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

Question

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

Answer 1

3x^2-12x-12=0<=>x^2-4x-4=0<=>(x-2)^2=0=>\\=>x=2" alt="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" align="absmiddle" class="latex-formula">