# Math Help - Absolute Value Piecewise Problem

1. ## Absolute Value Piecewise Problem

I'm hoping that someone can help me understand why I'm getting this wrong. I think I understand the rules, but maybe not.

Thanks,
Mike

2. ## Re: Absolute Value Piecewise Problem

Moving 2 units to the right: x -> x - 2, giving |x| -> |x - 2|

Reflecting in the x axis: f(x) -> -f(x), giving |x - 2| -> -|x - 2|.

So your function is

\displaystyle \begin{align*} f(x) &= -|x - 2| \\ &= -\begin{cases} x - 2 \textrm{ if } x - 2 \geq 0 \\ -(x - 2) \textrm{ if } x - 2 < 0 \end{cases} \\ &= -\begin{cases} x - 2 \textrm{ if } x \geq 2 \\ -x + 2 \textrm{ if } x < 2 \end{cases} \\ &= \begin{cases} -x + 2 \textrm{ if } x \geq 2 \\ x - 2 \textrm{ if } x < 2 \end{cases} \end{align*}

3. ## Re: Absolute Value Piecewise Problem

Originally Posted by slice38
I'm hoping that someone can help me understand why I'm getting this wrong. I think I understand the rules, but maybe not.
You almost have it.
$f(x) = \left\{ {\begin{array}{*{20}{c}}{ - x + 2,}&{x \ge 2}\\{x - 2,}&{x < 2}\end{array}} \right.$