Прошу помощи по несложной задачке)

Решите задачу:

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