The point of minimum of the function f(x)=4x^2-x|x-2|, x ∈ [0, 3] is

$\frac{dy}{dx}=\left\{\begin{matrix}12x^2-2x+2&;&2<x≤3\\12x^2+2x-2&;&0≤x<2\end{matrix}\right.$; i.e., $2(2x+1)(3x-1)$ for $0≤x<2$

f(x) is continuous but not differentiable at x = 2 and $\frac{dy}{dx}=0$ at $x=\frac{1}{3}$ decreasing in $\left(0,\frac{1}{3}\right)$ and increasing for $\left(\frac{1}{3},2\right)∪(2,3)$. Minima occurs at $x=\frac{1}{3}$