The function $$ f(x) = \frac{x^2-1}{x-1} $$ is not defined at x = 1. Even if I simplify the function, I need to keep in mind where I've came from: $$f(x) = x + 1 , x \neq 1$$ But if the original function looks like f(x) = x + 1, then there is no problem if I first expand the fraction so that x - 1 is divisor, and then again simplifying the divisor away.
Why is there a difference?
The function $$g(x) = \frac{\sqrt{x}-3}{x-9}$$ is defined for x = 9 because you can simplify the function; g(9) = 1/6. I get that (x - 1) is here the last factor with x in it, but I don't see why that should mean we treat the functions differently, as to their defined domain of x values.
Would anyone care to explain this to me?
The function $f(x)=\frac{x^2-1}{x-1}$ is not defined at $1$ since $\frac00$ is not defined. Thus we assume the domain of $f$ is $\Bbb R\setminus\{1\}$. The domain of the function is part of the function's definition, and never changes. Even if you simplify $f(x)=x+1$, that statement is still only true on the domain of $f$, which is $\Bbb R\setminus\{1\}$.
The function $h(x)=x+1$ is defined on all $\Bbb R$, so we assume this is the domain of $h$. The statement $h(x)=\frac{x^2-1}{x-1}$ is true whenever $x\neq 1$, but it is not true when $x=1$. Now the fact that $h(x)=\frac{x^2-1}{x-1}$ when $x\neq 1$ by itself does not imply $h(1)=2$. However, this is true regardless, since we begin with the definition $h(x)=x+1$.
The function $g(x)=\frac{\sqrt x-3}{x-9}$ has domain $\Bbb R\setminus \{9\}$. It is also true that $g(x)=\frac{1}{\sqrt x+3}$ whenever $x\neq 9$, but this doesn't change the fact that $9$ is not in the domain of $g$.
Really, we should acknowledge that "functions" are usually not defined rigorously in lower level math classes. This is mostly just to avoid confusion, since people already have a hard time understanding functions. Formally, a function $f$ is the data of
If a function is only defined with the equation $f(x)=\frac{1}{x-1}$, for example, then we make some reasonable assumptions about what $D$ and $C$ are. We can assume $C=\Bbb R \setminus \{0\}$, and assume that $D\subset \Bbb R$ is the set of all $x$ for which the equation $f(x)=\frac{1}{x-1}$ is defined.