Cho hàm số $y=f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$ có đồ...

$f'\left( x \right)=3a{{x}^{2}}+2bx+c$
Từ đồ thị hàm số $y=f\left( x \right)$ ta có:
$\left\{ \begin{aligned}
& f\left( -2 \right)=0 \\
& f\left( 3 \right)=5 \\
& f'\left( -1 \right)=0 \\
& f'\left( 3 \right)=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -8a+4b-2c+d=0 \\
& 27a+9b+3c+d=0 \\
& 3a-2b+c=0 \\
& 27a+6b+c=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-\dfrac{1}{5} \\
& b=\dfrac{3}{5} \\
& c=\dfrac{9}{5} \\
& d=-\dfrac{2}{5} \\
\end{aligned} \right.\Rightarrow f\left( x \right)=-\dfrac{1}{5}{{x}^{3}}+\dfrac{3}{5}{{x}^{2}}+\dfrac{9}{5}x-\dfrac{2}{5}$.
Ta có $f'\left( x \right)=\dfrac{-3}{5}{{x}^{2}}+\dfrac{6}{5}x+\dfrac{9}{5}\Rightarrow f''\left( x \right)=\dfrac{-6}{5}x+\dfrac{6}{5}\Rightarrow g\left( x \right)=-\dfrac{3}{5}x-\dfrac{3}{5}$.
$f'\left( x \right)=g\left( x \right)\Leftrightarrow \left[ \begin{aligned}
& x=4 \\
& x=-1 \\
\end{aligned} \right.$. Diện tích hình phẳng là:
$S=\int\limits_{-1}^{4}{\left| -\dfrac{3}{5}{{x}^{2}}+\dfrac{6}{5}x+\dfrac{9}{5}+\dfrac{3}{5}x+\dfrac{3}{5} \right|}=\dfrac{25}{2}$.