Answer
Absolute maximum: $f(-1)=8$
Absolute minimum: $f(2) = -19$
Work Step by Step
Using closed interval method, as $f$ in continuous over the given range, let's first calculate the value of $f$ at critical numbers followed by values of $f$ at endpoints of the given range.
For critical numbers, $f'(x)=0$ or $f'(x)$ should not exist. Let's differentiate $f(x)=2x^3-3x^2-12x+1$ with respect to $x$ keeping $[-2,3]$ as interval.
$$\frac{df}{dx}=6x^2-6x-12$$ $\frac{df}{dx}$ exists on the given interval. Thus, $\frac{df}{dx}=0$, $$6x^2-6x-12=0\implies x^2-x-2=0$$ $$x=-1,2.$$
Thus, $f(-1)=2(-1)^3-3(-1)^2-12(-1)+1=8$. Similarly, $f(2)=-19$. At endpoints,
$f(-2)=-3$ & $f(3)=-8$. Thus, absolute maximum $=f(-1)=8$ and absolute minimum $=f(2)=-19$.
