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The function $f(x) = x$ is point symmetric. But what's with $g(x) = (x^2 - x)/(x - 1)$?
$\mathbb{D}_g = \mathbb{R} \backslash \{1\}$
$g(x) = -g(-x)$ is true for every $x \in \mathbb{R} \backslash \{-1;1\}$ but not for $x = -1$ or $x = 1$ because $g(1)$ is not defined.
Is this a reason to say that $g$ is not point symmetric or is the singularity ignored?

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1 Answer 1

up vote 1 down vote accepted

What you call "point symmetric" is commonly referred to as "odd", e.g. $f$ is an odd function.

And yes, $g$ is not odd because $g(x)=-g(-x)$ is not true for all $x$ in the domain of $g$ even though $f$ is odd and agrees with $g$ on its domain.

And though we can extend $g$ continuously to all of, say, $\mathbb R$ such that it equals $f$, and we have that $f=g$ as fractions of polynomials, this still doesn't make $g$ an odd function.

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