The absolute maximum of y = x3 - 3x + 2 in [0, 2] is

$y=x^3-3 x+2 \Rightarrow \frac{d y}{d x}=3 x^2-3=0 \Rightarrow x= \pm 1$

∴ $\begin{array}{llll}x & 0 & 1 & 2 \\ y & 2 & 0 & 4\end{array}$

∴ absolute maximum is 4.