Let $y=2 x^{3}-9 x^{2}+12 x+4$
On differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=6 x^{2}-18 x+12$
$f^{\prime}(x)=0$
$\therefore 6 x^{2}-18 x+12=0$
$\Rightarrow x^{2}-3 x+2=0$
$\Rightarrow (x-2)(x-1)=0$
$x=2$ or 1
Now, $\frac{d^{2} y}{d x^{2}}=12 x-18>0$,
for $x=2$
$\therefore x=2$ is a point of local
$y_{\min }=16-36+24+4=8$