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?

  • $\begingroup$ $f\circ g$ is even if $f(g(-x))=f(g(x))$ and odd if $f(g(-x))=-f(g(x))$ $\endgroup$ Oct 1 at 5:52
  • $\begingroup$ But why exactly is that the case? Because in general a function f is even if f(-x) = f(x) so from here it seems to just be minus the input, so if I have a composite function f(g(x)) isn't that f(-g(x))? minus the input? Because g(x) is the input here? $\endgroup$
    – Tom Miller
    Oct 1 at 6:03
  • 4
    $\begingroup$ $g(x)$ is "the new variable" at an intermediate step of a computation of $f(g(x))$, but the "input" of $f\circ g$ in $(f\circ g)(x)$ is $x$. May be you will see it better if you name $h:=f\circ g$ and write the condition for $h$ to be even. $\endgroup$ Oct 1 at 6:18

1 Answer 1


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.


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