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Math Help - Domain and Range Functions

  1. #1
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    Domain and Range Functions

    I have this math problem that needs me to find the Domain:

    F(x)=2x^3-x^2+3

    How do you find the Domain or Range

    I noticed some range or domains look something like this:

    (2, infinity]
    [2, infinity]
    (infinity, 1) U (1, infinity)<------ this one confuses me a bit
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by snakeman11689 View Post
    I have this math problem that needs me to find the Domain:

    F(x)=2x^3-x^2+3

    How do you find the Domain or Range

    I noticed some range or domains look something like this:

    (2, infinity]
    [2, infinity]
    (infinity, 1) U (1, infinity)<------ this one confuses me a bit
    The first one is a function that is defined everywhere after two, but everywhere before and at two make the function undefined
    consider
    \sqrt{x-2}\ln^2(x-2)

    For the second one it is just a function that is undefined everywhere less than two. consider
    \sqrt{x-2}

    For the hred one it woul be someting defined for all numbers except one

    like

    \frac{1}{1-x}
    Last edited by Mathstud28; June 4th 2008 at 04:52 PM.
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    The first one is a function that is defined everywhere after two, but everywhere before and at two make the function undefined
    consider
    \sqrt{x-2}\ln(x-2)

    For the second one it is just a function that is undefined everywhere less than two. consider
    \sqrt{x-2}

    For the hred one it woul be someting defined for all numbers except one

    like

    \frac{1}{1-x}
    I still don't quite understand that
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    The first one is a function that is defined everywhere after two, but everywhere before and at two make the function undefined
    consider
    \sqrt{x-2}\ln(x-2)

    For the second one it is just a function that is undefined everywhere less than two. consider
    \sqrt{x-2}

    For the hred one it woul be someting defined for all numbers except one

    like

    \frac{1}{1-x}
    Quote Originally Posted by snakeman11689 View Post
    I still don't quite understand that
    Ok for the first one we see that for any value less than two say x=1 if we put it in the function we get

    \sqrt{1-2}\ln^2(1-2)=\sqrt{-1}\cdot\ln^2(-1)=i(\pi{i})^2=-\pi^2i

    and this is obviously not a real number, so any numberless than two will cause a non-real result, therefore it is out of the domain of these numbers


    For the second one the same thing applies

    if lets say x=0 we have \sqrt{0-2}=\sqrt{-2}=\sqrt{2}i

    once again not a real number, the same thing occurs with any other number less than two

    For the third one everything on the reals makes the function defined except x=1

    the reason being if you put one in we get

    \frac{1}{1-1}=\frac{1}{0}

    and as we both know we cannot divide by zero
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    Ok for the first one we see that for any value less than two say x=1 if we put it in the function we get

    \sqrt{1-2}\ln^2(1-2)=\sqrt{-1}\cdot\ln^2(-1)=i(\pi{i})^2=-\pi^2i

    and this is obviously not a real number, so any numberless than two will cause a non-real result, therefore it is out of the domain of these numbers


    For the second one the same thing applies

    if lets say x=0 we have \sqrt{0-2}=\sqrt{-2}=\sqrt{2}i

    once again not a real number, the same thing occurs with any other number less than two

    For the third one everything on the reals makes the function defined except x=1

    the reason being if you put one in we get

    \frac{1}{1-1}=\frac{1}{0}

    and as we both know we cannot divide by zero
    gotcha!! thanks!!
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