\(P\left(x\right)=3x^2-\left[3f\left(x\right)+1\right]x+3-f\left(x\right)=0\left(1\right)\)
Phương trình (1) có nghiệm thuộc \(\left(0;\frac{2}{3}\right)\) khi:
\(\hept{\begin{cases}\Delta=9f^2\left(x\right)+18f\left(x\right)-35\ge0\\P\left(0\right)=3-f\left(x\right)>0\\P\left(\frac{2}{3}\right)=\frac{11}{3}-3f\left(x\right)>0\end{cases}\Leftrightarrow\hept{\begin{cases}f\left(x\right)\le\frac{-3-2\sqrt{11}}{3}\left(h\right)f\left(x\right)\ge\frac{-3+2\sqrt{11}}{3}\\f\left(x\right)< 3\\f\left(x\right)< \frac{11}{9}\end{cases}}}\)
\(\Rightarrow f\left(x\right)\in(-\infty;\frac{-3-2\sqrt{11}}{3}]\)U\([\frac{-3+2\sqrt{11}}{3};\frac{11}{9})\)
Dễ thấy \(f\left(x\right)>0\forall x\in\left(0;\frac{2}{3}\right)\). Suy ra \(\frac{-3+2\sqrt{11}}{3}\le f\left(x\right)< \frac{11}{9}\)
Vậy \(minf\left(x\right)=\frac{-3+2\sqrt{11}}{3}\), đạt được khi \(x=\frac{-1+\sqrt{11}}{3}.\)