## Calculus (3rd Edition)

Minimum value is $f(4)=-24$ and maximum value is $f(6)=8$.
$f'(x)=3x^{2}-12x$ $f$ 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 $f(6)=8$.