$f'(x) = {e^{ - x}} - {e^{ - x}}(x + 2)$
$f'(x) = - {e^{ - x}}[x + 1]$
For increasing, $ - {e^{ - x}}(x + 1) > 0$ or ${e^{ - x}}(x + 1) < 0$
${e^{ - x}} > 0$ $(x + 1) < 0$
$x \in ( - \infty ,\,\infty )$ and $x \in ( - \infty , - 1)$
$\therefore x \in ( - \infty , - 1)$
Hence, the function is increasing in $( - \infty ,\, - 1)$
For decreasing, $ - {e^{ - x}}(x + 1) < 0$ or ${e^{ - x}}(x + 1) > 0$, $x \in ( - 1,\,\infty )$
Hence the function is decreasing in $( - 1,\;\infty )$.
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