The stationary points of $f\left(x\right)=x^3–9x^2+24x–12$ are

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-9{\mathrm{x}}^2+24\mathrm{x}-12 \\ {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-18\mathrm{x}+24 \end{gathered}$$

To evaluate minimum or maximum

$$\begin{gathered} {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=0 \\ 3{\mathrm{x}}^2-18\mathrm{x}+24=0 \\ {\mathrm{x}}^2-6\mathrm{x}+8=0 \\ \left(\mathrm{x}-2\right)\left(\mathrm{x}-4\right)=0 \\ \mathrm{x}=2/4 \end{gathered}$$

$$\begin{gathered} f\left(2\right)=8-9\times 4+24\times 2-12 \\ =8-36+48-12 \\ =8 \end{gathered}$$

$$\begin{gathered} f\left(4\right)=64-9\times 16+96-12 \\ =4 \end{gathered}$$

$\therefore$ Stationary points

$\left(2,8\right) \left(4,4\right)$