Let  $f\left(x\right)=\left\{\begin{array}{ll}-2x+{\log }_{\frac{1}{2}}({\lambda }^2-6\lambda +8),& -2\le x<-1\\ x^3+3x^2+4x+1,& -1\le x\le 3\end{array}\right.$ 

then the sum of all possible positive integer(s) in the range of $\lambda$ such that f(x) has smallest value at x = -1 is

$$\begin{gathered} f(-1)\le f(-1-h) \\ \Rightarrow 2+{\log }_{\frac{1}{2}}({\lambda }^2-6\lambda +8)\ge -1 \\ \Rightarrow {\lambda }^2-6\lambda +8\le 8\Rightarrow {\lambda }^2-6\lambda \le 0 \\ \Rightarrow \lambda (\lambda -6)\le 0\Rightarrow 0\le \lambda \le 6 \\ Again{\lambda }^2-6\lambda +8>0\Rightarrow (x-2)(\lambda -4)>0 \\ \begin{array}{l}\Rightarrow \lambda >4or\lambda <2\\ \Rightarrow \lambda \in [0,2)\cup (4,6]\end{array} \\ \begin{array}{l}\Rightarrow \lambda =0,1,5,6\\ sum=12\end{array} \end{gathered}$$