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Homework Help: Finding the domain for inequality

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data
    The problem is:
    for all 0≤a≤1

    so i need to find the domain

    2. Relevant equations

    3. The attempt at a solution

    I tried it like this:

    yet my solution is wrong,i am not so sure why.
    wolfram gives me this;

  2. jcsd
  3. Feb 7, 2012 #2
    Here are some problems:

    1) In the second step you wrote [itex]\frac{1}{a}[/itex]. This is only well-defined if a is nonzero. So you should look at the case a=0 seperately.

    2) In the last step you essentially did [itex]x=\sqrt{x^2}[/itex]. This is true only for positive x. Indeed: if x=-1, then [itex]x^2=1[/itex] and [itex]\sqrt{x^2}=-1[/itex]. So you need to look at the case where x is positive and x is negative.

    Moderator note: since this has nothing to do with calculus, I'm moving it to precalculus.
  4. Feb 8, 2012 #3
    Just to make it simple to solve for now, i am not going to look at case a=0 (only for now)

    so i tired to do this:


    Again my solution is wrong, i am now covering the entire number line (-inf,+inf), i feel like i am messing up basic algebra. Please help
  5. Feb 8, 2012 #4
    No, you again did [itex]\sqrt{x^2}=x[/itex]. This is not true!!!!!

    You need to consider two cases:

    1) x is positive.

    2) x is negative

    And please don't use things like [itex]\pm[/itex], it's confusing. Just taking the square root (which BY DEFINITION is positive) will suffice.
  6. Feb 8, 2012 #5
    I think i got it now:, Also when i take a square root of a number, isn't that i should take the positive and negative root?. I understand for
    Code (Text):
    , because its basically
    Code (Text):
  7. Feb 8, 2012 #6
    No. And in fact, taking the negative square root is wrong. Indeed, we have that 4>1, but if we take the negative square root of both sides then we have [itex]-\sqrt{4}>-\sqrt{1}[/itex]. This is the same as -2>-1 which is not true.

    So if a and b are positive than a<b implies [itex]\sqrt{a}<\sqrt{b}[/itex], but it doesn't imply that [itex]-\sqrt{a}<-\sqrt{b}[/itex].
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