I feel so dumb for asking this question but domain of a rational function help

[dumbQuestion]

Hello, I have a simple question that has bothered me forever.When I am given a rational function, say,

f(x) = (x²+x-6)/(x+3)say I want to look at the domain of this function. Well the first thing that catches my eye is the fact that if I plug in x=-3, the denominator will be 0, which would be undefined. So I immediately think, the domain of this function is all real numbers except -3. But then when I factor the numerator, I end up realizing that f can be written as:f(x) = (x-2)(x+3)/(x+3)I think to myself, ok, I can just cancel the (x+3) from the denominator and numerator, and now I just have the function f(x)=x-2Great! But that's not the case. I mean at least my calculus book gives me the impression that its not. In my book when introducing limits, they give that f(x) I listed as an example of a function where the function is not defined at a point because of a zero in the denominator. But why can't I just factor this out and cancel the (x+3) from the denominator? It makes me feel like whenever I'm doing ANY calculation with rational functions I shouldn't cancel out any of these (x+c) terms that are common in the numerator and denominator because I feel I'll be losing important information about the function (it's domain)! In other words: I no longer feel I can say (x+3)(x-2)/(x+3) = (x-2) because the function on the left doesn't have the same domain as the function on the right, so they aren't the same function!I just don't understand.

 

[haruspex]

Whenever you cancel terms that are potentially zero you ought to write "either (the term) equals zero or ...".
f(x), as defined, specifies a sequence of operations to map from x to f(x). The final step is the division. If that leads to 0/0 then f(x) is undefined for that x.
Cancellation produces a different function that happens to the same as f everywhere except at x=-3, and is continuous everywhere.

 

[dumbQuestion]

so in say a college algebra book when you are doing exercises about factoring and canceling, like say you have a problem f(x) = (x^2+x-6)/(x+3)

and then the problem says: reduce the function, and you end up with an answerf(x) = (x-2) its not really true? I mean... these really aren't the same function?

 

[HallsofIvy]

No, they are not the same function. They are the same everywhere except at x= -3. (And any good textbook would specify that exception.) The first function is undefined at x= -3 while the second has the value f(-3)= -5. If we call f(x)= (x^2+ x- 6)/(x+3) and the second g(x)= x- 2, then "f(x)= g(x) for all x except x= -3".

 

[Bacle2]

This is actually not a dumb question, it results from a good observation on your part.

 

[Hurkyl]

Sometimes, the difference between the function x-2 and the partial function (x²+x-6)/(x+3) is relevant.

Sometimes, the difference doesn't matter.Sometimes, they aren't different -- e.g. if you're defining functions on the domain of real numbers that aren't -3, or are doing algebra with formal rational functions.

 

[Peppino]

$$f(x) =\frac{x^2+x-6}{x+3}$$
$$g(x) = x-2$$

When using limits, f(-3) and g(-3) will return the same value

$$\lim_{x \to -3}f(x) = -5$$
$$\lim_{x \to -3}g(x) = -5$$

 

[chiro]

$$f(x) =\frac{x^2+x-6}{x+3}$$
$$g(x) = x-2$$

When using limits, f(-3) and g(-3) will return the same value

$$\lim_{x \to -3}f(x) = -5$$
$$\lim_{x \to -3}g(x) = -5$$

The big issue is that f(-3) isn't well defined, even though its limit is.

You have a special criteria for continuity which says that if the limit at a point equals the value of the function at that point then the function mapping is continuous. But in this case, although the limit is defined, the limit and the value of the function at this point are not the same (you get an ill-defined 0/0 for the actual value and not the limit).

Remember the limit is only equal to the value of the mapping in special cases (i.e. in continuous cases across the right interval).