Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ và có...

Ta có $g\left( x \right)=f\left( {{x}^{2}}-2x+1-\sqrt{{{\left( x-1 \right)}^{2}}} \right)$
có ${g}'\left( x \right)=\left( 2x-2-\dfrac{x-1}{\left| x-1 \right|} \right){f}'\left( {{x}^{2}}-2x+1-\left| x-1 \right| \right)=\left( x-1 \right)\left( 2-\dfrac{1}{\left| x-1 \right|} \right){f}'\left( {{x}^{2}}-2x+1-\left| x-1 \right| \right)$
Suy ra ${g}'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x-1=0 \\
& \left| x-1 \right|=\dfrac{1}{2} \\
& {{x}^{2}}-2x+1-\left| x-1 \right|=-1 \\
& {{x}^{2}}-2x+1-\left| x-1 \right|=0 \\
& {{x}^{2}}-2x+1-\left| x-1 \right|=1 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x=\dfrac{3}{2} \\
& x=\dfrac{1}{2} \\
& {{\left| x-1 \right|}^{2}}-\left| x-1 \right|+1=0\ (vn) \\
& {{\left| x-1 \right|}^{2}}-\left| x-1 \right|=0 \\
& {{\left| x-1 \right|}^{2}}-\left| x-1 \right|-1=0 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x=\dfrac{3}{2} \\
& x=\dfrac{1}{2} \\
& x=1(k) \\
& x=0 \\
& x=2 \\
& x=\dfrac{3+\sqrt{5}}{2} \\
& x=\dfrac{1-\sqrt{5}}{2} \\
\end{aligned} \right.$
Ta có bảng xét dấu ${g}'\left( x \right)$ :

Vậy hàm số $g\left( x \right)$ có 7 cực trị.