# Writing an absolute value function as a piecewise function

## Homework Statement

Write f(x) = |x2-x-12| as a piecewise function.

## The Attempt at a Solution

-x2+x+12 where x>1/2
x2-x-12 where x$\geq$0

-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..

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SammyS
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## Homework Statement

Write f(x) = |x2-x-12| as a piecewise function.

## The Attempt at a Solution

-x2+x+12 where x>1/2
x2-x-12 where x$\geq$0

-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..

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​

phinds
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2019 Award
-x2+x+12 where -3<x<4

2x2-5x-3, x$\leq$1/2
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?