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Thread: Domain of (26x-42)/(x^2 -2x - 63) ?

  1. #1
    s3a
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    Domain of (26x-42)/(x^2 -2x - 63) ?

    What is the domain of:

    (26x-42)/(x^2 -2x - 63)

    and how do I find it?

    (My graphing software shows me something that I would have never thought of)

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    Hi s3a.

    Quote Originally Posted by s3a View Post
    What is the domain of:

    (26x-42)/(x^2 -2x - 63)

    and how do I find it?

    (My graphing software shows me something that I would have never thought of)

    Any help would be greatly appreciated!
    Thanks in advance!
    You need to find the zeros of the denominator
    (x^2 -2x - 63) = 0

    in this case the zeros are $\displaystyle x_1 = 9$ and $\displaystyle x_2 = -7$. You know how to solve (x^2 -2x - 63) = 0, right?

    Therefore the denominator is $\displaystyle D = \mathbb{R} \backslash \{-7,9 \}$


    Yours
    Rapha
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by s3a View Post
    What is the domain of:

    (26x-42)/(x^2 -2x - 63)

    and how do I find it?

    (My graphing software shows me something that I would have never thought of)

    Any help would be greatly appreciated!
    Thanks in advance!
    Hell0:
    thie function is difinid $\displaystyle x^2 - 2x - 63 \ne 0
    $
    Solve : $\displaystyle x^2 - 2x - 63 = 0

    $
    $\displaystyle \left\{ \begin{array}{l}
    a = 1,b = - 2,c = - 63 \\
    \Delta = b^2 - 4ac \\
    \end{array} \right.
    $


    $\displaystyle \Delta = 256$
    $\displaystyle \begin{array}{l}
    x_1 = \frac{{ - b + \sqrt \Delta }}{{2a}} \\
    x_2 = \frac{{ - b - \sqrt \Delta }}{{2a}} \\
    \end{array}
    $
    $\displaystyle \left\{ \begin{array}{l}
    x_1 = 9 \\
    x_2 = - 7 \\
    \end{array} \right.
    $

    $\displaystyle D = \left] { - \infty , - 7} \right] \cup \left] { - 7,9} \right] \cup \left] {9, + \infty } \right]$
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