# Absolute Values and piecewise functions

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• Aug 26th 2008, 03:18 PM
Kingreaper
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)
• Aug 26th 2008, 03:50 PM
JaneBennet
Quote:

Originally Posted by Kingreaper
Rewrite the following into a piecewise function without any absolute values:

y = $\displaystyle \mid x-3\mid+\mid5+x\mid$

There are three “pieces” to the function:
• $\displaystyle x<-5$

• $\displaystyle -5\le x\le3$

• $\displaystyle x\ge3$

Consider each interval separately.
• Aug 26th 2008, 03:55 PM
Kingreaper
So how would I graph those?
• Aug 26th 2008, 04:01 PM
JaneBennet
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 PM
Kingreaper
Quote:

Originally Posted by JaneBennet
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 $\displaystyle y=\begin{cases}\ldots & x\le-5\\\\ \ldots & -5\le x\le 3\\\\ \ldots & x\ge3\end{cases}$.

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.
• Aug 26th 2008, 04:11 PM
Plato
Get your self a good graphing utility.
• Aug 26th 2008, 04:21 PM
Kingreaper
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 PM
Soroban
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 PM
Kingreaper
thank you all so much