Answer
Minimum value is $ f(4)=-24$ and maximum value is $ f(6)=8$.
Work Step by Step
$ f'(x)=3x^{2}-12x $
$ f $ is differentiable everywhere. So, the only critical point we get is $ c $ such that $ f'(c)=0$.
That is, $3x^{2}-12x=0$$\implies 3x^{2}=12x\implies x=4$
To find maximum and minimum values, we compare the values at the critical point and end points.
$ f(4)=(4)^{3}-6(4)^{2}+8=-24$
$ f(1)=(1)^{3}-6(1)^{2}+8=3$
$ f(6)=(6)^{3}-6(6)^{2}+8=8$
Minimum value is $ f(4)=-24$ and maximum value is $ f(6)=8$.
