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Math Help - Absolute Values and piecewise functions

  1. #1
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    Absolute Values and piecewise functions

    Rewrite the following into a piecewise function without any absolute values:

    y = [tex]\mid x-3\mid+\mid5+x\mid[\math]

    i don't know if i did latex right, so here it is in regular text

    y=(absolute value x-3) + (absolute value 5+x)
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  2. #2
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Kingreaper View Post
    Rewrite the following into a piecewise function without any absolute values:

    y = \mid x-3\mid+\mid5+x\mid
    There are three “pieces” to the function:
    • x<-5

    • -5\le x\le3

    • x\ge3

    Consider each interval separately.
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  3. #3
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    So how would I graph those?
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  4. #4
    Senior Member JaneBennet's Avatar
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    Are you required to draw the graph as well? I thought you were only asked to rewrite the function without using absolute values.

    Write the function in the form y=\begin{cases}\ldots & x\le-5\\\\<br />
\ldots & -5\le x\le 3\\\\<br />
\ldots & x\ge3\end{cases}.
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  5. #5
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    Quote Originally Posted by JaneBennet View Post
    Are you required to draw the graph as well? I thought you were only asked to rewrite the function without using absolute values.

    Write the function in the form y=\begin{cases}\ldots & x\le-5\\\\<br />
\ldots & -5\le x\le 3\\\\<br />
\ldots & x\ge3\end{cases}.
    So for example, the first one, y = ... x less than or equal to -5. What would that look like in slope intercept form? usually it would have a number in front of the x's instead of those dots.
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  6. #6
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    Get your self a good graphing utility.
    Attached Thumbnails Attached Thumbnails Absolute Values and piecewise functions-ineq2.gif  
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  7. #7
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    Thank you so much, but another problem has just come up that I have to do.

    Solve the following:

    \sqrt{2x+1}+\sqrt{2x+6}=5
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  8. #8
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    Hello, Kingreaper!

    Solve: . \sqrt{2x+1} + \sqrt{2x+6} \:=\:5

    Isolate a radical: . \sqrt{2x+1} \;=\;5 - \sqrt{2x+6}

    Square both sides: . \left(\sqrt{2x+1}\right)^2 \;=\;\left(5 - \sqrt{2x+6}\right)^2

    . . . . . . . . . . . . . . . . . 2x + 1 \;=\;25 - 10\sqrt{2x+6} + 2x + 6

    . . . . . . . . . . . . . . . 10\sqrt{2x+6} \:=\:30

    . . . . . . . . . . . . . . . . \sqrt{2x+6} \:=\:3

    Square both sides: . 2x + 6 \:=\:9\quad\Rightarrow\quad 2x \:=\:3 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{3}{2}}

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  9. #9
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    thank you all so much
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