The function f defined by f(x) = (x + 2)e^{-x | vedclass

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(D) Given $f(x) = (x + 2)e^{-x}$.To find the intervals of increase and decrease,we find the derivative $f'(x)$ using the product rule:$f'(x) = (1)e^{-

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

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(D) Given $f(x) = (x + 2)e^{-x}$.To find the intervals of increase and decrease,we find the derivative $f'(x)$ using the product rule:$f'(x) = (1)e^{-x} + (x + 2)(-e^{-x})$$f'(x) = e^{-x}(1 - x - 2)$$f'(x) = -e^{-x}(x + 1)$.For the function to be increasing,$f'(x) > 0$:$-e^{-x}(x + 1) > 0$Since $e^{-x} > 0$ for all $x$,we must have $-(x + 1) > 0$,which implies $x + 1 Thus,the function is increasing in $(-\infty, -1)$.For the function to be decreasing,$f'(x) $-e^{-x}(x + 1) Since $e^{-x} > 0$ for all $x$,we must have $-(x + 1)  0$,so $x > -1$.Thus,the function is decreasing in $(-1, \infty)$.Therefore,the function is increasing in $(-\infty, -1)$ and decreasing in $(-1, \infty)$.

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

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