1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    k0y3a.jpg

    so i need to find the domain



    2. Relevant equations
    N/A


    3. The attempt at a solution

    I tried it like this:
    9huw7p.jpg

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

    20h8pjr.jpg
     
  2. jcsd
  3. Feb 7, 2012 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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:

    2nlvihw.jpg

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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):
    \sqrt{x^2}=x
    , because its basically
    Code (Text):
     \abs{x}
    mkfcl0.jpg
     
  7. Feb 8, 2012 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding the domain for inequality
  1. Finding domain (Replies: 6)

Loading...