The absolute maximum of $f\left(x\right)=x^3–3x+2$ in [0, 2] is

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3\mathrm{x}+2 \\ {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-3 \\ =3\left(\mathrm{x}-1\right)\left(\mathrm{x}+1\right) \end{gathered}$$

$\mathrm{f}\left(\mathrm{x}\right)$ is minimum or maximum $\Rightarrow {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=0$

$$\begin{gathered} \mathrm{x}=1,-1 \\ \mathrm{x}=1\in \left[0,2\right] \end{gathered}$$

$$\begin{gathered} \mathrm{f}\left(0\right)=0-0+2=2 \\ \mathrm{f}\left(1\right)=1-3+2=0 \\ \mathrm{f}\left(2\right)=8-6+2=4 \end{gathered}$$

$\therefore$ absolute maximum = 4