Cho $f\left( x \right)$ là hàm đa thức và cho hàm đa thức bậc ba...

$g\left( x \right)=f\left( x+1 \right)\Rightarrow {g}'\left( x \right)={f}'\left( x+1 \right)$.
$\left( x-1 \right){g}'\left( x+3 \right)=\left( x+1 \right){g}'\left( x+2 \right)$ hay $\left( x-1 \right){f}'\left( x+4 \right)=\left( x+1 \right){f}'\left( x+3 \right)$.
Cho$\left\{ \begin{aligned}
& x=1 \\
& x=-1 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {f}'\left( 4 \right)=0 \\
& {f}'\left( 3 \right)=0 \\
\end{aligned} \right.\Rightarrow {f}'\left( x \right)=a\left( x-3 \right)\left( x-4 \right) $. $ y=f\left( 2{{x}^{2}}-4x+5 \right)\Rightarrow {y}'=\left( 4x-4 \right).{f}'\left( 2{{x}^{2}}-4x+5 \right)$
${y}'=0\Leftrightarrow \left[ \begin{aligned}
& 4x-4=0 \\
& {f}'\left( 2{{x}^{2}}-4x+5 \right)=0 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& 2{{x}^{2}}-4x+5=4 \\
& 2{{x}^{2}}-4x+5=3 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& 2{{x}^{2}}-4x+1=0 \\
& 2{{x}^{2}}-4x+2=0 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x=\dfrac{2-\sqrt{2}}{2} \\
& x=\dfrac{2+\sqrt{2}}{2} \\
& x=1 \\
\end{aligned} \right.$
Vậy hàm số có 3 cực trị