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

[1+1=2]

Homework Statement

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

Homework Equations

?

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?

 

[jedishrfu]

Isn't the range simply the possible y values ?

 

[Robert1986]

Isn't the range simply the possible y values ?

Not $y=0$

 

[1+1=2]

Isn't the range simply the possible y values ?

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?

 

[Villyer]

Your idea of making this a peacewise function is helpful.

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.

 

[eumyang]

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?

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?

 

[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?

 

[Mark44]

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.

 

[skiller]

Not $y=0$

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

 

[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.
Can you use that to help you break your function up into a piece wise function?

 

[HallsofIvy]

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

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

 

[Robert1986]

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

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. :)

 

[STEMucator]

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.

 

[Ray Vickson]

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?

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

 

[1+1=2]

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. :)

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

 

[1+1=2]

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.

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?

 

[1+1=2]

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.

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.

 

[eumyang]

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} <- incorrect
R = {y=1} ( Since for all possible x values, f(x) = 1 )

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.

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

No. Please read my previous post.

Case #2 : x<0

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

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

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.

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}

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, ∞).

 

[HallsofIvy]

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.

 

[eumyang]

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.

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).

 

[eumyang]

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.

But x > 3, which further restricts the range. y will never be less than or equal to 1, so the range for this piece is y > 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.

x < 3, so again, the range is restricted. y will never be less than or equal to -1, so the range for this piece is y > -1.

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

Therefore, the range of f(x) is the union of the range of the two pieces, namely y > -1. (When x = 1.5, y = 1, so y = 1 is clearly in the range of f(x)).

I ended up making a table of values in a spreadsheet to determine the range. It seems that the fact that the function is fractional with an absolute value is throwing us off.EDIT: expanded this post a bit.

 

[1+1=2]

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.

Domain: {xl x≠0}
Range: [1,-1] or just 1 and -1,

In f(x), all y values are either 1 or -1. The range contains no vertical y values greater than or less than 1 and -1. So maybe I should write this range as {yl y = 1 and -1} or [1]U[-1]? How would you specify that the range of a function consists of 2 values, but no other values?

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.

Thanks, I have started to memorize these, but now I will pay attention to their domains, ranges, and how to traslate them into other 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, ∞).

I had not thought about this approach. It looks as if two methods accurately yield range of this function: Method 1) isolate x and solve for the newly formulated denominator as in post # 16. Method 2) Identify the function's parent function, and use translational techniques to determine the range based on the graph of the function. These seem like equally acceptable methods, but would both of these methods not work when absolute values are present in a rational function?

 

[1+1=2]

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).

________________________________________________
Firstly, the method of piecewise has caused something to click for me in regard to the original function f(x) = x/lxl

The original method proposed by Zondrina is to create a piecewise of the form:

{ x, x>0
{-x, x<0

This piecewise incorporates the denominator of the function f(x) = x/lxl into a piecewise.

Yet, in the recent case of f(x) = x/|x-3|, you include the entire function in the piecewise set.

So that would mean I could also solve the original function by including the entire function in a piecewise set like so:

{ x/x, x>0
{-(x/x), x<0

This piecewise seems to yield the same domain and range values as Zondrina denominator-only piecewise set up.

Is it advisable to include the entire function into the piecewise even though the denoninator-only piecewise will also yield a correct domain and range?
___________________________________________________


Secondly, my attempt at long division of x/(x-3) is:

(1)
1+ (-3/x) ...making the domain {xl x≠0} and range is {yl y≠0}

(2)
I don't know.

I don't know if (1) is correct either. There is no long division LaTex code in the reference to show my work on this. I have not had much (if any) practice on dividing polynomials. I understand it to be similar to long division. Is (1) correct by any chance?

 

[1+1=2]

This also leads to another question. In thinking about the process of inverting a function and finding a function's range.


Using the function $y=\frac{1}{x-3}$


The inverse is gotten by switching the x and y values and solving for y.

