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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?

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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.

Here's an analogy which might help explain the two different viewpoints that lead to two different answers here. Say you're standing on a street with a friend, and are looking at a car. He says "That care looks exactly like mine". You answer "No, it doesn't. The license plates have different numbers on them". Who's correct? Literally speaking, you are - two cars with different license plates obviously don't look exactly the same. But the difference is trivial, and since the license plates will obviously be different since it's not actually his car, it's clear that he was referring to everything else being equal.

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  • $\begingroup$ So, are there several correct answers? $\endgroup$ – Andrey Yankin Dec 5 '13 at 14:14
  • $\begingroup$ Nitpicking: you should $1/(x-6)$ not call $f$. $\endgroup$ – Michael Hoppe Dec 5 '13 at 14:18
  • $\begingroup$ And there is only one answer: the $f\colon\boldsymbol{R}\setminus $\endgroup$ – Michael Hoppe Dec 5 '13 at 14:20
  • $\begingroup$ @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. $\endgroup$ – Daniel Fischer Dec 5 '13 at 14:20
  • $\begingroup$ @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... $\endgroup$ – fgp Dec 5 '13 at 14:21
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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.

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  • $\begingroup$ Yes. My set is larger. And function is defined on -4. What's wrong with my answer? $\endgroup$ – Andrey Yankin Dec 5 '13 at 14:16
  • $\begingroup$ But the expression is not defined in $-4$ since the denominator becomes $0$ and the fraction $x/y$ is not defined when $y=0$. $\endgroup$ – Emanuele Paolini Dec 5 '13 at 14:20
  • $\begingroup$ So, we can only find domain for the expression but not for the function? $\endgroup$ – Andrey Yankin Dec 5 '13 at 14:24
  • $\begingroup$ @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 $\endgroup$ – Hakim Jul 19 '14 at 0:19
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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

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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 $x=-4$.

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