Решите уравнение f'(t)=0 если f(t)=1/8 (t+3) (t-3)^2

[tex]f(t)= \frac{(t+3)(t-3)^2}{8} =\frac{(t^2-9)(t-3)}{8} =\frac{t^3-3t^2-9t+27}{8} \\ f'(t)= \frac{3t^2-6t-9}{8} \\ \frac{3t^2-6t-9}{8}=0 \\ 3t^2-6t-9=0 \\3(t-3)(t+1)=0 \\ t_1=3;t_2=-1[/tex]