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We can know whether a function is even or odd by substituting using F(-x) but what if the function has a single hole like this

f(x) = $\frac{x(x-2)}{x-2}$

is such a function considered odd or neither? since substitution in the original function will confirm that it is neither while substitution after simplification will confirm that it is odd

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  • $\begingroup$ Why is $\dfrac {x (x - 2)} {x - 2}$ not odd? Surely $\dfrac {x - 2} {x - 2}$ is always $1$ except where $x = 2$, unless you disallow the term for odd-except-for-a-hole. $\endgroup$ Commented Nov 27, 2021 at 8:20
  • $\begingroup$ Substituting with f(-x) it will be $\frac{-x(-x-2)}{-x-2}$ which is clearly not equal to -f(x) $\endgroup$ Commented Nov 27, 2021 at 8:23
  • $\begingroup$ The hole at the 2 is not reflected across the origin if you got what I mean @PrimeMover $\endgroup$ Commented Nov 27, 2021 at 8:26
  • $\begingroup$ Replace the word "hole" by "pole". $\endgroup$
    – Jean Marie
    Commented Nov 27, 2021 at 9:20
  • $\begingroup$ Wouldn't it be better to replace "hole" by "singularity" as a pole is a special kind of singularity? (In our case it's a removable one.) $\endgroup$ Commented Nov 27, 2021 at 9:59

1 Answer 1

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It depends upon how you defined odd function and even function. Let $D\subset\Bbb R$.

  • If you say that a function $f\colon D\longrightarrow\Bbb R$ is odd if $x\in D\implies-x\in D$ and $f(-x)=-f(x)$ (this would be my definition), then $f$ is not odd (since $-2$ belongs to its domain, but $2$ doesn't).
  • If you say that a function $f\colon D\longrightarrow\Bbb R$ is odd if, whenever both $x$ and $-x$ belong to $D$, then $f(-x)=-f(x)$, then your function is odd.
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  • $\begingroup$ I've edited my answer. Thank you. $\endgroup$ Commented Nov 27, 2021 at 10:01

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