Answer
The minimum value is $ f(0)=0$ and the maximum value is $ f(1)=10$
Work Step by Step
$ f'(x)=3x^{2}-24x+21$ is differentiable everywhere. The critical points of the function are the solutions of $3x^{2}-24x+21=0$.
$3x^{2}-24x+21=0$ when $ x=7$ or when $ x=1$.
The critical points of $ y $ are 7 and 1.
7 is outside the given interval. Therefore, the minimum and maximum value occurs either at $ x=1$ or at the end points.
$ f(1)=(1)^{3}-12(1)^{2}+21(1)=10$
$ f(0)=(0)^{3}-12(0)^{2}+21(0)=0$
$ f(2)=(2)^{3}-12(2)^{2}+21(2)=2$
The minimum value is $ f(0)=0$ and the maximum value is $ f(1)=10$
