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Ok, so I got an answer wrong on my exam because my teacher says that the function $f(x)=\frac{(x+2)x}{x+2}=x$ but I insist that it isn't defined for x=-2. If it was then $\frac{x}{x}=1$ for all reals and so $\frac{0}{0}=1$. However this doesn't seem to do the trick with my teacher. How can I use the latter fact to prove something outrageous and convince her?

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Ask her what the domains of the functions are. She may be thinking "where both functions are defined." – Ted Shifrin Aug 28 '13 at 0:33
Yes, but did you give your teacher the complete picture on the exam? As in, did you say that $f(x) = x$ except when $x = -2$? Otherwise I think you don't have a case. – Evan Aug 28 '13 at 0:40
More prudent to read the exact text of the exam and the exact text of your answer before excoriating the teacher. – Did Aug 28 '13 at 5:55
What exactly was the question? Was it true/false? What was your answer? How many points did you lose? You are correct that '' $\frac{(x+2)x}{x+2}$ '' and '' $x$ '' are not the same function, because they have different domains (if no more context or information is given). – Stefan Smith Aug 29 '13 at 1:20
up vote 22 down vote accepted

$$\begin{align}\frac00&=\frac{2\times0}{1\times0}=\frac21=2\\\\\\=\frac00&=\frac{3\times0}{1\times0}=\frac31=3\\\\\\\implies2&=3\end{align}$$ NOPE.

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Unless you're a physicist, in which case 2 = 3 for extremely large values of 2 and 3. – David H Aug 28 '13 at 0:44
Being trained as a physicist, I'm pretty sure we won't treat 2 = 3. We usually only treat expression involving $\pi$ as 1. e.g $2\pi = 1$ in quantum physics and $4\pi = 1$ in electrodynamics. – achille hui Aug 28 '13 at 4:12

Here's an old chestnut, customized for your example $$\begin{align} x&=-2\\ x^2&=(-2)^2=4\\ x^2+2x&=4+2x\\ (x+2)x&=2(x+2)\\ \frac{(x+2)x}{x+2}&=2\\ x&=2 &\longleftarrow\text{ OOPS!} \end{align}$$ but we assumed $x=-2$ in the first line, so -2 = 2.

The problem here is going from the line just before to the line labeled OOPS. As you said, you can't conclude $$ \frac{(x+2)x}{x+2}=x $$ when the denominator is zero (i.e., when $x=-2$), though to be fair to your teacher the statement above is indeed true if you include the clause whenever both sides of the equality are defined, which is perhaps what she was thinking.

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On your second line $x^2=4 \implies x=\pm 2$ – user80551 Aug 28 '13 at 9:15
@user80551. Right, but the first line specified that $x$ was $-2$. – Rick Decker Aug 28 '13 at 13:18
Yeah, but you are going a longer way to prove the same thing. The proof that follows is reducible to $x^2=4\implies x=\pm 2$ on the second line. You could just say that $x^2=4\implies x=+2$ using the second line and achieve the same result. – user80551 Aug 28 '13 at 18:31
If $x^2=4$ then, as you noted, either $x=2$ or $x=-2$. Obviously the value of $x$ cannot simultaneously be $2$ and $-2$. We chose $x=-2$ so it cannot be that $x=2$ and so if the subsequent steps yield $x=2$ we must have made an invalid inference somewhere. It happens that all of the inferences are correct except where I "cancelled" the $x+2$ factors to get to the "OOPS!" line. – Rick Decker Aug 29 '13 at 14:36

It depends on how the question was formulated. It is customary in real analysis not to explicitly mention the domain and/or codomain of a function given by a formula. The convention then is that the domain is the largest one for which the formula give well-defined values and the codomain is assumed to be the reals. If that was the case, then it is indeed the case that the function is equal to $x$, meaning for whatever $x$ for which the function is defined, its value is $x$.

If, on the other hand, the question included a domain that contains the number $-2$ then the question is ill-posed and you have a case.

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By convention, we assume that the domain of such functions is the larges subset of $\mathbb{R}$ for which the function is defined. So, the domain of $f(x)=x\dfrac{(x+2)}{(x+2)}$ is $\mathbb{R}\setminus\{-2\}$. Additionally, for all $x\ne -2$, we can simplify the function to $f(x)=x$. But we still have the domain of this function as $\mathbb{R}\setminus \{-2\}$. You may try asking your teacher whether they think that $f(x)=\dfrac{x+2}{x+2}$ is defined at $x=-2$ or not.

Another thing that might help:

Ask them whether they think the function $f(x)=\dfrac{x^5-x^2+3x-3}{x^3+2x^2-3}$ is defined at $x=1$. If they say no, then tell them that the function is equivalent to $\dfrac{(x^4+x^3+x^2+3)(x-1)}{(x^2+3x+3)(x-1)}$ and so from their line of reasoning, the function should equal $\dfrac67$ at $x=1$.

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Division by zero is a domain violation; your function is defined when $x\not = -2$. The domain of the division operator $(a,b)\mapsto a/b$ is the set of all $(a,b)$ so $b \not= 0$. You are done here.

She would have to additionally define $f(-2) = -2$; if you use a case-wise definition, then $f(x) = x$ for all $x$. Otherwise, no mas.

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