Antiderivatives, working backwards. Sorry if my title is a bit confusing. I have the question:
A function f defined for all real numbers satisfies: $f'(1) = 1$, $f(0) = 4$, and $f''(x) = 12x^2-12x$. Find the value of $f(2)$.
I left out some unimportant parts which I should be capable of doing myself after I understand this bit, my question is am I doing this correctly?

$$f'(x) = 4x^3-6x^2 + C$$
$$4(1)^3-6(1)^2 + C = 1$$
$$4-6+C = 1$$
$$C = 3$$

$$f(x) = x^4-2x^3 +C$$
$$f(0) = 0^4-2(0)^3+C = 4$$
$$C= 4$$

$$f(2) = 2^4-2(4)^3=C$$
$$16-2(64)=C$$
$$16-128=C$$
$$-112=C$$

Is this what I am being asked to do, or am I going about this the wrong way?

Your first step (finding $f^\prime(x)$) looks good, but from there you miss a few things.  After finding the first $C$, you have 
$$
f^\prime(x)=4x^3-6x^2+3
$$   Thus, when we integrate, we have 
$$
f(x)=x^4-2x^{3}+3x+C
$$  where here $C$ is a different constant than the first one.  Now we use $f(0)=4$ to find this $C$, after which you can plug in $x=2$.
