Question

Найдите экстремумы функции: f(x)=x^2(x-3)

Answer 1

$f(x)'=x^2*(x-3) \to 2x*(x-3)+x^2 \\ 2x^2-6x+x^2=0 \\ 3x^2-6x=0 \\ 3x(x-2)=0 ; x = 0 ; x = 2 \\ x_{max} = 0 ; x_{min} = 2$