Tìm giá trị nhỏ nhất của C(x) = 3x² + x -1

C(x) = 3x² + x – 1

$\begin{array}{l}C\left(x\right)=3\left(x^2+\frac{1}{3}x-\frac{1}{3}\right)=3\left(x^2+2\cdot \frac{1}{6}x+\frac{1}{36}-\frac{1}{36}-\frac{1}{3}\right)\\ =3{\left(x+\frac{1}{6}\right)}^2-\frac{3}{36}-\frac{3}{3}=3{\left(x+\frac{1}{6}\right)}^2-\frac{13}{12}\end{array}$

Ta thấy:  $3{\left(x+\frac{1}{6}\right)}^2\ge 0\Rightarrow 3{\left(x+\frac{1}{6}\right)}^2-\frac{13}{12}\ge -\frac{13}{12}$

Dấu “ = ” xảy ra khi  $x+\frac{1}{6}=0\iff x=-\frac{1}{6}$

Vậy min_C(x) =  $-\frac{13}{2}$ khi x =  $-\frac{1}{6}$