# Writing an absolute value function as a piecewise function

1. Dec 30, 2011

### thornluke

1. The problem statement, all variables and given/known data
Write f(x) = |x2-x-12| as a piecewise function.

2. Relevant equations

3. The attempt at a solution
-x2+x+12 where x>1/2
x2-x-12 where x$\geq$0

According to the answer book the answers are
-x2+x+12 where -3<x<4
x2-x-12 where x$\geq$4
2x2-5x-3, x$\leq$1/2

I am really bad at functions when it comes to domain and range..

2. Dec 30, 2011

### SammyS

Staff Emeritus
Start by writing |u| as a piecewise function:
$\displaystyle \left|u\right|=\left\{ \matrix{\ u\,,\ \text{ if }\ u\ge 0\\-u\,,\ \text{ if }\ u<0}\right.$​

Doing that for |x2-x-12|, gives:
$\displaystyle \left|x^2-x-12\right|=\left\{ \matrix{\ x^2-x-12\,,\ \text{ if }\ x^2-x-12\ge 0\\-(x^2-x-12)\,,\ \text{ if }\ x^2-x-12<0}\right.$​

That leaves you to solve:
x2-x-12 ≥ 0​
and
x2-x-12 < 0​

3. Jan 8, 2012

### phinds

Let me ask you a question.

In the quoted solutions, the first is for the range from -3 to +4. I choose to pick the point -1, which is in this range, and so I know that at the point x=-1, the equation to use is the first of the two.

In the second answer, the range is anything where x is less than +1/2. I choose the point -1, which is certainly less than 1/2, and so I know that for the point x=-1, the second answer is the correct equation.

Now the two equations are not equivalent but the stated answers give them both as the solution for the point x=-1.

What do you think about that?