Question

Прошу помощи по несложной задачке)

Answer 1

$y=x^3-4x^2+4x+3; ,; ; xin [-1,3, ]\\y'=3x^2-8x+4=0\\D/4=16-12=4\\x_1=frac{4-2}{3}=frac{2}{3}; ,; ; x_2=frac{4+2}{3}=2\\y(-1)=-1-4-4+3=-6\\y(frac{2}{3})=frac{8}{27}-frac{16}{9}+frac{8}{3}+3=frac{32}{27}+3=frac{113}{27}=4frac{5}{27}\\y(2)=8-16+8+3=3\\y(3)=27-36+12+3=6\\y_{naibol.}=6\\y_{naimen.}=-6$