Odd and Even Functions What are odd and even functions? 
An interpretation with an example will be appreciated.
Thanks in advance.
 A: Let $f:A\to\Bbb R$ be defined on a set $A\subset\Bbb R$ s.t $-a\in A\iff a\in A$ ($A$ is symmetric wrt. origin).
We have two conditions $f(-x)=-f(x)$ and $f(-x)=f(x)$. So, $f(x)=0$ for all $x\in A$.
A: Functions who are odd and even have to satisfy two properties
$$(even) \qquad f(x)=f(-x)$$
$$(odd) \qquad f(x)=-f(-x)$$
from which you can deduce that, for all $x$, $f(x)=-f(x)$ which, in most contexts, means that $f(x)=0$. (for example, if you are in the reals, $0$ is the only odd and even function)
A: And odd function is a function $f$ satisfying 
$$\forall x,\quad f(x)=-f(-x).$$
Examples.


*

*$f(x)=\sin(x)$





*

*$f(x)=x^3$



Remark: and odd function $f$ always satisfies $f(0)=0$, and it's graph have a central-symmetry with the origin.
And even function is a function $f$ satisfying 
$$\forall x,\quad f(x)=f(-x).$$
Examples.


*

*$f(x)=\cos(x)$





*

*$f(x)=x^2$



Remark: the graph of an even function is symmetric to the line $x=0$.

Edit.
If the function is odd and even, you can deduce from what I have previously said about the symmetries of the graph that $f(x)=0$ for all $x$.
You can also deduce that from an algebraic point of view using $f(x)=-f(-x)=f(-x)$, so $f(-x)=0$ for all $x$, so $f(x)=0$ everywhere.
