If $-4\le \mathrm{x}\le 4$ then the critical point of ${\text{⨍ (x)=x}}^2\text{-6}\left|\mathrm{x}\right|\text{+4}$ are

$$\begin{gathered} -4\le \mathrm{x}\le 4 \\ {\text{⨍ (x)=x}}^2\text{-6}\left|\mathrm{x}\right|\text{+4} \\ ={\left|\mathrm{x}\right|}^2-6\left|\mathrm{x}\right|+4 \\ {\text{⨍ }}^1\text{(x)=2}\left|\mathrm{x}\right|\left(\frac{\left|\mathrm{x}\right|}{\mathrm{x}}\right)\text{-6}\frac{\left|\mathrm{x}\right|}{\mathrm{x}} \\ \text{ =}\frac{\left|\mathrm{x}\right|}{\mathrm{x}}\left(2\left|\mathrm{x}\right|-6\right) \end{gathered}$$

at $\mathrm{x}=0 {\text{⨍ }}^1\text{(x)}$ not defined $\text{⨍ (x)}$ is min or max

$$\begin{gathered} {\text{⨍ }}^1\text{(x)=0} \\ 2\left|\mathrm{x}\right|=6\Rightarrow \left|\mathrm{x}\right|=3 \\ \Rightarrow \mathrm{x}=\pm 3 \end{gathered}$$

$\therefore$ critical points are

$\mathrm{x}=0, -3, 3$