# Thread: Absolute Values and piecewise functions

1. ## 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)

2. 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.

3. So how would I graph those?

4. 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 $\displaystyle y=\begin{cases}\ldots & x\le-5\\\\ \ldots & -5\le x\le 3\\\\ \ldots & x\ge3\end{cases}$.

5. 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.

6. Get your self a good graphing utility.

7. Thank you so much, but another problem has just come up that I have to do.

Solve the following:

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

8. Hello, Kingreaper!

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

9. thank you all so much