Cho hàm số $y = f\left( x \right) =  - 2{x^2}$. Tìm $b$ biết $f\left( b \right) \le  - 5b + 2$.

Ta có $f\left( b \right) \le  - 5b + 2$ $\Leftrightarrow  - 2{b^2} \le  - 5b + 2 \Leftrightarrow 2{b^2} - 5b + 2 \ge 0$$\Leftrightarrow 2{b^2} - 4b - b + 2 \ge 0 \Leftrightarrow 2b\left( {b - 2} \right) - \left( {b - 2} \right) \ge 0 \Leftrightarrow \left( {2b - 1} \right)\left( {b - 2} \right) \ge 0$

$\Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}2b - 1 \ge 0\\b - 2 \ge 0\end{array} \right.\\\left\{ \begin{array}{l}2b - 1 \le 0\\b - 2 \le 0\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}b \ge \dfrac{1}{2}\\b \ge 2\end{array} \right.\\\left\{ \begin{array}{l}b \le \dfrac{1}{2}\\b \le 2\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}b \ge 2\\b \le \dfrac{1}{2}\end{array} \right.$

Vậy $\left[ \begin{array}{l}b \le \dfrac{1}{2}\\b \ge 2\end{array} \right.$ là giá trị cần tìm.