8. Find all minimum points and maximum points of the function $f(x)=\left(x^{2}-3 x-9\right) \cdot \sqrt{x^{2}-6 x+9}$. $x_{\min } \in\{0 ; 4\}, x_{\max } \in\{3\}$.
Solution. (1) $\left(x^{2}-3 x-9\right) \cdot \sqrt{x^{2}-6 x+9}=\left(x^{2}-3 x-9\right) \cdot \sqrt{(x-3)^{2}}=\left(x^{2}-3 x-9\right) \cdot |x-3|$ $=\left\{\begin{array}{c}-\left(x^{2}-3 x-9\right)(x-3) \text { for } x \leqslant 3, \\ \left(x^{2}-3 x-9\right)(x-3) \text { for } x \geqslant 3\end{array}=\left\{\begin{array}{r}-\left(x^{3}-6 x^{2}+27\right) \text { for } x \leqslant 3, \\ x^{3}-6 x^{2}+27 \text { for } x \geqslant 3 .\end{array}\right.\right.$

(2) Let $g(x)=x^{3}-6 x^{2}+27$. Then $g^{\prime}(x)=3 x^{2}-12 x=3 x(x-4)$,

- $g^{\prime}(x)>0$ on the interval $x \in(-\infty ; 0)$,
- $g^{\prime}(x)<0$ on the interval $x \in(0 ; 4)$,
- $g^{\prime}(x)>0$ on the interval $x \in(4 ;+\infty)$.

Therefore, the function $g(x)$

- increases on the interval $x \in(-\infty ; 0]$,
- decreases on the interval $x \in[0 ; 4]$,
- increases on the interval $x \in[4 ;+\infty)$.

(3) Since $f(x)=\left\{\begin{array}{r}-g(x) \text { for } x \leqslant 3, \\ g(x) \text { for } x \geqslant 3 .\end{array}\right.$ the function $f(x)$

- decreases on the interval $x \in(-\infty ; 0]$,
- increases on the interval $x \in[0 ; 3]$,
- decreases on the interval $x \in[3 ; 4]$,

Interregional Multidisciplinary Olympiad 2010. Second stage

- increases on the interval $x \in[4 ;+\infty)$.

Therefore

- $x=0$ is a point of minimum.
- $x=3$ is a point of maximum,
- $x=4$ is a point of minimum

of the function $f(x)$.