The largest value of  $f\left(x\right)=2x^3-3x^2-12x+5$  for $-2\le x\le 4$  occurs at  $x=$

$f\left(x\right)=2x^3-3x^2-12x+5\text{ and }-2\le x\le 4$

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=6x^2-6x-12=0\\ \Rightarrow x^2-x-2=0\end{array}$

$\begin{array}{r}\Rightarrow x^2-2x+x-2=0\\ \Rightarrow x(x-2)+1(x-2)=0\\ (x-2)(x+1)=0\\ x=2,-1\end{array}$

$\begin{array}{r}{f}^{\prime \prime }\left(x\right)=12x-6\\ f^{\prime \prime }\left(2\right)=18>0\end{array}$

$$\begin{gathered} \therefore At x=2,value of f\left(x\right) is minimum \\ f\text{'}\text{'}\left(-1\right)=-18<0 \end{gathered}$$

$\begin{array}{r}f(-2)=-16-12+24+5=1\\ f\left(2\right)=16-12-24+5=-15\\ f(-1)=-2-3+12+5=12\\ f\left(4\right)=128-48-48+5=37\end{array}$

$f^{\mathrm{\prime }}\left(x\right)=0\Rightarrow x=-1,2\Rightarrow$ Largest value at $x=4$