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Math Help - Implied Domain

  1. #1
    Senior Member Stroodle's Avatar
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    Implied Domain

    How do I work out the implied domains for f(x)=\frac{x^2-1}{x+1} and also for f(x)=\sqrt\frac{x-1}{x+2}

    Thanks heaps
    Last edited by Stroodle; June 14th 2009 at 02:29 AM.
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  2. #2
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    Quote Originally Posted by Stroodle View Post
    How do I work out the implied domains for f(x)=\frac{x^2-1}{x+1} and also for f(x)=\sqrt\frac{x-1}{x+2}

    Thanks heaps
    For the first one you require all values of x such that x + 1 \neq 0.

    For the second one, you require all values of x such that \frac{x - 1}{x + 2} \geq 0.
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  3. #3
    Senior Member Stroodle's Avatar
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    Thanks. I get the first one now, but my problem is solving \frac{x-1}{x+2}\geq0

    I'm getting 1-\frac{3}{x+2} so is the domain R\{-2} ?

    THanks for your help

    What's the command for squiggly brackets in latex btw?
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  4. #4
    AMI
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    I think for the first one there are no restrictions, because \frac{x^2-1}{x+1}=x-1 so the implied domain should be \mathbb{R}.
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  5. #5
    Senior Member Stroodle's Avatar
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    For the first one I'm getting \frac{x^2-1}{x+1}=2+\frac{-2}{x+1}
    Last edited by Stroodle; June 14th 2009 at 04:15 AM.
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  6. #6
    AMI
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    Quote Originally Posted by Stroodle View Post
    For the first one I'm getting \frac{x^2-1}{x+1}=2+\frac{-2}{x+1}
    That's not correct.
    2+\frac{-2}{x+1}=\frac{2x}{x+1}
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  7. #7
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    hi
    for the first one : \mathbb{R}
    for the second : ]-\infty,-2[ \cup [1,+\infty[
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  8. #8
    Senior Member Stroodle's Avatar
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    Hmm. I'm definitely doing something wrong here. I'm getting \frac{x^2-1}{x+1}=\frac{(x+1)(x-1)}{x+1}=1+\frac{x-1}{x+1}=1+\frac{(x+1)-2}{x+1}=2+\frac{-2}{x+1}
    Which must be wrong. But the answer is definitely R/{-1}
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  9. #9
    Senior Member Stroodle's Avatar
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    Quote Originally Posted by Raoh View Post
    hi
    for the first one : \mathbb{R}
    for the second : ]-\infty,-2[ \cup [1,+\infty[
    Thanks Raoh.
    How'd you get the second answer?
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  10. #10
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    Quote Originally Posted by AMI View Post
    I think for the first one there are no restrictions, because \frac{x^2-1}{x+1}=x-1 so the implied domain should be \mathbb{R}.
    Incorrect. If x = -1 then you have just committed the cardinal sin of dividing by zero when you do this. The domain is all real numbers EXCEPT x = -1. The graph of y = \frac{x^2 - 1}{x + 1} has a 'hole' at x = -1.

    Quote Originally Posted by Raoh View Post
    hi
    for the first one : \mathbb{R}
    [snip]
    Incorrect. See above.

    Quote Originally Posted by Raoh View Post
    [snip]
    for the second : ]-\infty,-2[ \cup [1,+\infty[
    Your bracket notation needs some minor corrections:

     {\color{red} (}-\infty,-2{\color{red})} \cup [1,+\infty{\color{red})}
    Last edited by mr fantastic; June 14th 2009 at 04:33 AM.
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  11. #11
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    Quote Originally Posted by Stroodle View Post
    Hmm. I'm definitely doing something wrong here. I'm getting \frac{x^2-1}{x+1}=\frac{(x+1)(x-1)}{x+1} Mr F says: Correct. Now cancel the common factor of (x + 1) to get x - 1. You can do this provided {\color{red}x \neq -1} (otherwise you're dividing by zero).

    =1+\frac{x-1}{x+1}=1+\frac{(x+1)-2}{x+1}=2+\frac{-2}{x+1} Mr F says: Wrong.
    Which must be wrong. But the answer is definitely R/{-1}
    Quote Originally Posted by Stroodle View Post
    Quote Originally Posted by Raoh View Post
    hi

    for the first one : \mathbb{R}

    for the second : ]-\infty,-2[ \cup [1,+\infty[
    Thanks Raoh.

    How'd you get the second answer?
    Quote Originally Posted by mr fantastic View Post
    [snip]
    For the second one, you require all values of x such that \frac{x - 1}{x + 2} \geq 0.
    Case 1: x - 1 \geq 0 AND x + 2 > 0 \Rightarrow x \geq 1.

    Case 2: x - 1 \leq 0 AND x + 2 < 0 \Rightarrow x < -2.
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  12. #12
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    Smile

    \mathbb{R}/\left \{ -1 \right \} (sorry about later)
    for the second one : study the sign of x-1 and x+2 then study \frac{x-1}{x+2}
    and see where \frac{x-1}{x+2} is positive.
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  13. #13
    Senior Member Stroodle's Avatar
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    Awesome. Thanks. I made some pretty silly errors there
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  14. #14
    AMI
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    Quote Originally Posted by mr fantastic View Post
    Incorrect. If x = -1 then you have just committed the cardinal sin of dividing by zero when you do this. The domain is all real numbers EXCEPT x = -1. The graph of y = \frac{x^2 - 1}{x + 1} has a 'hole' at x = -1.
    My judgement was that we don't actually divide by x+1 since the final expression is x-1, but I understand what you mean.
    So, for example, \frac{x^2}{x} will not be defined in 0 even though it's equal to x??
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  15. #15
    AMI
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    Oh, nevermind, I think I got it.
    I think I was saying something like "we don't need any restrictions for the function x\mapsto (\sqrt{x})^2 because it's equal to x", which is obviously false
    Sorry about that..
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