## Calculus (3rd Edition)

The minimum value is $f(0)=0$ and the maximum value is $f(1)=10$
$f'(x)=3x^{2}-24x+21$ 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 $f(1)=10$