Question

найдите экстремумы функции y= x^3-6x^2+12x

Answer 1

$y=x^3-6x^2+12x\y'=3x^2-12x-12\y'=0=>3x^2-12x-12=0<=>x^2-4x-4=0<=>(x-2)^2=0=>\=>x=2$