Instead of relying on the second derivative, track the sign of $f'(x) = 12x(x+2)(x-1)$ across the number line to see where the function is rising or falling. The three roots of $f'(x)$ split the real line into four intervals: $x<-2$, $-2<x<0$, $0<x<1$, and $x>1$.
Picking a test value in each: at $x=-3$, $f'(-3)=12(-3)(-1)(-4) = -144 < 0$, so $f$ is decreasing on $x<-2$. At $x=-1$, $f'(-1)=12(-1)(1)(-2)=24>0$, so $f$ is increasing on $-2<x<0$. At $x=0.5$, $f'(0.5)=12(0.5)(2.5)(-0.5)=-7.5<0$, so $f$ is decreasing on $0<x<1$. At $x=2$, $f'(2)=12(2)(4)(1)=96>0$, so $f$ is increasing for $x>1$.
This sign pattern (down, up, down, up) shows $f$ has minima at $x=-2$ and $x=1$ (where it switches from falling to rising) and a maximum at $x=0$. Computing the function values at the two minimum points: $f(-2) = 48 - 32 - 48 + 6 = -26$ and $f(1) = 3+4-12+6 = 1$. Since $f$ falls all the way from $+\infty$ down to $-26$ before rising again, and the other dip only reaches $1$, the overall lowest value the function ever takes is $-26$.
\[\boxed{f(x)_{min} = -26}\]