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Math Help - Finding the Domain (really simple)

  1. #1
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    Finding the Domain (really simple)

    What is the domain of the functions:

    g(x)= square root (t - 2), over regular (t - 3)

    g(x)= square root (x - 4) plus square root (r) minus 2

    I don't know how to use latex, so this is the best I can do. Sorry if it's unclear.
    I don't understand how to find the domain.
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  2. #2
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    It's interesting that you took the time to find out that mathematical typesetting is done using LaTeX but you didn't bother to at least learn some...

    Anyway

    g(t) = \frac{\sqrt{t-2}}{t-3}.

    You can't have the square root of a negative number, so t - 2 \geq 0.

    Therefore t \geq 2.

    Also you can't have a denominator of 0, so that means

    t - 3 \neq 0

    t \neq 3.


    Putting it together gives

    t \in [2, 3) \cup (3, \infty).
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    is your second question this??
    g(x) = \sqrt{x-4} + \sqrt{x} - 2 ??


    if this was your second question then the answer is  x \geq 4  and x \geq 0 therefore the final answer is x \in [4,\infty)

    if your question is  g(x) = \sqrt{x-4} + \sqrt{r} - 2 ??

    then the answer is  x \geq 4 and  r \geq 0

    the domain for x is  x \in [4,\infty)
    and the domain for r is r \in [0,\infty)
    Last edited by grgrsanjay; September 5th 2010 at 04:37 AM.
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    Quote Originally Posted by grgrsanjay View Post
    is your second question this??
    g(x) = \sqrt{x-4} + \sqrt{x} - 2 ??


    if this was your second question then the answer is  x \geq 4  and x \geq 0 therefore the final answer is x \in [0,\infty)
    Actually it would be x \in [4, \infty).
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  5. #5
    Member grgrsanjay's Avatar
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    ah just was a silly mistake
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