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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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    Note $t(s(x))$ is also even.2011-09-10

1 Answers 1

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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.

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    Thank you both Brian M. Scott and Austin Mohr for your help. It is greatly appreciated.2011-09-10