Which of the following is true for the function \[f\left( x \right) = \left( {x + 2} \right){e^{ - x}}\]?
A. Decreasing for all \[x\]
B. Decreasing in \[\left( { - \infty , - 1} \right)\] and increasing in \[\left( { - 1,\infty } \right)\]
C. Increasing for all \[x\]
D. Decreasing in \[\left( { - 1,\infty } \right)\] and increasing in \[\left( { - \infty , - 1} \right)\]

Hint: To check a function increasing or decreasing, first-order derivative test is used. Find the differentiation of the given function and equate it with zero to find the critical points. Use the concept that a function is increasing in the interval on which the first-order derivative of the function is positive. Otherwise, it is decreasing.

Formula used
Quotient Rule: If \[f\left( x \right)\] and \[g\left( x \right)\] be two functions of \[x\] then the differentiation of the function \[\dfrac{{f\left( x \right)}}{{g\left( x \right)}}\] is \[\dfrac{d}{{dx}}\left\{ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right\} = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\} - f\left( x \right)\dfrac{d}{{dx}}\left\{ {g\left( x \right)} \right\}}}{{{{\left\{ {g\left( x \right)} \right\}}^2}}}\], provided \[g\left( x \right) \ne 0\]
\[\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\]
\[\dfrac{d}{{dx}}\left( c \right) = 0\], where \[c\] is a constant.

Complete step by step solution
The given function is \[f\left( x \right) = \left( {x + 2} \right){e^{ - x}}\]
Simplifying the given function and we get
\[ \Rightarrow f\left( x \right) = \left( {x + 2} \right) \times \dfrac{1}{{{e^x}}}\]
\[ \Rightarrow f\left( x \right) = \dfrac{{x + 2}}{{{e^x}}}\] ………….(1)
Differentiating the function (1) with respect to \[x\], we get
\[f'\left( x \right) = \dfrac{{\left( {{e^x}} \right)\dfrac{d}{{dx}}\left( {x + 2} \right) - \left( {x + 2} \right)\dfrac{d}{{dx}}\left( {{e^x}} \right)}}{{{{\left( {{e^x}} \right)}^2}}}\]
Now, \[\dfrac{d}{{dx}}\left( {x + 2} \right) = \dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( 2 \right) = 1 - 0 = 1\]
and \[\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\]
So, \[f'\left( x \right) = \dfrac{{\left( {{e^x}} \right)\left( 1 \right) - \left( {x + 2} \right)\left( {{e^x}} \right)}}{{{{\left( {{e^x}} \right)}^2}}}\]
Simplify the expression.
Take the term \[\left( {{e^x}} \right)\] as common from the numerator.
\[ \Rightarrow f'\left( x \right) = \dfrac{{\left( {{e^x}} \right)\left\{ {1 - \left( {x + 2} \right)} \right\}}}{{{{\left( {{e^x}} \right)}^2}}}\]
\[ \Rightarrow f'\left( x \right) = \dfrac{{\left( {{e^x}} \right)\left( {1 - x - 2} \right)}}{{{{\left( {{e^x}} \right)}^2}}}\]
Cancel out the term \[{e^x}\] from the numerator.
\[ \Rightarrow f'\left( x \right) = \dfrac{{ - x - 1}}{{{e^x}}}\]
\[ \Rightarrow f'\left( x \right) = - \left( {x + 1} \right){e^{ - x}}\]
Now \[\dfrac{{dy}}{{dx}} = 0\] gives \[x = - 1\]
For the interval \[\left( { - \infty , - 1} \right)\], \[f'(x)\] gives positive values. i.e., \[f'(x) > 0\]
For the interval \[\left( { - 1,\infty } \right)\], \[f'(x)\] gives negative values. i.e., \[f'(x) < 0\]
Therefore, the given function \[f\left( x \right) = \left( {x + 2} \right){e^{ - x}}\] increasing in the interval \[\left( { - \infty , - 1} \right)\] and decreasing in the interval \[\left( { - 1,\infty } \right)\] .

Hence option D is correct.

Note: Many students get confused about the condition for a function to be increasing and decreasing. They should remember that a function is increasing if the first order derivative of the function is positive for all real values of \[x\] but if the first order derivative of a function is negative for all real values of \[x\] then the function is decreasing function.