The function f(x) = (x + 2) e^{-x} is | vedclass

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(C) Given $f(x) = (x + 2) e^{-x}$.To determine the intervals of increase and decrease,we find the derivative $f'(x)$:$f'(x) = (x + 2) \frac{d}{dx}(e^{

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increasing in $(-\infty, -1)$ and decreasing in $(-1, \infty)$

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(C) Given $f(x) = (x + 2) e^{-x}$.To determine the intervals of increase and decrease,we find the derivative $f'(x)$:$f'(x) = (x + 2) \frac{d}{dx}(e^{-x}) + e^{-x} \frac{d}{dx}(x + 2)$$f'(x) = (x + 2)(-e^{-x}) + e^{-x}(1)$$f'(x) = e^{-x} [-(x + 2) + 1] = e^{-x}(-x - 1) = -e^{-x}(x + 1)$.Since $e^{-x} > 0$ for all $x \in \mathbb{R}$,the sign of $f'(x)$ depends on $-(x + 1)$.For $f(x)$ to be increasing,$f'(x) > 0 \Rightarrow -(x + 1) > 0 \Rightarrow x + 1 Thus,$f(x)$ is increasing in $(-\infty, -1)$.For $f(x)$ to be decreasing,$f'(x)  0 \Rightarrow x > -1$.Thus,$f(x)$ is decreasing in $(-1, \infty)$.Therefore,the correct option is $C$.

The function $f(x) = (x + 2) e^{-x}$ is

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