# Domain and range of f(x) = x/lxl

1. Jul 20, 2012

### 1+1=2

1. The problem statement, all variables and given/known data

Find the domain and range of the function: F(x) = $\frac{x}{lxl}$

2. Relevant equations

?

3. The attempt at a solution

domain is all x such that x ≠0

range: do i have to make this into a piecewise function, if so, how?

2. Jul 20, 2012

### Staff: Mentor

Isn't the range simply the possible y values ?

3. Jul 20, 2012

### Robert1986

Not $y=0$

4. Jul 20, 2012

### 1+1=2

Yes, jedish, that is what I understand the definition of range to be, thus, so how do I find those possible y values?

Robert, why not y = 0?

5. Jul 20, 2012

### Villyer

Since the purpose of making it peacewise is to remove the absolute value function, look to see where the absolute value changes the number inside and when it doesn't. The point where it's effect on the value inside is the point at which you want to break up your function.

6. Jul 20, 2012

### eumyang

Think about it. If y = 0, you'll need x-value(s) that make it so.
$$0 = \frac{x}{\left| x \right|}$$
Now, is it possible to find an x-value such that y = 0?

7. Jul 20, 2012

### 1+1=2

eumyang, thanks, that makes sense. I follow the rationale.

It makes sense that 0 ≠ $\frac{x}{lxl}$.

Is there any analytical approach to figuring y = 0? Supposing a range of a more complex function is supposed to be found? How would I go about it?

Villyer, thanks, how can I turn this into a piecewise function?

8. Jul 20, 2012

### Staff: Mentor

I think that you're making this more complicated than it really is. There really are only two cases: x < 0 and x > 0.

If x < 0, x/|x| = ?
If x > 0, x/|x| = ?

You already know that x can't equal 0.

9. Jul 20, 2012

### skiller

I'm confused as to why you have singled out that one particular value for $y$.

10. Jul 20, 2012

### Villyer

The purpose of making it piecewise is to remove the absolute value from the function.

abs(f(x)) returns either f(x) or -f(x) depending on whether or not f(x) is positive or negative.

11. Jul 21, 2012

### HallsofIvy

Staff Emeritus
Considering how many values y is not equal to, I kind of wondered that myself!

12. Jul 21, 2012

### Robert1986

Ahh, yes. Because for some reason, when I read the question, I (thought I) read $y=\frac{x^2}{|x|}$ and then somehow interpreted this as being the same as the graph of $y=x$ with a hole at $x=0$.

So, if you misread the question (as I did) and then make the embarrassing error that the square of a real number can be negative (as I did) then my response was quite reasonable. :)

13. Jul 21, 2012

### Zondrina

Just simply remember the definition.....

|x| =
{x if x≥0
{-x if x<0

So break it up into two cases... ill show you one of them for an example :

Case #1 : x≥0

So f(x) = x/|x| = x/x = 1

D = {x$\in$ℝ}
R = {y=1} ( Since for all possible x values, f(x) = 1 )

I leave the rest to you.

14. Jul 21, 2012

### Ray Vickson

What is stopping you from getting some insight by trying a few values, such as x = 1, x = 2, x = -1, x = -2, etc?

RGV

15. Jul 21, 2012

### 1+1=2

Robert, appreciate the clarification, so would mean that y = 0 is in fact an accepted value for range?

16. Jul 21, 2012

### 1+1=2

Case #2 : x<0

So f(x) = x/|x| = x/(-x) = -1

D = {x$\in$ℝ}
R = {y=-1} ( Since for all possible x values, f(x) = -1 )

This seems fairly easy (with advice and thought), but it seems unlikely that correct range values would be as easily obtained (as plugging in a few x values and noticing a pattern) with more complex functions.

For instance, f(x) = $\frac{1}{x-3}$

The domain is simply the denominator set equal to 0, {xl x≠3}

However, range is found by solving for (isolating x to one side) and setting the denominator equal to zero:

x = $3+\frac{1}{y}$

So range is {xl x≠0}

This is a systematic method that I assume is the only way to find the range. I also assume this method works for all rational functions. However, I don't see how it is possible to apply this approach with the given function: f(x) = $\frac{x}{lxl}$

Is it possible to isolate x to one side in such a way that taking the denominator on the resulting function would yield a range value?

17. Jul 21, 2012

### 1+1=2

Okay, and suppose I have the function:

f(x) = $\frac{x}{|x-3|}$

In my head, I can simply say that Domain: x≠3

But, finding the range seems more complex. Do I automatically set up a piecewise form?

I'm not even sure how I'd go about creating a peicewise function for this type of function.

18. Jul 21, 2012

### eumyang

The above is not completely correct. Case 1 should be x > 0, not x ≥ 0. x cannot equal zero. That also means the the domain is not ℝ.

You're not done yet. You still have yet to describe the domain and range of the function
$$f(x) = \frac{x}{\left| x \right|}$$
What we have above are the domains (which are incorrect) and ranges of the two pieces. Put them together.

No, it would be {y | y ≠ 0}.

I find it helpful to memorize the domains and ranges of some of the parent functions. Since many of the functions we encounter are based on the parent functions, you can adjust the domains and ranges appropriately. The functions that I consider as parent functions are these:
$f(x) = x$
$f(x) = x^2$
$f(x) = x^3$
$f(x) = \sqrt{x}$
$f(x) = \frac{1}{x}$
$f(x) = e^x$
$f(x) = \ln x$
$f(x) = \frac{1}{1+e^x}$
$f(x) = \left| x \right|$
$f(x) = int(x)$
... plus the six trig functions.

In the case of $f(x) = \frac{1}{x-3}$, the parent function is $f(x) = \frac{1}{x}$. The domain and range of this parent function is (-∞, 0) U (0, ∞). $f(x) = \frac{1}{x-3}$ is a translation of the parent function 3 units to the right. The range is the same as the parent function, but the domain changes to (-∞, 3) U (3, ∞).

19. Jul 21, 2012

### HallsofIvy

Staff Emeritus
You are right that determining the domain is typically easier than determining the range- so reverse domain and range.

If x> 3 then $$y= \frac{x}{|x- 3|}$$ becomes $$y= \frac{x}{x- 3}$$ so y(x- 3)= xy- 3y= x. xy- x= x(y- 1)= 3y, and $$x= \frac{3y}{y- 1}$$. Now it is clear that y can be any positive number except 1.

If x< 3 then $$y= \frac{x}{|x-3|}$$ becomes $$y= -\frac{x}{x- 3}$$ so y(x- 3)= xy- 3y= -x. xy+ x= x(y- 1)= 3y, and $$x= \frac{3y}{y+ 1}$$. Now it is clear that y can be any negative number except -1.

That is, the range of all real numbers except 1 and -1.

Last edited by a moderator: Jul 21, 2012
20. Jul 21, 2012

### eumyang

Yes, you set up as a piecewise function. You have two cases:
(1) x > 3:
$f(x) = \frac{x}{x - 3}$
(2) x < 3:
$f(x) = \frac{x}{3 - x}$

For case (1), divide the polynomials (yes, the numerator and denominator are polynomials), and rewrite the function as
$quotient + \frac{remainder}{divisor}$
You'll see that this form is a variation of the basic function $f(x) = \frac{1}{x}$. From there, you can determine the range.
Repeat for case (2).