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The title pretty much explains it:

Given that $s(x)$ is even and $t(x)$ is odd, and both are defined on the real line $\mathbb R$, is $s(t(x))$ even or odd?

Some things I found out on my own:
I found that $s(x) t(x)$ is odd because:

$$st(-x) = s(-x) \cdot t(-x)$$

$$st(-x) = s(x) \cdot (-t(x))$$ (This is because for an even function, $f(-x) = f(x)$, and for an odd function, $f(-x) = -f(x)$.)

$$st(-x) = -st(x)$$

Therefore, by definition, $s(x) t(x)$ is odd.

I still can't figure out $s(t(x))$ though...

Any help is appreciated.
Thank you in advanced.

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Take it in steps. You want to know whether $s(t(-x))$ is always $s(t(x))$ ($s\circ t$ is even), always $-s(t(x))$ ($s\circ t$ is odd), or neither. $t$ is odd, so $t(-x)=-t(x)$; and $s$ is even, so $s(-t(x)) =$? If it helps, call $t(x)$ $y$ for short; $s(-t(x))=s(-y)=$? – Brian M. Scott Sep 10 '11 at 23:05
I would believe even. Substituting $-t(x)$ in your previous statement for $u$, for the sake of argument, $s(-u)$ = $s(u)$, since $s$ is even. Substituting back in, $s(t(-x)) = (s(t(x))$, making it even, correct? – Mike Gates Sep 10 '11 at 23:10
Note $t(s(x))$ is also even. – Henry Sep 10 '11 at 23:30
up vote 5 down vote accepted

Your definitions for even and odd are both correct. The next step is to "plug and chug" for $s(t(x))$.

$$ \begin{align*} s(t(-x)) &= s(-t(x)) \text{ (since t is odd)}\\ &= s(t(x)) \text{ (since s is even)}. \end{align*} $$

So, $s(t(-x)) = s(t(x))$, which means $s(t(x))$ is even.

share|cite|improve this answer
Thank you both Brian M. Scott and Austin Mohr for your help. It is greatly appreciated. – Mike Gates Sep 10 '11 at 23:16

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