Question

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

Answer 1

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