Question
The function defined by $$f\left( x \right) = \left( {x + 2} \right){e^{ - x}}$$ is
A.
decreasing for all $$x$$
B.
decreasing in $$\left( { - \infty , - 1} \right)$$ and increasing $$\left( { - 1,\infty } \right)$$
C.
increasing for all $$x$$
D.
decreasing in $$\left( { - 1,\infty } \right)$$ and increasing in $$\left( { - \infty , - 1} \right)$$
Answer :
decreasing in $$\left( { - 1,\infty } \right)$$ and increasing in $$\left( { - \infty , - 1} \right)$$
Solution :
$$\eqalign{
& f'\left( x \right) = - \left( {x + 2} \right){e^{ - x}} + {e^{ - x}} = - \left( {x + 1} \right){e^{ - x}} = 0 \Rightarrow x = - 1 \cr
& {\text{For}}\,x \in \left( { - \infty , - 1} \right),f'\left( x \right) > 0\,{\text{and}}\,{\text{for}}\,x \in \left( { - 1,\infty } \right),f'\left( x \right) < 0 \cr
& \therefore f\left( x \right){\text{ is increasing on}}\,\left( { - \infty , - 1} \right)\,{\text{and}}\,{\text{decreasing on}}\,\left( { - 1, - \infty } \right) \cr} .$$