In the interval $[-2,4],$ the absolute maximum of $f\left(x\right)=2x^3-3x^2-12x+5$ occurs at $x=\mathrm{..........}$

$f\left(x\right)=2x^3-3x^2-12x+5\Rightarrow f\text{'}\left(x\right)=6x^2-6x-12.$

$f\text{'}\left(x\right)=0\Rightarrow 6x^2-6x-12=0\Rightarrow x^2-x-2=0\Rightarrow (x-2)(x+1)=0\Rightarrow x=2or-1.$

$f(-1)=-2-3+12+5=12,f\left(2\right)=16-12-24+5=-15.$

$f(-2)=-16-12+24+1,f\left(4\right)=128-48-48+5=37.$

$\therefore f\left(x\right)$ has absolute maximum at $x=4.$