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

Đặt $g\left( x \right)=f\left( 1-3x \right)+1$.
Ta có:
$\bullet $ ${{\left( \left| g\left( x \right) \right| \right)}^{\prime }}=0\Leftrightarrow \dfrac{g\left( x \right)}{\left| g\left( x \right) \right|}\cdot {g}'\left( x \right)=0$.
$\bullet $ $g\left( x \right)=0\Leftrightarrow f\left( 1-3x \right)=-1\Leftrightarrow \left[ \begin{aligned}
& 1-3x=a , \left( a<-1 \right) \\
& 1-3x=b , \left( -1<b<3 \right) \\
& 1-3x=c , \left( x>3 \right) \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=\dfrac{1-a}{3} , \left( \dfrac{1-a}{3}>\dfrac{2}{3} \right) \\
& x=\dfrac{1-b}{3} , \left( -\dfrac{2}{3}<\dfrac{1-a}{3}<\dfrac{2}{3} \right) \\
& x=\dfrac{1-c}{3} , \left( \dfrac{1-c}{3}<-\dfrac{2}{3} \right) \\
\end{aligned} \right.$.
$\bullet $ ${g}'\left( x \right)=0\Leftrightarrow -3{f}'\left( 1-3x \right)=0\Leftrightarrow \left[ \begin{aligned}
& 1-3x=-1 \\
& 1-3x=3 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=\dfrac{2}{3} \\
& x=-\dfrac{2}{3} \\
\end{aligned} \right.$.
Và $g\left( -\dfrac{2}{3} \right)=f\left( 3 \right)+1=-3+1=-2$ ; $g\left( \dfrac{2}{3} \right)=f\left( -1 \right)+1=5+1=6$.
Bảng biến thiên
Vậy hàm số $y=\left| g\left( x \right) \right|=\left| f\left( 1-3x \right)+1 \right|$ có $5$ điểm cực trị.