Is one form of a function more 'true' than another? Here's a function: $f(x) = \frac{x^2}{x}$
Now, if we were to look for the $0$ value, we would end up with a division by zero situation. 
By simplifying it to an equivalent function: $f(x) = x$, this is no longer a problem.
Does this mean that the latter function is, in a sense, a more 'true' form of the function? Do both of these forms have the same domain? 
 A: The function $f(x) = \dfrac{x^2}{x}$ has largest domain $\mathbb{R}\setminus\{0\}$ whereas the function $g(x) = x$ has largest domain $\mathbb{R}$. For all $x \neq 0$, $\dfrac{x^2}{x} = x$; as $g|_{\mathbb{R}\setminus\{0\}} = f$ we can say that $g$ is an extension of $f$. But for any $a \in \mathbb{R}$,
$$h_a(x) = \begin{cases}
x &\ \text{if}\ x\neq 0\\
a &\ \text{if}\ x = 0
\end{cases}$$
is also an extension of $f$. However, there is only one way to extend $f$ to a continuous function, and that extension is $h_0 = g$.
In the context of your question, $f(x) = \dfrac{x^2}{x}$ and $g(x) = x$ are not 'equivalent functions' unless we take as understood that $g$ has domain $\mathbb{R}\setminus\{0\}$. The function $g$ is not a 'truer' form of $f$, but rather its unique continuous extension. 
You may think that this is incredibly pedantic; why don't we automatically extend functions continuously, just like we automatically take the largest domain of a function if it is not specified? One reason is that a continuous extension doesn't always exist. For example, $f(x) = \dfrac{1}{x}$ has largest domain $\mathbb{R}\setminus\{0\}$ but has no continuous extension defined on $\mathbb{R}$.
A: No. The domain of the $f(x)$  you gave does NOT include $0$.
A: $\frac{x^2}{x}$ and $x$ are equal as polynomials, but not as functions.
A: Good question.
Firstly, let me say that the concept of an "implicit domain" is one of my pet peeves! The day we universally reject such antiquated conventions will be a glorious day for mathematics.
Thus, my answer is that we should never define a function with just with a formula; always define it by specifying a particular domain and codomain.
For example:

Let $f$ denote the unique function $\mathbb{R}\setminus\{0\} \rightarrow \mathbb{R}$ subject to the following condition. $$f(x) = \frac{x^2}{x}.$$

This is a (not-at-all ambiguous, and therefore perfectly acceptable) shorthand for the following, more long-winded formalization.

Let $f$ denote the unique function $\mathbb{R}\setminus\{0\} \rightarrow \mathbb{R}$ such that for all $x \in \mathbb{R}\setminus\{0\}$, we have the following. $$f(x) = \frac{x^2}{x}.$$

Now lets focus our attention on the expression $x$. There's (at least!) two ways of defining a function using this formula. One way defines a function $g : \mathbb{R} \setminus \{0\} \rightarrow \mathbb{R}$, the other a function $h : \mathbb{R} \rightarrow \mathbb{R}$. These are different functions. The first coincides with $f$; that is, $g=f$. The second does not.
That's basically all there is to it.
Actually, there's one more thing that should really be addressed. Both $f$ and $g$ (they're the same function, after all) can be extended in a unique manner to a continuous function $\mathbb{R} \rightarrow \mathbb{R},$ and this function turns out to be $h$.
