The function $f(x) = x^3-4a^2x$ has primitive function $F(x)$ . Find the constant $a$ so that $F(x)$ has the minimum value $0$ and $F(2)=4$. 
The function $f(x) = x^3-4a^2x$ has primitive function $F(x)$ . Find the constant $a$ so that $F(x)$ has the minimum value $0$ and $F(2)=4$.

This is what I tried below.
I am not sure how to use the minimum value information for the $F(x)$ function.
I tried to find $c$ in terms of $a$ to help simplify the problem, but I don't think it helps.
I think I might have to set up some sort of system of equations with the two pieces of information.


Following your work with $F(2) = 4 \implies c = 8a^2$. Thus: $F(x) = \dfrac{x^4}{4} -2a^2x^2+8a^2= \left(\dfrac{x^2}{2} - 2a^2\right)^2+8a^2 - 4a^4\ge 8a^2-4a^4$. This minimum value equals to $0$. So $8a^2 - 4a^4 = 0\implies 4a^2(2-a^2)=0\implies a = 0,\pm \sqrt{2}$.
