[tex]C=2x^2+x=2(x^2+x)=2(x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16})[/tex]
[tex]=2[(x+\frac{1}{4})^2-\frac{1}{16}]=[tex]2(x+\frac{1}{4})^2-\frac{1}{8}[tex]\geq[/tex]\frac{-1}{8}[/tex]
Dấu " = " khi [TEX]2(x+\frac{1}{4})^2=0\Leftrightarrow x=\frac{-1}{4}[/TEX]
Vậy MIN C = -1/8 khi x = -1/4[/tex]