- Thread starter
- #1

#### paulmdrdo

##### Active member

- May 13, 2013

- 386

this is how i do the problem,

I removed the absolute value bars first

f(x)= x+3-x+3 = 6

now i don't know how to define it piecewise. can you show me how define it correctly. thanks!

- Thread starter paulmdrdo
- Start date

- Thread starter
- #1

- May 13, 2013

- 386

this is how i do the problem,

I removed the absolute value bars first

f(x)= x+3-x+3 = 6

now i don't know how to define it piecewise. can you show me how define it correctly. thanks!

- Admin
- #2

On the interval:

i) \(\displaystyle (-\infty,-3)\)

we have:

\(\displaystyle x+3<0\,\therefore\,|x+3|=-(x+3)\)

\(\displaystyle x-3<0\,\therefore\,|x-3|=-(x-3)\)

and so \(\displaystyle f(x)=(-(x+3))-(-(x-3))=-6\)

Can you try the other two intervals?