Ta có: \(F=x^4-2x^3+3x^2-2x+2\)
\(=\left(x^4-2\cdot x^2\cdot x+x^2\right)+\left(2x^2-2x+2\right)\)
\(=\left(x^2-x\right)^2+2\left(x^2-x+1\right)\)
\(=\left(x^2-x\right)^2+2\left(x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)\)
\(=\left(x^2-x\right)^2+2\cdot\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\)
\(=\left(x^2-x\right)^2+2\left(x-\frac{1}{2}\right)^2+2\cdot\frac{3}{2}\)
\(=\left(x^2-x\right)^2+2\left(x-\frac{1}{2}\right)^2+3\)
Ta có: \(\left(x^2-x\right)^2\ge0\forall x\)
\(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x^2-x\right)^2+2\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x^2-x\right)^2+2\left(x-\frac{1}{2}\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}\left(x^2-x\right)^2=0\\2\left(x-\frac{1}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-x=0\\\left(x-\frac{1}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x\left(x-1\right)=0\\x-\frac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\x=\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: GTNN của đa thức \(F=x^4-2x^3+3x^2-2x+2\) là 3 khi \(x\in\left\{0;1;\frac{1}{2}\right\}\)
