If the absolute maximum and absolute minimum value of the function $f\left(x\right)=x^3-2x^2+x-3$ defined on [0,2] are M and m respectively, then M+m

Given $f\left(x\right)=x^3-2x^2+x-3$
$f^{\text{'}}\left(x\right)=3x^2-4x+1$
For absolute maxima or minima put $f^{\text{'}}\left(x\right)=0$
$\begin{array}{l}\therefore 3x^2-4x+1=0\\ \Rightarrow 3x^2-3x-x+1=0\\ (3x-1)(x-1)=0\\ x=\frac{1}{3}\text{ or }x=1\end{array}$
Clearly,the critical point of f(x)in [0,2] is x=1/3 or 1
Now, $f\left(0\right)=-3,f\left(\frac{1}{3}\right)=-\frac{77}{27}\approx -2.851$
$f\left(1\right)=-3,f\left(2\right)=-1$
Therefore absolute minimum value i.e m=-3 
and absolute maximum value i.e M=-1
Now, M+m=-1-3=-4