\(y=x^4+4x^3+4x^2-x^2-2x+2\)
\(y=\left(x^2+2x\right)^2-\left(x^2+2x\right)+2\)
Đặt \(x^2+2x=t\)
Với \(x\in\left[-2;4\right]\Rightarrow t\in\left[-1;24\right]\)
Xét \(f\left(t\right)=t^2-t+2\) trên \(\left[-1;24\right]\)
Ta có \(-\frac{b}{2a}=\frac{1}{2}\)
\(f\left(-1\right)=4\) ; \(f\left(\frac{1}{2}\right)=\frac{7}{4}\) ; \(f\left(24\right)=554\)
\(\Rightarrow y_{max}=554\) khi \(t=24\Leftrightarrow x=4\)
\(y_{min}=\frac{7}{4}\) khi \(t=\frac{1}{2}\Leftrightarrow x=\frac{-2\pm\sqrt{2}}{2}\)