# Need Help finding range of function with absolute value in denominator

• Jan 21st 2011, 01:20 PM
jester469
Need Help finding range of function with absolute value in denominator
I am completely stuck on this question, thanks for all help.

I am supposed to find the domain and range of the function:

$\displaystyle f(x)=\frac{x}{|x|}$

I have solved the domain as $\displaystyle (-\infty,0)\cup(0,\infty)$

It's been quite a few years since I've worked with domain and ranges, then again one with an absolute value in the denominator.

I know that from $\displaystyle (-\infty,0)$ it's a horizontal line at y=-1 and that from $\displaystyle (o,\infty)$ it's a horizontal line at y=1. How do I write the range for this?
• Jan 21st 2011, 01:21 PM
dwsmith
Quote:

Originally Posted by jester469
I am completely stuck on this question, thanks for all help.

I am supposed to find the domain and range of the function:

$\displaystyle f(x)=\frac{x}{|x|}$

I have solved the domain as $\displaystyle (-\infty,0)\cup(0,\infty)$

It's been quite a few years since I've worked with domain and ranges, then again one with an absolute value in the denominator.

I know that from $\displaystyle (-\infty,0)$ it's a horizontal line at y=-1 and that from $\displaystyle (o,\infty)$ it's a horizontal line at y=1. How do I write the range for this?

You just wrote it.
• Jan 21st 2011, 01:22 PM
dwsmith
Quote:

Originally Posted by dwsmith
You just wrote it.

The range is the y values.
• Jan 21st 2011, 01:25 PM
jester469
Thank you, but how do I write the answer properly? Is it written the same as like a piecewise function?
• Jan 21st 2011, 01:27 PM
dwsmith
Quote:

Originally Posted by jester469
Thank you, but how do I write the answer properly? Is it written the same as like a piecewise function?

Yes, that is fine.
• Jan 21st 2011, 02:47 PM
Krahl
It's fine how you wrote it. If you wanted to write it in compact form you use set notation,
$\displaystyle R_{f}: \{1\} U \{-1\}$
• Jan 21st 2011, 02:57 PM
HallsofIvy
Even more compactly, {1, -1}!(Happy)
• Jan 21st 2011, 02:59 PM
skeeter
$\displaystyle y = \pm 1$
• Jan 21st 2011, 04:10 PM
Krahl
lol trust you guys to make a sport of compactness
• Jan 21st 2011, 04:40 PM
skeeter
Quote:

Originally Posted by Krahl
lol trust you guys to make a sport of compactness

... it's a living.