Cho hàm số f(x) thỏa mãn $f''\left( x \right)=12{{x}^{2}}+6x-4$ và...

Phương pháp:
Ta có: $f'\left( x \right)=\int f''\left( x \right)dx;f\left( x \right)=\int f'\left( x \right)dx.$
Cách giải:
Ta có $:f''\left( x \right)=12{{x}^{2}}+6x-4$
$\Rightarrow f'\left( x \right)=\int \left( 12{{x}^{2}}+6x-4 \right)~dx=4{{x}^{3}}+3{{x}^{2}}-4~x+~C$
$\Rightarrow f\left( x \right)=\int f'\left( x \right)dx=\int ~\left( 4{{x}^{3}}+3{{x}^{2}}~-4~x+~C \right)~dx~$
$={{x}^{4}}+{{x}^{3}}-2{{x}^{2}}~+Cx+~{{C}^{'}}~$
Lại có: $\left\{ \begin{aligned}
& f\left( 0 \right)=1 \\
& f\left( 1 \right)=3 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{C}^{'}}=1 \\
& 1+1-2+C+{{C}^{'}}=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{C}^{'}}=1 \\
& C=2 \\
\end{aligned} \right.$
$f\left( x \right)={{x}^{4}}+{{x}^{3}}-2{{x}^{3}}+2x+1$
$f\left( -1 \right)={{\left( -1 \right)}^{4}}+{{\left( -3 \right)}^{3}}-2{{\left( -1 \right)}^{2}}+2\left( -1 \right)+1=-3$