Let $g(x)=6x^2-18x+8$
If $f_1=|g(x)|$,$f_2(x)=|f_1(x)-P_1|$, $f_3(x)=|f_2(x)-P_2|$, $f_4(x)=|f_3(x)-P_3|$.
If $P_1$=7 , the range of $P_2$ such that $f_3(x)$ has exactly 10 points of non-differentiability.
Let $g(x)=6x^2-18x+8$
If $f_1=|g(x)|$,$f_2(x)=|f_1(x)-P_1|$, $f_3(x)=|f_2(x)-P_2|$, $f_4(x)=|f_3(x)-P_3|$.
If $P_1$=7 , the range of $P_2$ such that $f_3(x)$ has exactly 10 points of non-differentiability.
Hint:
$y=f_1(x)$ has two such points. The local maximum of $f_1$ has $y<7$. Hence, $f_2$ has four such points.
$f_3$ needs ten such points, i.e. six more points. This can be obtained if $P_3$ if between $0$ and the local maximum.