Solving $A(x) = 2A(x/2) + x^2$ Using Generating Functions Suppose I have the recurrence:
$$A(x) = 2A(x/2) + x^2$$ with $A(1) = 1$.
Is it possible to derive a function using Generating Functions? I know in Generatingfunctionology they shows show to solve for recurrences like $A(x) = 2A(x-1) + x$. But is it possible to solve for the above recurrence as well?

I am a little confused by the way you worded this question (it seems that you have a functional equation rather than a recurrence relation), so I interpreted it in the only way that I could make sense of it. If this is not what you are looking for, then please clarify in your original question or in a comment.
Let's assume that $A(x)$ is a formal power (or possibly Laurent) series, $A(x) = \sum_n a_n x^n$. Plugging this into your equation, we get
$$
\sum_n a_n x^n = 2 \sum_n a_n \frac{x^n}{2^n} + x^2
$$
For $n\neq 2$, we get
$a_n = 2^{1-n} a_n$, so if $n \neq 1,2$ we get $a_n = 0$. For $n=2$, we get
$a_2 = a_2/2 + 1$, so $a_2 = 2$. Finally, the condition $A(1) = 1$ gives $a_1 = -1$, so we have
$$
A(x) = -x + 2x^2
$$
Check:
$$
2 A(x/2) + x^2 = 2( -x/2 + x^2/2) + x^2 = -x + 2x^2 = A(x)
$$
