What is the Domain of f(x) = x + 1? Exploring a Math Problem

[Kys91]

Suppose you have f(x) = ( x^2 + 4x + 3 ) / (x+3), making a scan in that function you can say that D : R / x =! -3, because x with -3 would give a division by zero.

But, suppose you simplify that function by factorizing, you get f(x) = x + 1

It would represent the same value, but my teacher says that in f(x) = x + 1, the domain would be the same as the other function, as far as I understand, it would for the first function, but not in the second, because even though they represent the same, are different functions.

So f(x) = x + 1 domain would be: D : R / x =! 0

Unless, she's still taking about the first function, and just using the simplification of the function just to make it easier to graph.

What is your take on this?

Correct me if I am wrong.

Thanks

 

[Elucidus]

The domain of any ration function [itex]\frac{p(x)}{q(x)}[/itex] where p and q are polynomials is [itex]\{ x : q(x) \neq 0 \}[/itex]

In your example [itex]f(x) = \frac{x^2+4x+3}{x+3}[/itex], the domain is all x other than -3 (as was mentioned).

However, [itex]f(x) = x + 1[/itex] for every x in the domain of f since

[tex]\frac{x^2+4x+3}{x+3} = \frac{(x+1)(x+3)}{x+3} = x+1.[/tex]

But x + 1 is not the same as f, since f is not defined for x = -3 whereas x + 1 is. The problem is that the cancellation of the factor (x + 3) is only possible if the factor is nonzero.

So [itex]f(x) = x+1,x \neq 0.[/itex]

I hope this is helpful.

--Elucidus

 

[Дьявол]

The domain of any ration function [itex]\frac{p(x)}{q(x)}[/itex] where p and q are polynomials is [itex]\{ x : q(x) \neq 0 \}[/itex]

In your example [itex]f(x) = \frac{x^2+4x+3}{x+3}[/itex], the domain is all x other than -3 (as was mentioned).

However, [itex]f(x) = x + 1[/itex] for every x in the domain of f since

[tex]\frac{x^2+4x+3}{x+3} = \frac{(x+1)(x+3)}{x+3} = x+1.[/tex]

But x + 1 is not the same as f, since f is not defined for x = -3 whereas x + 1 is. The problem is that the cancellation of the factor (x + 3) is only possible if the factor is nonzero.

So [itex]f(x) = x+1,x \neq 0.[/itex]

I hope this is helpful.

--Elucidus

You mean [itex]f(x)=x+1, x \neq -3[/itex], right?

 

[Elucidus]

You mean [itex]f(x)=x+1, x \neq -3[/itex], right?

Yeh. :| Sorry for any confusion.

--Elucidus

 

[Kys91]

But x + 1 is not the same as f, since f is not defined for x = -3 whereas x + 1 is. --Elucidus

Wow thank you, so, they are not an equivalent function, they can't be simplified, it is literal. You could find something that would be equivalent, and maybe "simpler" to manage with the same domain, range, and express the same thing, but they would be still be different.

A function is whateaver it is told it is, but if we are doing a computer program for temperature conversion, 2 programmers might use different mathematical function, it might "mean" the same to us, but in mathematics, it would be a different function.

What do you think?

 

[Elucidus]

Wow thank you, so, they are not an equivalent function, they can't be simplified, it is literal. You could find something that would be equivalent, and maybe "simpler" to manage with the same domain, range, and express the same thing, but they would be still be different.

A function is whateaver it is told it is, but if we are doing a computer program for temperature conversion, 2 programmers might use different mathematical function, it might "mean" the same to us, but in mathematics, it would be a different function.

What do you think?

Technically, two functions are equal if they have the same domain, the same codomain, and have the same value for the same argument. i.e.:

For [itex]f:A \rightarrow B, g: C \rightarrow D[/itex] we say f = g if

[itex]A=C,B=D, \text{ and } f(x)=g(x) \text{ for all } x \in A (=C)[/itex]

So [itex]f= (x \mapsto (x+1)(x-1)):\mathbb{R} \rightarrow \mathbb{R}[/itex] and [itex]g=(x \mapsto x^2-1):\mathbb{R} \rightarrow \mathbb{R}[/itex] are equal, but

[itex]h=(x \mapsto (x^3-x)/x):\mathbb{R}-\{0\} \rightarrow \mathbb{R}[/itex] is not equal to f (or g) since h is not defined at x = 0.

So variations in the structure of the rule are permissable as long as the function value is the same on the common domain.

Note, some texts define equality without worrying about codomain, but this is being marginally sloppy.

--Elucidus

 

[HallsofIvy]

If two functions have the same domain and have the same value for the same argument, how can they NOT have the same codomain?

 

[Elucidus]

If two functions have the same domain and have the same value for the same argument, how can they NOT have the same codomain?

[itex]f=(x \mapsto x^2):\mathbb{R} \rightarrow \mathbb{R}[/itex]

and

[itex]g=(x \mapsto x^2):\mathbb{R} \rightarrow [0, \infty)[/itex]

agree on their domains, but have different codomains. There are technically not the same function, since they belong to different Cartesian product spaces. Some texts do not worry about such distinctions use a looser sense of "equivalence."

--Elucidus

 

[Kys91]

Thanks for the responses Elucidus. My textbook doesn't include "codomain", it's an Spanish one. I am Mexican but I am glad that I have been understood even when I have to talk math concepts using English because I have never tried so, nor had a math class in English.

I really admire you sir, for the ones interested in the class, I am really sure you inspire them. You do for me, even online.

I have general questions, but I will leave them for later, I will first browse around the forum.

Thanks.

 

[HallsofIvy]

Thanks, Elucidus, for the example. Kys91, perhaps your textbook uses the term "range" which is not exactly the same as "codomain". In the examples Elucidus gave in response to my question, the "ranges" would be the same: [itex][0,\infty)[/itex] though the "codomains" are, as he says, different.

 

[srijithju]

I just don't get the concept of codomain .. why at all is anything of the sort defined .

If we have a function f(x) defined such that x belongs to the set A and f(x) belongs to the set B ( and there are no elements of B that cannot be expressed as f(x) such that x belongs to A), then what role does the codomain have to play ? From what Elucidus says what i can see is that the codomain would be a superset of B -- let as say C.
Why is it required , there would be elements in C which are not the image of any point in A.

I mean for a function to be defined is it not enough just to define a set A , and a set B and define some mapping from points in A to points in B , such that all points in A and all points in B are covered ? Why would we want to make a set C by adding some elements to B , and then say the function defined as the mapping from A to B is not the same as the function defined by the mapping from A to C. what sense does it make ??

 

[HallsofIvy]

Yes, that's true and is one reason why I, seeing "co-domain", carelessly thought "range" and asked for an example! However, there are times when we want to be able to say "f is a function from A to B" where f is not "onto" B- that is, f(A) is a proper subset of B.

 

[Elucidus]

It is very useful to discuss collections (or families) of functions that have the same domain and codomian - for example - all continuous functions from R to R. Now f(x) = x³ - x and g(x) = x² - 4 are both members of this collection and are both polynomials. But the range of f is R, while the range of g is [1, infinity). It would seem weird to say that f and g are unrelated because their ranges are different.

The concept of codomain (which is a superset of the range) allows us to collect things more elegantly. A codomain is the set of values to could be assigned to an argument, but may not actually be assigned.

The distinction between one codomain and another becomes important when considering functions as relations between sets and when considering functions of functions (like the derivative).

To add even more complexity, most people discuss functions like sqrt(x) in the same breath as f or g, but sqrt(x) is really a partial function since ts domain is not all of R.

--Elucidus