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What is the domain of this function?

  1. Oct 5, 2006 #1
    suppose x is a real variable. what is the domain of the function [tex]\sqrt{{x-3}\over{x-1}}[/tex]?

    here [tex]{{x-3} \over {x-1}} \geq 0[/tex]
    from this we get the domain to be [tex]\left(-\infty, 1\right) \cup \left[3, \infty\right)[/tex]

    but if we write the function [tex]\sqrt{{x-3}\over{x-1}} = \frac{\sqrt{x-3}}{\sqrt{x-1}}[/tex]

    then [tex]x-3 \geq 0 \Rightarrow x \geq 3[/tex]
    and [tex]x-1 > 0 \Rightarrow x > 1[/tex]

    from these two, we get the domain [tex]\left[3, \infty\right)[/tex]

    why do i get two different domains?
     
    Last edited: Oct 6, 2006
  2. jcsd
  3. Oct 5, 2006 #2

    matt grime

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    Because in the first instance you let yourself evaluate a/b first and if the answer is positive we can take the square root. In the second you want to take the square roots of a and b first. Obviously a/b is not only positive when a and b are.
     
  4. Oct 5, 2006 #3

    arildno

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    Furthermore, the maximal domain for the last function starts at 3, not 1.
     
  5. Oct 6, 2006 #4
    you are right. i have edited my previous post now.
     
  6. Oct 6, 2006 #5
    but how can the same function has different domains? can anybody please explain?
     
  7. Oct 6, 2006 #6

    arildno

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    Because it isn't the same function! :smile:
     
  8. Oct 6, 2006 #7

    matt grime

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    The idea that 'the domain of a function' is what you can work out by 'making sure we don't divide by zero or square root a negative number' at any point is completely anad utterly wrong. It is not you who is at fault but a moronic misapprehension of the definition of function that is propogated throughout the education system of many parts of the supposedly educated world.

    The domain of a function is part of its definition. What you wrote out are two different expressions which when interpreted in some manner define functions that have different domains according to some particularly idiotic ideas of what a domain is, and that, when evalauate on the intersection of the 'domians' agree.
     
    Last edited: Oct 6, 2006
  9. Oct 6, 2006 #8

    matt grime

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    Actually, this is yet another case of something that a lot of your posts embody: often you perform some manipulation and get an answer that contradicts your intuition. You then ask us why the answer is what it is when really you should not be asking us anything but asking yourself 'is my intuition at all correct'.
     
  10. Oct 7, 2006 #9
    well, i know that i am a moron. but i want to learn. thats why i ask stupid questions. i have learnt a lot from this forum by asking stupid questions. not everyone is as skilled in maths as you are. so try to be a little patient with the unfortunate (or dumb) ones. :smile:
     
  11. Oct 7, 2006 #10
    well...if you take the square root of a ngative over a negative you'd end up with a bunch of i which would cancel making the function real again.

    [tex]\sqrt{\frac{a} {b}} = \frac{\sqrt{a}} {\sqrt{b}}[/tex] regardless of a and b as long as a/b >0 or a=0
     
  12. Oct 8, 2006 #11

    matt grime

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    I didn't call you a moron. I didn't insult you at anypoint. I said two things.

    1. The concept of domain is mistaught a hell of a lot
    2. Ask yourself 'why did I expect something different each time'
     
  13. Oct 8, 2006 #12
    i never said that you insulted me in any way. i am sorry for any misunderstandings.
     
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