 ≥ x2 ... leading to my original answer of sqrt(2) ≥ x

    Why is this approach wrong?

    Thanks for all the replies by the way :)
  5. Jan 23, 2013 #4


    User Avatar
    Homework Helper

    Dividing through by x before or after moving the x3 to the other side is an equivalent procedure.

    What you should do is draw the graph of [itex]y=x^3-2x[/itex] and notice where y>0. You should know that [itex]y=x(x^2-2)[/itex] is the same graph. Now draw just [itex]y=x^2-2[/itex]. Although the shape is completely different, the part where it crosses the x-axis at [itex]x=\pm\sqrt{2}[/itex] is still the same with both graphs, but it's missing some extra information, such as it doesn't cross the x-axis at x=0.

    Why's this? Well, when you divided [itex]x^3-2x[/itex] by x, this assumes that [itex]x\neq 0[/itex] because you can't divide by zero, but x can be equal to zero, so you've changed the graph.

    By the way, if [itex]x^2\leq 2[/itex] then the solution set for this inequality is [-2,2]. Again, draw the graph of [itex]y=x^2-2[/itex] and find where y<0.
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