Ta có: \(f\left(x\right)=-3x^2+10x-2=-3\left(x^2-\dfrac{10}{3}x+\dfrac{2}{3}\right)\)
\(=-3\left(x^2-2.x.\dfrac{10}{6}+\dfrac{100}{36}-\dfrac{19}{9}\right)\)
\(=-3\left(x-\dfrac{10}{6}\right)^2+\dfrac{19}{3}\le\dfrac{19}{3}\forall x\in R\)
Dấu = xảy ra khi: \(-3\left(x-\dfrac{10}{6}\right)^2=0\Leftrightarrow x=\dfrac{10}{6}=\dfrac{5}{3}\)
Vậy GTLN của f(x) \(=\dfrac{19}{3}\) khi x = \(\dfrac{5}{3}\)
\(=-3\left(x^2-\dfrac{10}{3}x+\dfrac{2}{3}\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{19}{9}\right)\)
\(=-3\left(x-\dfrac{5}{3}\right)^2+\dfrac{19}{3}< =\dfrac{19}{3}\)
Dấu = xảy ra khi x=5/3
\(f\left(x\right)=-3\left(x^2-\dfrac{2.5x}{3}+\dfrac{25}{9}-\dfrac{25}{9}\right)-2=-3\left(x-\dfrac{5}{3}\right)^2+\dfrac{19}{3}\le\dfrac{19}{3}\)
Dấu ''='' xảy ra khi x = 5/3
