For generic $f(x)$, can we determine if $f(x^2)$ is even or odd, and how do we interpret the change from $x$ to $x^2$? This is on problem set 1 on MITOCW's 18.01SC course. The solution says that for a generic function f(x), $f((-x)^2) = f(x^2)$ therefore $f(x^2)$ is even.
I believe I may have misunderstood what $f(x^2)$ means.
My line of reasoning is:
Take for example
$$f(x) = 3x^2$$
$$u = x^2$$
$$f(x^2) = f(u) = 3u$$
$f(u)$ is linear in $u$ and has a domain of $[0, + \infty)$. Therefore, if negative $u$ values are not available, can we say the function $f(u)$ is neither odd nor even?
Of course we have examples where the change of variable generates an even function: $$f(x) = x^4, u = x^2 \Rightarrow f(x) = u^2, \text{ which is even.}$$
 A: We cannot simultaneously have $f(x) = 3x^2$ and $f(u) = 3u$ (unless the domain of $f$ is a subset of $\{0,1\}$), so you're either misunderstanding function notation or you just have a typo.  I assume you meant $f(x^2) = 3x^2$, so that by setting $u = x^2$, we have $f(u) = 3u$.
That's fine, but you're looking at it backwards.  The function $f(u) = 3u$ is the "generic function" which doesn't have to be odd or even or even have a domain on which we can meaningfully talk about odd/even (if we assume $u \in [0,\infty)$)
The function that is necessarily even is the one you started with: $f(x^2) = 3x^2$.  The implied domain for the $x$-values will be symmetric about $0$, so we can always make sense of even/odd symmetry unless the domain of $f$ is trivial (e.g. $\{0\}$)
A: For a general function f(x), if $f(-x)=f(x)$, then it’s an even function. Not sure why your solution says otherwise. In fact, $f((-x)^2)=f(x^2)$ is always true if $x^2$ is in the domain of the function .
In your example, $f(x)=3x^2$ is equivalent to $f(u)=3u^2$. Using a different variable name does not change the function itself, f(u) is still an even function, you have $f(-u)=f(u)$.  If $u=x^2$, you replace u with $x^2, f(x^2)=3x^4$.
Hope this clarifies some confusion.
