## Calculus (3rd Edition)

Maximum value of $f$ on [0, 2] is 5 and the minimum value is 0. Maximum value of $f$ on [0, 3] is 9 and the minimum value is 0.
$f'(x)=6x^{2}-18x+12=0\implies$ $x=\frac{18±\sqrt {18^{2}-4\times6\times12}}{2\times6}$ That is, $x=2$ or $x=1$. Thus 1 and 2 are the critical points of $f$. We compare the values of the function at the critical points and at the end points. $f(2)=2(2)^{3}-9(2)^{2}+12(2)=4$ $f(1)=2(1)^{3}-9(1)^{2}+12(1)=5$ $f(0)=2(0)^{3}-9(0)^{2}+12(0)=0$ $f(3)=2(3)^{3}-9(3)^{2}+12(3)=9$ The maximum value of $f$ on [0, 2] is $f(1)=5$ and the minimum value is $f(0)=0$. The maximum value of $f$ on [0, 3] is $f(3)=9$ and the minimum value is $f(0)=0$.