Cho hàm số bậc ba y = f(x) có f'(x) = 3x^2 + 2x - m + 1

Ta có \(f\left( x \right) = \int {f'\left( x \right){\rm{d}}x} \)\( = \int {\left( {3{x^2} + 2x - m + 1} \right){\rm{d}}x} \)\( = {x^3} + {x^2} + \left( {1 - m} \right)x + C\).

Theo giả thiết \(\left\{ \begin{array}{l}f\left( 2 \right) = 1\\f\left( 1 \right) =  - 3\end{array} \right.\)\(\)\( \Rightarrow \left\{ \begin{array}{l}{2^3} + {2^2} + \left( {1 - m} \right).2 + C = 1\\1 + 1 + \left( {1 - m} \right) + C =  - 3\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l} - 2m + C =  - 13\\ - m + C =  - 6\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}m = 7\\C = 1\end{array} \right.\).

Suy ra \(f\left( x \right) = {x^3} + {x^2} - 6x + 1\). Vậy \(f\left( { - 1} \right) = 7\).