Domain, codomain, and range

This question isn't typically associated with the level of math that I'm about to talk about, but I'm asking it because I'm also doing a separate math class where these terms are relevant. I just want to make sure I understand them because I think I may end up getting answers wrong when I'm over thinking things.

In my first level calculus class, we're now talking about critical values and monotonic functions. In one example, the prof showed us how to find the critical values of a function $$f(x)=\frac{x^2}{x-1}$$ He said we have to find the values where $f' (x)=0$ and where $f'(x)$ is undefined.$$f'(x)=\frac{x^2-2x}{(x-1)^2}$$

Clearly, $f'(x)$ is undefined at $x=1$, but he says that $x=1$ is not in the domain of $f(x)$, so therefore $x=1$ is not a critical value. Here's where my question comes in:

Isn't the "domain" of $f(x)$ $\mathbb{R}$, or $(-\infty,\infty)$? If my understanding of Domain, Codomain, and Range is correct, then wouldn't it be the "range" that excludes $x=1$?

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If $f:X\to Y$ is a function from $X$ to $Y$, we usually call $X$ the domain, $Y$ the codomain, and $f(X)$ the range. Some authors call $Y$ the range, in which case $f(X)$ is called the image.

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I'm still confused - does this mean then that it's correct to say that x=1 is not in the domain, as opposed to not in the range? My mental image of domain and range has been totally messed up thanks to my introduction to the formal definition of functions. –  agent154 Nov 11 '12 at 1:04
Oh, ok.. I think I may understand now. If we want to get fussy, then $1/x$ is not really a function, because it is undefined at some points in the domain... thus if we exclude 0 from the domain, $1/x$ becomes a function. For this reason it is implicitly understood that 0 is not in the domain? Otherwise, would this be called a partial function? –  agent154 Nov 11 '12 at 1:07
@agent154: Yes, the function $f(x)=\frac1x$ is a partial function from $\Bbb R$ to $\Bbb R$ and a function from $\Bbb R\setminus\{0\}$ to $\Bbb R$. In the context of your calculus class it is indeed understood that $0$ is excluded from the domain. –  Brian M. Scott Nov 11 '12 at 1:13

$x=1$ is not in the domain because when $x=1$, $f(x)$ is undefined. And by definition, strictly speaking, a function defined on a domain $X$ maps every element in the domain to one and only element in the codomain.

The domain and codomain of a function depend upon the set on which $f$ is defined and the set to which elements of the domain are being mapped; both are usually made explicit by including the notation $f: X \to Y$, e.g. along with defining $f(x)$ for $x\in X$.

$X$ is then taken to be the domain of $f$, and $Y$ the codomain of $f$, though you'll find that some people interchange the terms "codomain" and "range". So "range" is a bit ambiguous, depending on the text used and how it is defined, because "range" is sometimes defined to be the set of all values $y$ such that there is some $x \in X$ for which it is true that $f(x) = y$, i.e. $f[X]$.

One way to circumvent any ambiguity related to use of "range" to refer to $f[X]$ is to note that many prefer to define $f[X]$ to be the "image" of $X$ under $f$, often denoted by $\text{Im}f(x)$, with the understanding that $f[X] = \text{Im}f(x) \subseteq Y.\;\; f[X]=\text{Im}f(x) = Y$ when $f$ is onto $Y$.

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Strictly speaking, a function is a subset $f$ of $(\hbox{domain}) \times (\hbox{codomain})$ such that if $(a, b), (a, c) \in f$, then $b = c$. That is, a specification of the domain and codomain are part of the definition of a function. But this is usually considered too technical for a calculus course, which is what the original poster was talking about, so ...
... in the typical (American) calculus course, you consider real-valued functions defined on a subset of the reals, where a "function" is given by an equation like "$f(x) = \dfrac{x^2}{x - 1}$'', as in the original post. Then the convention is that "domain" means "the largest subset of the reals on which the expression in the defining equation is defined", unless there is a statement to the contrary. (I've heard this referred to as the "natural domain".) Thus, for this $f$, the (natural) domain by this convention is $\{x \mid x \ne 1\}$. And so $x = 1$ is not a critical point, since it's not in the "domain".
If for some reason I want a smaller subset, I must say so explicitly. For example, I might say "the function $f(x) = x^2$ defined on $[0, \infty)$".
Thus, in (calculus, precalculus, college algebra) textbooks you will see questions like: "What is the domain of $f(x) = \dfrac{1}{x}$?" By the strict definition of a function, the question makes no sense, since you haven't defined the function unless you've stated what the domain is up front. But by the convention described above, the domain is the nonzero reals.