Chọn C

Xét hàm số $t\left(x\right)=\left|x^2-8x+7\right|+x^2-3$ có TXĐ là R

Ta có: $t\left(x\right)=\left\{\begin{array}{l}2x^2-8x+4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}khi\hspace{0.17em}}x\le 1,x\ge 7\\ 8x-10\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}khi }1<x<7\end{array}\right.$

Có $t\text{'}\left(x\right)=\left\{\begin{array}{l}4x-8\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}khi\hspace{0.17em}}x\le 1,x\ge 7\\ 8\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}khi }1<x<7\end{array}\right.$

Ta có:

Suy ra hàm số t(x) có 1 cực trị tại x = 1

Suy ra hàm số $y=\left|f\left(t\right)\right|$ có 7 điểm cực đại.