Illusionary singularities in functions

Question

If given the function $$f(x) = \frac{-x^3 + 1}{x^2 - 1}$$ one can clearly see that it is not defined when $x = \pm1$.

We can rewrite the equation by factoring out $(x-1)$ in both the numerator and the denominator $$f(x) = \frac{-(x-1)(x^2+x+1)}{(x+1)(x-1)} = - \frac{x^2 +x + 1}{x+1}$$

Now $f(x)$ is defined for $x = 1$. Why? How can this be? How can a function change properties when simplifying it? Is it only the most simple version of a function that defines its properties? Please help me understand.

Answer 1

When dividing both numerator and denominator with $x-1$ you get a different function. This is because you can divide the numerator and denominator with $x-1$ only if $x-1\ne 0$. These functions are the same when $x-1\ne 0$, but there is no reason to assume that the functions will be the same at $x=1$. In fact one these two functions is not defined at all at $x=1$.

Answer 2

The two expressions are equivalent everywhere they are both defined.

For a simpler example, consider the expressions "$\frac xx$" and "$1$". They're the same, right? Nope. They are the same for all nonzero $x$.