Is it possible for the sum of even and neither odd nor even function to be even or odd? Consider the functions $f(x)$, $g(x)$, $h(x)$, where $f(x)$ is neither odd nor even, $g(x)$ is even and $h(x)$ is odd. Is it possible for  $f(x) + g(x)$ to be 


even;
odd?


For the second case I can imagine for example $f(x) = x - 1$ and $g(x) = 1$. Then $f$ is neither even nor odd and $g$ is even but their sum is odd, hence it's possible to get odd function from the sum of neither odd nor even and even function.

It feels like $f(x) + g(x)$ can never be even, but I couldn't manage to prove that.

I've tried to do it the following way:
Let $f(x) = - g(x) - h(x)$, which doesn't contradict the initial statement. Then we can express $g(x)$ and $h(x)$ and see whether the facts that they are either even or odd holds, but this always leads to valid equations:

$$
h(x) = \frac{f(-x) - f(x)}{2} \;\;\; \text{is an odd function} \\
g(x) = \frac{-f(x) - f(-x)}{2} \;\;\; \text{is an even function}
$$

I'm stuck at that point.


  How can I prove/disprove that $f(x) + g(x)$ may be even?
If $f+g$ is even, then $(f+g)-g$ is the difference of even functions, hence even
f(x) = f(-x)

$g(x) \ne g(-x)$ for some x

z(x) = f(x) + g(x)

then $z(-x) =  f(-x) + g(-x) \ne f(x) + g(x) $ for some x