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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.

Martin Sleziak
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user347002
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  • This apparent issue is readily dismissed if you factor in the concept of quantification. Do you know about quantifiers? – Git Gud Jun 10 '16 at 19:38
  • Pedants and teachers get excited about the function being different after factoring out the $x-1$. But for all practical purposes it is the same. If a function has a removable discontinuity (ie if you can define it to have a value equal to the limit there and get a continuous function), then most people just regard it as having that value. There is no good reason for not taking $f(1)$ to be $-\frac{3}{2}$. – almagest Jun 10 '16 at 19:42
  • I'm afraid that the OP may not know what the word "(dis)continuity" mean. OP: If you draw graphs of both functions, they will be the same graphs except of one point ($x=1$). You are advised to count graphs differing in a finite number of points (this time just one point) as if they would be the same – porton Jun 10 '16 at 19:44
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    You can only factor out non-zero terms. (x-1) is not non-zero at (x-1) so you can not factor out (x-1) at x=1. The function resulting is a different function entirely and is not merely a "simplification". – fleablood Jun 10 '16 at 19:51
  • There are mainly three types of singularities, cf. https://en.m.wikipedia.org/wiki/Isolated_singularity – Michael Hoppe Jun 10 '16 at 19:55
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    This question appears to be a duplicate of "Why are removable discontinuities even discontinuities at all?". – Will R Jun 10 '16 at 20:36

2 Answers2

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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 $x=1$.

porton
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  • this is the real analysis point of view. in complex analysis, it is more subttle, in most cases it is more convenient to consider analytic continuation of functions instead – reuns Jun 11 '16 at 05:51
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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 $x$.

MPW
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