Solution:
\( \displaylines{f^{\prime}\left(x\right)=6x^2-30x+36=0\implies6\left(x^2-5x+6\right)=0\\ \left(x-3\right)\left(x-2\right)=0\implies x=2,3\\ f\left(-1\right)=-51\\ f\left(2\right)=30\\ f\left(3\right)=29\\ f\left(4\right)=34\\ f\text{ has a maximum of 34 at x=4, and a minimum of -51 at x=-1}} \)