$y=\frac{1}{x-3}$ => $x=\frac{1}{y-3}$

$f^{-1} = 3 + \frac{1}{x}$

This is the inverse, but it is also seems to be the same steps to finding the range (according to the previous method used). Simply setting the denominator of 1/x equal to zero yields the range.

So it must be accurate that finding the range of a rational function involves the same steps as finding the inverse of a function.

 

[eumyang]

Secondly, my attempt at long division of x/(x-3) is:

(1)
1+ (-3/x) ...making the domain {xl x≠0} and range is {yl y≠0}

(2)
I don't know.

I don't know if (1) is correct either. There is no long division LaTex code in the reference to show my work on this. I have not had much (if any) practice on dividing polynomials. I understand it to be similar to long division. Is (1) correct by any chance?

Nope. It should be $1 + \frac{3}{x - 3}$. It looks like you divided $\frac{x - 3}{x}$ instead.
So $f(x) = \frac{x}{x - 3} = \frac{3}{x - 3} + 1$.

In general for rational functions in the form
$f(x) = \frac{a}{x- h} + k$
the domain will be all real numbers except at x = h, and the range will be all real numbers except at y = k. If you have a graphing calculator, plug in some random values of a, h, and k and see how they differ from $f(x) = \frac{1}{x}$.

 

[eumyang]

This also leads to another question. In thinking about the process of inverting a function and finding a function's range.

Using the function $y=\frac{1}{x-3}$

The inverse is gotten by switching the x and y values and solving for y.

$y=\frac{1}{x-3}$ => $x=\frac{1}{y-3}$

$f^{-1} = 3 + \frac{1}{x}$

This is the inverse, but it is also seems to be the same steps to finding the range (according to the previous method used). Simply setting the denominator of 1/x equal to zero yields the range.

If a function is one-to-one, then the domain of the function is the range of its inverse, and vice versa. If we take
$f(x) = \frac{1}{x - 3}$,
the domain is (-∞, 3) U (3, ∞), and the range is (-∞, 0) U (0, ∞). This function is one-to-one.

The inverse, as you found out, is
$f^{-1}(x) = \frac{1}{x} + 3$,
so the domain of f^-1(x) is (-∞, 0) U (0, ∞) and the range is (-∞, 3) U (3, ∞).

 

[1+1=2]

If a function is one-to-one, then the domain of the function is the range of its inverse, and vice versa. If we take
$f(x) = \frac{1}{x - 3}$,
the domain is (-∞, 3) U (3, ∞), and the range is (-∞, 0) U (0, ∞). This function is one-to-one.

The inverse, as you found out, is
$f^{-1}(x) = \frac{1}{x} + 3$,
so the domain of f^-1(x) is (-∞, 0) U (0, ∞) and the range is (-∞, 3) U (3, ∞).

So if a function is one-to-one, then to find its range, one can simply invert and set its denominator equal to zero. However, if the function is not one-to-one, I must use translational techniques in order to determine range.

P.S. I will investigate polynomial division and post #25 further on Monday.

 

[Bohrok]

I've got a quick question:

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:
$f(x) = x$
$f(x) = x^2$
$f(x) = x^3$
$f(x) = \sqrt{x}$
$f(x) = \frac{1}{x}$
$f(x) = e^x$
$f(x) = \ln x$
$f(x) = \frac{1}{1+e^x}$

I'm curious why you've included this last one, 1/(1+e^x), in the list above. I've seen it pop up only once or twice, not nearly as much as the other ones. Is it an important function to be familiar with like the others?

 

[eumyang]

I'm curious why you've included this last one, 1/(1+e^x), in the list above. I've seen it pop up only once or twice, not nearly as much as the other ones. Is it an important function to be familiar with like the others?

The logistic function? Actually, it's arbitrary what the parent functions are. In one precalculus textbook, only two of the 6 trig functions are included (sine and cosine) but not the others. In another textbook, the logistic function is not included, but 3 of the 6 inverse trig functions (arcsin, arccos, arctan) are. You'll notice that I wrote above a list of "what I consider" to be parent functions.