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Math Help - Can someone help me refreshen up my memories on piecewise functions?

  1. #1
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    Can someone help me refreshen up my memories on piecewise functions?

    Can someone help me refreshen up my memories on piecewise functions?-peicewise-functions.png
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  2. #2
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    Re: Can someone help me refreshen up my memories on piecewise functions?

    Quote Originally Posted by chewydrop View Post
    Click image for larger version. 

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    let $f(x)=\dfrac{n(x)}{d(x)}$ where $n(x)=x-4$ and $d(x)=x+8$

    There are 3 regions of note

    i) $n(x), d(x)<0$

    ii) $n(x)<0, 0 <d(x)$

    iii) $0<n(x), 0<d(x)$

    $n(x) < 0 \Rightarrow x<4$

    $d(x)<0 \Rightarrow x<-8$

    so

    i) is $x < -8$

    ii) is $-8 < x < 4$

    iii) is $4 < x$

    In region 1 the negative signs cancel each other so we are left with the original expression.

    In region 2 we have the negative of the original expression

    In region 3 we have no negative signs so we have the original expression.

    So we have the original expression on $(x<-8) \cup (4<x)$

    and the negative of the original expression on $-8 < x < 4$

    Finally note that $x \neq -8$ as this would lead to division by zero, but x=4 is allowed. So we refine our intervals as

    Original expression on $(-\infty, -8) \cup (4, \infty)$

    Negative expression on $(-8, 4]$

    so we have finally

    $f(x) = \begin{cases}
    \dfrac{x-4}{x+8} &(-\infty, -8)\cup (4,\infty)\\
    \dfrac{4-x}{x+8} &(-8,4]\end{cases}$

    which is answer #1
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  3. #3
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    Re: Can someone help me refreshen up my memories on piecewise functions?

    Hello, chewydrop!

    Express f(x) \:=\:\left|\frac{x-4}{x+8}\right| as a piecewise function
    by eliminating the absolute value signs.

    There are two cases to consider.

    (1) The fraction is positive: . \frac{x-4}{x+8} \:>\:0

    Then there are two more cases to consider.

    . . (a) The numerator and denominator are both positive.
    . . . . . \begin{Bmatrix}x-4 \,>\,0 &\Rightarrow & x \:>\:4 \\ x+8 \:>\:0 &\Rightarrow & x \:>\:\text{-}8\end{Bmatrix} \quad \Rightarrow\quad x \:>\:4

    . . (b) The numerator and denominator are both negative.
    . . . . . \begin{Bmatrix}x-4\,<\,0 & \Rightarrow & x \,<\,4 \\ x+8 \,<\,0 & \Rightarrow & x \,<\,\text{-}8\end{Bmatrix} \quad\Rightarrow\quad x \,<\,\text{-}8

    Hence, the fraction is positive for: . x < \text{-}8\;\cup\;x > 4

    And the fraction is negative for: . \text{-}8 < x < 4

    Therefore: . f(x) \;=\;\begin{Bmatrix}\dfrac{x-4}{x+8} & \text{ for }(\text{-}\infty,\text{-}8) \cup [4,\,\infty) \\ \\[-3mm]\dfrac{4-x}{x+8} & \text{ for }(\text{-}8,\,4) \end{Bmatrix}

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