Cho hàm số $y=f\left( x \right)$ bậc ba có bảng biến thiên sau...

Ta có ${g}'\left( x \right)={f}'\left( x \right)\left[ f\left( x \right)-1 \right]{f}'\left( \dfrac{1}{2}{{f}^{2}}\left( x \right)-f\left( x \right) \right)$ nên
${g}'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& {f}'\left( x \right)=0; f\left( x \right)=1 \\
& {f}'\left( \dfrac{1}{2}{{f}^{2}}\left( x \right)-f\left( x \right) \right)=0. \\
\end{aligned} \right.$
Xét phương trình ${f}'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=0 \\
& x=\dfrac{3}{2}. \\
\end{aligned} \right.$
Xét phương trình $f\left( x \right)=1\Leftrightarrow \left[ \begin{aligned}
& x=0 \\
& x=a \left( a>\dfrac{3}{2} \right). \\
\end{aligned} \right.$
Xét phương trình
${f}'\left( \dfrac{1}{2}{{f}^{2}}\left( x \right)-f\left( x \right) \right)=0.\Leftrightarrow \left[ \begin{aligned}
& \dfrac{1}{2}{{f}^{2}}\left( x \right)-f\left( x \right)=0 \\
& \dfrac{1}{2}{{f}^{2}}\left( x \right)-f\left( x \right)=\dfrac{3}{2} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& f\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=\dfrac{3}{2} \\
& x=b \left( b<0 \right) \\
\end{aligned} \right. \\
& f\left( x \right)=2\Leftrightarrow x=c \left( c>a \right) \\
& f\left( x \right)=3\Leftrightarrow x=d \left( d>c \right) \\
& f\left( x \right)=-1\Leftrightarrow x=e \left( e<b \right). \\
\end{aligned} \right.$
Vậy phương trình ${g}'\left( x \right)=0$ có $7$ nghiệm phân biệt.