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

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

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

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

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

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