Given,
f(x)=∣(x−1)(x+1)(x−3)∣+(x−3)
f(x)={\begin{matrix}(x-3)({x}^{2}), & 3\leq x\leq 4 \\ (x-3)(2-{x}^{2}), & 1\leq x<3 \\ (x-3)({x}^{2}), & 0<x<1\end{matrix}
{f}^{'}(x)={\begin{matrix}3{x}^{2}-6x, & 3<x<4 \\ -3{x}^{2}+6x+2, & 1<x<3 \\ 3{x}^{2}-6x, & 0<x<1\end{matrix}
f′(3+)>0f′(3−)<0→ Minimum
f′(1+)>0f′(1−)<0→ Minimum
For x∈(1,3),f′(x)=0 at one point → Maximum
For x∈(3,4),f′(x)=0
For x∈(0,1),f′(x)+0
So points of minima are 2, so m=2
and point of maxima is 1, so M=1
So, M+n=3