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First of all we know, $f(x)=sin(x)$ is an odd function because $f(x)=-f(-x)$

The question is if there is a discontinuity (single , interval ,..etc).

I will give you some examples to get what I mean:

ex1: $f(x)=sin(x)\times\frac{x^2-4}{x^2-4} :x\neq\pm2$

ex2: $f(x)=sin(x)\times\frac{x-1}{x-1} :x\neq1$

ex3: $f(x)=sin(x)\times\frac{(x-1)(x+2)}{(x-1)(x+2)} :x\neq1 ,x\neq-2$

ex4: $f(x)=sin(x)\times\frac{\sqrt{x^2-1}}{\sqrt{x^2-1}} :x\gt1$

I feel abit confused because I cant say $f(x)=-f(-x) $ for all $x$, but intuitively it feels ok. is that definition not accurate? is there is something I am missing up?

Mina Eskandar
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1 Answers1

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A function $f$ is odd if whenever both $x$ and $-x$ are in its domain $$ f(x) = -f(-x). $$

We don't usually include the italicized part of the previous definition since it's not usually an issue.

The formal definition of a function requires specifying its domain. Think about the domains of the functions you are trying to define with your formulas.

Ethan Bolker
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  • Interesting to note a function whose domain is limited to only the positive reals (for example $f:\mathbb R^+\to \mathbb R^+; x\mapsto \sqrt x$) is vacuously defined to be both even and odd. Likewise if we do something fairly artificial such as take a function, $f$ that naturally isn't either odd but restrict it to a domain of points only where $f(-x)=-f(x)$ the resulting restricted function was contrived to be odd. For example $f:{-3,3}\to {-6,6}; x\mapsto x^2 + 2x -9$ would be odd even though $f:\mathbb R\to \mathbb R; x\mapsto x^2 + 2x - 9$ clearly is not. – fleablood Dec 19 '21 at 17:08
  • So, If iam understanding correctly, ex2 & ex3 are not odd in their domain – Mina Eskandar Dec 19 '21 at 17:08
  • No. All those examples are odd. As for all $x$ where $x$ and $-x$ is in the domain we have $f(-x) = f(x)$. Ex 2 is odd because for all if $x$ and $-x$ are both in the domain then $x \ne 1$ and $-x \ne 1$ so $x \ne \pm 1$. And if $x \ne \pm 1$ then we have $f(-x) = -f(x)$. So it is odd. – fleablood Dec 19 '21 at 17:11
  • ...cont... it doesn't matter in ex2 that $f(-1) \ne -f(1)$ because $-f(1)$ is not defined and it's meaningless to talk about whether anything does or does not equal $-f(1)$ because $-f(1)$ is not a meaningful entity. – fleablood Dec 19 '21 at 17:21
  • Not everybody agrees with this definition. It's quite common to require the domain to be symmetric with respect to the origin. See here, for example: https://math.stackexchange.com/questions/748368/odd-and-even-functions. – Hans Lundmark Dec 19 '21 at 18:36
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    @HansLundmark Either definition is acceptable as long as context makes clear which is meant. The one I gave is the one in wikipedia and other reliable places. – Ethan Bolker Dec 19 '21 at 19:08