The function f defined by f(x)=(x+2) e^-x is

We have,

$f(x)=(x+2) e^{-x} \Rightarrow f^{\prime}(x)=e^{-x}-(x+2) e^{-x}=-(x+1) e^{-x}$

∴ $f^{\prime}(x)<0$

$\Rightarrow -(x+1) e^{-x}<0 \Rightarrow-(x+1)<0 \Rightarrow x+1>0 \Rightarrow x>-1$

Hence, f(x) is decreasing in $(-1, \infty)$ and increasing in $(-\infty,-1)$.