给出以下函数的导数并求解该函数的任一极值点: 


f(x) = x³ - 6x² + 9x + 3
\begin{aligned} \mathtt{f(x)} &= \mathtt{x^3 - 6x^2 + 9x + 3} \\ \frac{\mathrm{d}}{\mathrm{d}x} \mathtt{f(x)} &= \mathtt{3x^2 - 12x + 9} \\ \end{aligned}

将导数等于零得到：

\begin{aligned} 3x^2 - 12x + 9 &= 0 \\ x^2 - 4x + 3 &= 0 \\ (x - 1)(x - 3) &= 0 \end{aligned}

因此，极值点为1和3。