# Function domain and simplification: how to combine them?

What is the domain of the real-valued function:

$$f(x) = \frac{x+4}{(x+4)(x-6)}$$

Wolfram|Alpha says that:

$$\{ x \in \mathbb{R} : x \ne -4, x \ne 6 \}$$

I believe it should be more like this:

$$\{ x \in \mathbb{R} : x \ne 6 \}$$

I couldn't find any explanation of this on the Internet. Please, could you give some references on why the first answer should be considered as correct. Maybe I just do not understand what the function domain is?

Update

It seems that to understand explanations I should first see true mathematical rigorous definitions of function, expression, domain, etc on this topic. Could you include any of these in your answers, please?

That's mostly a matter of definition. Literally, $$f(x) = \frac{x+4}{(x+4)(x-6)}$$ is not defined for $x=-4$, because in general $\frac{0}{0}$ is undefined. You can, however, easily extend $f$ from $\mathbb{R}\setminus\{-4,6\}$ to $\mathbb{R}\setminus\{6\}$ while keeping properties like continuity and the like intact.
That extension is, in fact, so simple in the case of your $f$ that most people will automatically do it in their head, i.e. will cancel the troublesome $x+4$ terms, and thus actually work with $$f(x) = \frac{1}{x-6}$$ which of course is defined for $x=-4$.
Still, literally speaking, Wolfram Alpha is correct to say that your $f$ is undefined for both $x=-4$ and $x=6$. Even though that undefined-ness is pretty artificial.
• Nitpicking: you should $1/(x-6)$ not call $f$. – Michael Hoppe Dec 5 '13 at 14:18
• And there is only one answer: the $f\colon\boldsymbol{R}\setminus – Michael Hoppe Dec 5 '13 at 14:20 • @MichaelHoppe Why not? If fgp views it as a rational function (the most reasonable way to view it), we have $$\frac{x+4}{(x+4)(x-6)} = \frac{1}{x-6}$$ by defintion of rational functions. – Daniel Fischer Dec 5 '13 at 14:20 • @MichaelHoppe I don't think I did. What I wrote was that most people will read the definition as$f(x) = \frac{1}{x-6}$. That's not the same thing as me saying that they are equal - they're just often treated as equal. But it depends on the context of course... – fgp Dec 5 '13 at 14:21 When not specified elsewhere, the domain of a function given as an expression, is the largest set in which the expression makes sense. If the expression contains a fraction, the denominator must be different from zero. • Yes. My set is larger. And function is defined on -4. What's wrong with my answer? – Andrey Yankin Dec 5 '13 at 14:16 • But the expression is not defined in$-4$since the denominator becomes$0$and the fraction$x/y$is not defined when$y=0$. – Emanuele Paolini Dec 5 '13 at 14:20 • So, we can only find domain for the expression but not for the function? – Andrey Yankin Dec 5 '13 at 14:24 • @AndreyYankin You are confusing between an “expression” and a “function”. First of all functions are rules, they map one number$x$to another (which we denote by$f(x)$) according to that rule,$f$. We write$f:x\mapsto f(x)$. Some functions can be described by expressions, like yours$f:x\mapsto \tfrac{x+4}{(x+4)(x-6)}$. However, an expression doesn't have a domain. A domain is only a characteristic of a function, which we determine by collecting all the values for which the expression makes sense, i.e. for which the expression is defined, like when we have something of the form$x/0$... etc – Hakim Jul 19 '14 at 0:19 The ratio$\frac{0}{0}$can be found only as a limit. For$x=-4$you have the indeterminate form$\frac{0}{0\dot{}(-10)}$wich is unacceptable It stems from the difference between the functions$f(x)=\frac{1}{x-6}$and$f(x)=\frac{x+4}{(x+4)(x-6)}$. Notice that the former is defined on all values of the real line except at$x=6$, but the latter is not defined on$x=6$as well as$x=-4$. The graph of the second function will be almost the same as the graph of the first function, except for the fact that it contains a "hole" (undefined point) at$x=-4\$.