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

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

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

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(A) Given the function $f(x) = (x+2)e^{-x}$.To determine the intervals of increase and decrease,we find the derivative $f'(x)$ using the product rule:$f'(x) = \frac{d}{dx}(x+2) \cdot e^{-x} + (x+2) \cdot \frac{d}{dx}(e^{-x})$$f'(x) = 1 \cdot e^{-x} + (x+2) \cdot (-e^{-x})$$f'(x) = e^{-x}(1 - (x+2))$$f'(x) = e^{-x}(1 - x - 2)$$f'(x) = -e^{-x}(x+1)$Now,we analyze the sign of $f'(x)$:Since $e^{-x} > 0$ for all real $x$,the sign of $f'(x)$ depends on $-(x+1)$.$1$. If $x  0$. Thus,$f'(x) > 0$,and the function is monotonically increasing in $(-\infty, -1)$.$2$. If $x > -1$,then $(x+1) > 0$,so $-(x+1) Therefore,the function is increasing in $(-\infty, -1)$ and decreasing in $(-1, \infty)$.

The function $f(x)$ is defined by $f(x)=(x+2) e^{-x}$. Which of the following statements is true?

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