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.

Solution 1:

If $F(x)=\frac14x^4-2a^2x^2+c$, the minimum of $F$ is attained at a point $x$ such that $f(x)=0$, that is, it is attained at $\pm a$ or at $0$. Actually, it cannot be attained at $0$, since $F''(0)=f'(0)=-4a^2$, and therefore $F$ has a local maximum at $0$. On the other hand,$$F(\pm a)=c-\frac74a^4,$$and therefore $F(\pm a)=0\iff a=0\vee c=\frac74a^4$.

If $a=0$, then you just take $c=0$, so that $F(2)=4$. And if $c=\frac74a^4$, then$$F(x)=\frac14x^4-2a^2x^2+\frac74a^4,$$and then $F(2)=4-8a^2+\frac74a^4$. So, $F(2)=4\iff a=0\vee a=\pm4\sqrt{\frac27}$.