The next few questions ask about the symmetry properties of combinations of symmetric functions. For example, if $f$ and $g$ are both odd, then $f+g$ is also odd. You can see this as follows. We know that $f(x) = -f(-x)$ and $g(x)= - g(-x)$, since $f$ and $g$ are both odd. For the sum $h=f+g$ we get: $$\begin{array}{rcl} h(x) &=& (f+g)(x) \\ &=& f(x)+g(x) \\ &=& -f(-x)-g(-x) \\ &=& -(f+g)(-x) \\ &=&-h(-x),\\ \end{array}$$ and hence the sum is odd as well.

You can see similarly that if $f$ and $g$ are both even, then so is $f+g$.

On the other hand, if $f$ is even and $g$ is odd then usually (i.e., always, unless $f$ or $g$ is zero) $f+g$ will be neither even nor odd, as illustrated, for example by $f(x) =1$ and $g(x) = x$.

Note that this is counterintuitive: The sum of two odd integers is even, and the sum of and even and an odd integer is odd. In those two cases functions are unlike integers. On the other hand, the sum of two even integers is even, as is the sum of two even functions.

As always, you don't want to memorize rules like that. You don't even need to go through the above general argument each time you need a rule. Instead, just consider an example. For example, clearly $x+x^3$ is odd, $x^2+x^4$ is even, and $x+1$ is neither even nor odd. Those examples generalize, but if you are ever in doubt you can always revert to checking the actual definition, as above.

What happens when we subtract instead of add $f$ and $g$? Let $$h=f-g.$$ Then

$h$ is if $f$ and $g$ are both even,

$h$ is if $f$ is even and $g$ is odd, and

$h$ is if $f$ and $g$ are both odd.

(As before, and in the following problems, enter E if the function is even, O if it is odd, and N if it is neither even nor odd.)