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Math Help - Range of function

  1. #1
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    Range of function

    Find the range of  \frac{2x-3}{x-2}
     x> 2 x ∈ R

    I keep getting stuck on questions about finding the range, can someone please show me if there is a general rule? And how would I go about finding the range of this function?

    Thank you,
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Tweety View Post
    Find the range of  \frac{2x-3}{x-2}
     x> 2 x ∈ R

    I keep getting stuck on questions about finding the range, can someone please show me if there is a general rule? And how would I go about finding the range of this function?

    Thank you,
    Let y=\frac{2x-3}{x-2}.

    Interchange variables and then solve for y
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    Let y=\frac{2x-3}{x-2}.

    Interchange variables and then solve for y
    Thanks,

    Do you mean like this:

     x = \frac{2y-3}{y-2}

    and than solve for 'y'?
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Tweety View Post
    Thanks,

    Do you mean like this:

     x = \frac{2y-3}{y-2}

    and than solve for 'y'?
    Yes, and then take note of the implied domain of x. This is the range that you seek.
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  5. #5
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    Quote Originally Posted by VonNemo19 View Post
    Yes, and then take note of the implied domain of x. This is the range that you seek.
    Okay but I am quite sure what you mean, when I solve for y I get the same answer.
     y = \frac{2x-3}{x-2}
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Tweety View Post
    Okay but I am quite sure what you mean, when I solve for y I get the same answer.
     y = \frac{2x-3}{x-2}
    Damn, tweety. I'm sorry. I was absent minded just then. Don't exchange variables. just solve for x outright.

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  7. #7
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    Quote Originally Posted by VonNemo19 View Post
    Damn, tweety. I'm sorry. I was absent minded just then. Don't exchange variables. just solve for x outright.
    Okay thanks,
    so when I solve I get  x = \frac{3y-2}{y-2} so the range is all the values y can take on, would that mean the range is y>2 ?

    And also can I use this method for all functions involving a fraction like this?
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  8. #8
    Senior Member Stroodle's Avatar
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    Quote Originally Posted by Tweety View Post
    Okay thanks,
    so when I solve I get  x = \frac{3y-2}{y-2} so the range is all the values y can take on, would that mean the range is y>2 ?

    And also can I use this method for all functions involving a fraction like this?
    Another method is to rearrange the equation to make it easier to see the range:

    y=\frac{2x-3}{x-2}

    y=\frac{2(x-2)+1}{x-2}

    y=2+\frac{1}{x-2}

    *Edit - Remembering that x>2
    Last edited by Stroodle; January 18th 2010 at 04:55 AM.
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  9. #9
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    Smile

    Quote Originally Posted by Tweety View Post
    Okay thanks,
    so when I solve I get  x = \frac{3y-2}{y-2} so the range is all the values y can take on, would that mean the range is y>2 ?

    And also can I use this method for all functions involving a fraction like this?
    the range is (2,\infty ).
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  10. #10
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    Smile

    if we take y=2+\frac{1}{x-2}
    \lim_{x\to 2^+}\left (2+\frac{1}{x-2}  \right )=\infty
    and,
    \lim_{x\to \infty }\left (2+\frac{1}{x-2}  \right )=2
    wouldn't now be the range (\infty ,2).
    (i might be wrong though).
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