A function is even $\iff$ $f(-x) = f(x)$ and odd $\iff f(-x) = -f(x)$.
If I have some function $f$ that is even and some function $g$ that is even, their composition is $f(g(x))$, right?
When I'm trying to find if this function is even or odd, isn't $g(x)$ technically the $x$ here? so, wouldn't I check $f(-g(x))$, or am I misunderstanding something?
Because I feel like I am, since I am getting different answers than I'm supposed to. But I'm not sure why exactly this isn't okay, because $g(x)$ is just the new variable in this case, is it not?
If $g$ is odd and $f$ is even, then $(f\circ g)(x)$ is even.
$f(g(-x)) = f(-g(x)) = f(g(x))$
If $g$ is even, regardless of the function $f$, $(f\circ g)(x)$ is even.
$f(g(-x)) = f(g(x))$
If $f$ and $g$ are both odd, $(f\circ g)(x)$ is odd.
$f(g(-x)) = f(-g(x)) = -f(g(x))$
When in doubt, consider $f(x) = x$ as a simple example of an odd function and $f(x) = x^2$ as an example of an even function. While you can't prove a proposition is true by example, you can prove a proposition is false, and examples help build intuition.