Suppose $g$ is even and let $h=f \circ g$. Is $h$ always an even function? I came across one of the following problems in my homework set: 
$$ \text{Suppose} \, g \, \text{is even and let} \, h=f \circ g. \text{Is} \, h \, \text{always an even function?}$$
I came to the conclusion through examples that "yes" the answer was true. Rather than just random examples, I also tried thinking about in different way. If we have a function $f$ and we compose it with $g$ and perform the even function test then the even function, $g$ will always evaluate to itself and which makes entire function equivalent. For example, let $g$ be $x^2$ and let $f$ be $\sin x$ then $f \circ g$ is $\sin (x^2)$ and $f(x) = f(-x)$ because $\sin (x^2) = \sin ((-x)^2)$.
Is my second approach better than my first? Is there a different and more concrete way to do this? 
Thanks!
 A: Yes you are right. $f\circ g(-x)=f(g(-x))=f(g(x))=f\circ g(x)$ always.
A: Here's my take on a concrete-as-can-be proof of the assertion:
"Let $h$ be the composition $f\circ g$.  Then by the definition of even function, $h$ is an even function if $h(-x) = h(x)$ for all $x$.  Since $g$ is specified as an even function, we have that $g(-x) = g(x)$ for all $x$; therefore, since one property of a function is that $x=y\implies f(x)=f(y)$, by substitution we have $f(g(-x)) = f(g(x))$ for all $x$.  But written in terms of the composition $h$, this is the statement that $h(-x) = h(x)$ for all $x$; in other words, that $h$ is an even function."
This is essentially your argument made concrete; while various steps could be taken to put it into more formal proof syntax, this should be good enough for all intents and purposes.  You can be more concrete than this (e.g., this could be turned into a truly formal proof, one that could be verified by a theorem prover), but it's difficult to really imagine a different approach to the problem that wouldn't be worse; all this one uses are the relevant defining properties of 'function' and 'even function'.
As I mentioned in a comment, I encourage you to try and perform the analagous proof for the assertion that "if $g$ is an odd function and $f$ an arbitrary function, then $h=f\circ g$ is an odd function"; is the statement true or false, and why?  If the statement is false, can you find conditions on $f$ to make it true?  Working through this may give you a better sense for how to use these definitions in formal proofs.
