The function $f$ is defined by $f\left(x\right)=\left(x+2\right)e^{-x}$ is

The correct option is D

Decreasing in $\left(-1,\infty \right)$ and increasing in $\left(-\infty ,-1\right)$.

Explanation for the correct option

Given function, $f\left(x\right)=\left(x+2\right)e^{-x}$.

Differentiate the given function with respect to $x$.

$\begin{array}{rcl}\frac{\mathrm{d}}{dx}\left[f\left(x\right)\right]& =& \frac{d}{dx}\left[\left(x+2\right)e^{-x}\right]\\ & =& e^{-x}\frac{d}{dx}\left(x+2\right)+\left(x+2\right)\frac{d}{dx}\left(e^{-x}\right)\left[\because \frac{d}{d\mathrm{x}}(\mathrm{u}·\mathrm{v})=v\frac{d\mathrm{u}}{d\mathrm{x}}+u\frac{d\mathrm{v}}{d\mathrm{x}}\right]\\ & =& \left(e^{-x}\times 1\right)-e^{-x}\left(x+2\right)\\ & =& e^{-x}-x{e}^{-x}-2e^{-x}\\ & =& -e^{-x}\left(x+1\right)\end{array}$

Now, for increasing intervals $f\text{'}\left(\mathrm{x}\right)>0$.

$$\begin{gathered} \Rightarrow -e^{-x}\left(x+1\right)>0 \\ \Rightarrow e^{-x}\left(x+1\right)<0 \\ \Rightarrow \frac{x+1}{e^x}<0 \\ \Rightarrow x+1<0\times e^x\left[\because e^x>0\right] \\ \Rightarrow x<-1 \end{gathered}$$

Thus, the given function is increasing in $\left(-\infty ,-1\right)$.

Hence, the function will be decreasing in $\left(-1,\infty \right)$.

Therefore, the function $f\left(x\right)=\left(x+2\right)e^{-x}$ is decreasing in $\left(-1,\infty \right)$ and increasing in $\left(-\infty ,-1\right)$.

Hence, the function is decreasing in $\left(-1,\infty \right)$ and increasing in $\left(-\infty ,-1\right)$ and therefore the correct option is (D).