12.14 $y=x^{3}-3 x^{2}+3 x+2$ on $[-2,2]$.
12.14 First, we find the values of $y$ at the endpoints of the interval: $y(-2)=-24, y(2)=4$. Now let's find the critical points within the interval $[-2,2]$. We have $y^{\prime}=3 x^{2}-6 x+3=3(x-1)^{2} ;$ the equation $y^{\prime}=0$ has a root $x=1$, and $y(1)=3$. By comparing the values of $y(-2), y(2)$, and $y(1)$, we conclude that $y_{\text {min }}=-24$, $y_{\text {max }}=4$.

Answer: $y_{\text {min }}=-24, y_{\text {max }}=4$.