Can there be a function that's even and odd at the same time? I woke up this morning and had this question in mind. Just curious if such function can exist.
 A: Others have mentioned that $f(x)=0$ is an example. In fact, we can prove that it is the only example of a function from $\mathbb{R}\to \mathbb{R}$ (i.e a function which takes in real values and outputs real values) that is both odd and even. Suppose $f(x)$ is any function which is both odd and even. Then $f(-x) = -f(x)$ by odd-ness, and $f(-x)=f(x)$ by even-ness. Thus $-f(x) = f(x)$, so $f(x)=0.$
A: Suppose  $f$   odd  an  even.  Let  $x \in D$   ( D  is  set  definition  of  $f$)  then  you  have  :  $  f(x)=f(-x)=-f(x)$.  What can you   conclude about $f$ ?
A: If $K$ is a field of characteristic 2, every function $K\to K$ is both even and odd.
A: As other people have mentioned already, the real function $f(x)$ which maps every real number to zero (i.e.$f(x) = 0 \space \forall x \in \mathbb{R}$) is both even and odd because  $$f(x) - f(-x) = 0 \space \space ,  f(x)+f(-x) = 0\space \forall x \in \mathbb{R} .$$ Also it is the only function defined over $\mathbb{R}$
to possess this property.
A: Let $R$ be a Boolean ring and $X$ be an arbitrary set. Then every function $f:R\rightarrow X$ is both even and odd.  
A: I post this as an addendum to the awesome answers already present for this Q.
Most trivial example:
For $f(x)=0$ we have:
$f(x)= 0 =-f(-x)$ Hence, Odd
$f(x)= 0 =f(-x)$ Hence, Even
I was searching for an answer to my question.
$f(x) = \sin(x) + \sin(\pi + x)$
Here, the function is also even and odd at the same time (another example or representation of the same). This is due to the value resulting in zero over the entire domain. So, it can be concluded that all the functions that have their $Range=\{0\}$ should be both even and odd at the same time despite their notation is of a constant function($f(x)=0$) or not.
Another example,
$f:\{1,-1\}\to \mathbb{R}$
$f(x) = x^{2}-1$
$f(x) = 0$ for all values in the domain. So, it is both even and odd at the same time because while deciding even odd functions the domain to which the function is restricted by definition must be considered.
A: Yes. The constant function $f(x) = 0$ satisfies both conditions.
Even:
$$
f(-x) = 0 = f(x)
$$
Odd:
$$
f(-x) = 0 = -f(x)
$$
Furthermore, it's the only real function that satisfies both conditions:
$$
f(-x) = f(x) = -f(x) \Rightarrow 2f(x) = 0 \Rightarrow f(x) = 0
$$
A: Hint $\rm\ f\:$ is even and odd $\rm\iff f(x) = f(-x) = -f(x)\:\Rightarrow\: 2\,f(x) = 0.\:$ This is true if $\rm\:f = 0,\:$ but may also have other solutions, e.g. $\rm\:f = n\:$ in $\rm\:\mathbb Z/2n =\:$ integers mod $\rm 2n,$ where $\rm\: -n \equiv n.$
A: We could also think even and odd functions as the following to get $f(x) = 0$ is a function that is both odd and even and go for some different functions:
Odd functions have graphs that are symmetric with respect to origin.
Even functions have graphs that are symmetric with respect to $y$-axis.
So, graph of $f(x) = 0$ satisfies both conditions, therefore it is both odd and even. Furthermore, we can define some piecewise functions that satisfies this condition using their graphs. For example, let $f(x)$ not defined on intervals $(-1,-2)$ and $(1,2)$; and whenever it is defined, $f(x) = 0$. Then $f(x)$ is, again, both even and odd. And as long as we define a piecewise function by removing intervals that are symmetric with respect to $y$-axis, and rest of the function $= 0$ even though they are not continuous.
