Let $f(x)=x^3-6 x^2+9 x+18$, then f(x) is strictly decreasing in

$(-\infty, 1]$ $[3, \infty)$ $(-\infty, 1] \cup[3, \infty)$ $[1,3]$

$[1,3]$

$f(x) =x^3-6 x^2+9 x+18$ $f'(x) =3 x^2-12 x+9$ $=3\left(x^2-4 x+3\right)$ $=3(x-1)(x-3) \leq 0$ $x \in[1,3]$