The absolute maximum of $\mathrm{y}={\mathrm{x}}^3-3\mathrm{x}+2$  in $0\le \mathrm{x}\le 2$  is

$$\begin{gathered} \mathrm{Given} y=\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3\mathrm{x}+2,0\le x\le 2 \\ \mathrm{diff} \mathrm{w},\mathrm{r}.\mathrm{to}.\mathrm{x} \mathrm{on} \mathrm{both} \mathrm{sides} \\ {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-3 \\ Now for max or min {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=0 \\ \Rightarrow 3(x^2-1)=0 \\ \Rightarrow (x+1)(x-1)=0 \\ \Rightarrow x=-1 or x=1 \\ Now f\left(0\right)=2 \\ f\left(1\right)=0 \\ f\left(2\right)=4 \\ \therefore Absolute maximum=max\left\{f\left(0\right),f\left(1\right),f\left(2\right)\left. \right\}\right. \\ =4 \end{gathered}$$