Ta có \(y = f\left( x \right) = 3{x^2} + 1\). Do đó
\(f\left(\dfrac{1}{2} \right) = 3.{\left( \dfrac{1}{2} \right)^2} + 1 = 3.\dfrac{1}{4} + 1 = \dfrac{7}{4}\)
\(f\left( 1 \right) = {3.1^2} + 1 = 3.1 + 1 = 3 + 1 = 4\)
\(f\left( 3 \right) = {3.3^2} + 1 = 3.9 + 1 = 27 + 1 \)\(= 28.\)
