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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- Aug 26th 2008, 03:18 PMKingreaperAbsolute 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) - Aug 26th 2008, 03:50 PMJaneBennet
- Aug 26th 2008, 03:55 PMKingreaper
So how would I graph those?

- Aug 26th 2008, 04:01 PMJaneBennet
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 $\displaystyle y=\begin{cases}\ldots & x\le-5\\\\

\ldots & -5\le x\le 3\\\\

\ldots & x\ge3\end{cases}$. - Aug 26th 2008, 04:08 PMKingreaper
- Aug 26th 2008, 04:11 PMPlato
Get your self a good graphing utility.

- Aug 26th 2008, 04:21 PMKingreaper
Thank you so much, but another problem has just come up that I have to do.

Solve the following:

$\displaystyle \sqrt{2x+1}+\sqrt{2x+6}=5$ - Aug 26th 2008, 04:34 PMSoroban
Hello, Kingreaper!

Quote:

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

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

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

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

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

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

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

- Aug 26th 2008, 04:43 PMKingreaper
thank you all so much