Is continuity characteristic required for two equations to be equal The function below is discontinuous:
$$
f(x)=\frac{x^2-9}{x-3}
$$
But when it is further simplified, it becomes continuous:
$$
g(x)=x+3
$$
First question, is $f(x) = g(x)$ ? Second question, if presented with $f(x)$, is it mathematically incorrect to further simplify it?
 A: Short answer: The function $g$ is a continuous extension of $f$.
I will assume that the domain $D$ satisfies $3\in D$. The functions are not equal at $3$, $f(3)\neq g(3)$, since $f(3)$ is not defined. However $f$ has a removable discontinuity. In this case look at the Riemann extension theorem:

Let $D\subset \mathbb C$ be an open subset of the complex plane, $a\in D$ a point of $D$ and $f(a)$ holomorphic function defined on the set $D\setminus \{a\}$. The following are equivalent:

*

*$f$ is holomorphically extendable over $a$.

*$f$ is continuously extendable over $a$

*There exists a neighborhood of $a$ on which $f$ is bounded.

*$\lim\limits_{z\to a}(z - a) f(z) = 0$

We will trivially prove that our function meets the fourth condition:
$$\lim\limits_{x\to 3}\,(x - 3) \frac{x^2-9}{x-3}=\lim\limits_{x\to 3}x^2-9=3^2-9=0.$$
Now by the Riemann extension theorem we can conclude that $f$ is continuously (even holomorphically) extendable over $3$. And this extension is $g(x)=x+3$.
A: If the domain is the set of real numbers, no, $f(x) \neq g(x)$. Even though $f$ simplifies to $g$, $f$ is a rational function, and $g$ is a linear function. Because $(x - 3)$ is a factor on both numerator and denominator, $x = 3$ does not belong on the domain of $f$, and it does on the domain of $g$.
If you are graphing $f$ and $g$, they will look almost the same, but for $f$, $x = 3$ is a hole.
This is called a removable discontinuity. See this page from WolframAlpha for more information.
A: First question. For which $x$? We have something like
$$f : \mathbb R\setminus \{3\} \to \mathbb R\quad\mbox{and}\quad g:\mathbb R\to \mathbb R $$
If you ask if $f(x) = g(x)$ for every $x\in\mathbb R$, then that is false, because $f(3)$ is not defined. You can define $f(3)$ such that the extension is continuous. Just because there is a canonical way to extend $f$, does not mean you can't define $f(3) = 1000$, for instance.
Second question. Not incorrect. By definition of $f$ the quantity $x-3$ is always non-zero. Thus we can simplify.
Title question. Yes, two functions are equal if and only if they have the same domain $X$, same codomain and $f(x) = g(x)$ for every $x\in X$. Thus, if one of them is continuous, so must the other.
The functions $f$ and $g$ are equivalent, because they are equal almost everywhere. But this equivalence need not maintain continuity.
A: for the first question :
$\frac{x^2-9}{x-3}$=$x+3$
$x^2-9$ = $(x+3)(x-3)$
$x^2-9$= $x^2-9$
$f(x)$=$g(x)$ $∀ x≠ 3$
