Ta có:
\(\begin{array}{l}f(1) = 3.1 + 2 = 5;\\f(0) = 3.0 + 2 = 2\\f( - 2) = 3.\left( { - 2} \right) + 2 = - 4;\\f\left( {\dfrac{1}{2}} \right) = 3.\dfrac{1}{2} + 2 = \dfrac{7}{2};\\f\left( { - \dfrac{2}{3}} \right) = 3.\left( { - \dfrac{2}{3}} \right) + 2 = 0\end{array}\)