If the function f(x)=xe^{-x}, x in R attains | vedclass

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(B) Given function: $f(x) = x e^{-x}$.To find the maximum value, we first find the derivative $f'(x)$:$f'(x) = 1 \cdot e^{-x} + x \cdot (-e^{-x}) = e^

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$\left(1, \frac{1}{e}\right)$

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(B) Given function: $f(x) = x e^{-x}$.To find the maximum value, we first find the derivative $f'(x)$:$f'(x) = 1 \cdot e^{-x} + x \cdot (-e^{-x}) = e^{-x}(1-x)$.Setting $f'(x) = 0$ gives $e^{-x}(1-x) = 0$. Since $e^{-x} 
eq 0$ for all $x \in R$, we have $1-x = 0$, which implies $x = 1$.Now, we find the second derivative $f''(x)$ to check for maxima:$f''(x) = \frac{d}{dx}[e^{-x} - x e^{-x}] = -e^{-x} - (e^{-x} - x e^{-x}) = e^{-x}(x-2)$.At $x = 1$, $f''(1) = e^{-1}(1-2) = -e^{-1} = -\frac{1}{e} Since $f''(1) Thus, $\alpha = 1$.The maximum value $\beta = f(1) = 1 \cdot e^{-1} = \frac{1}{e}$.Therefore, $(\alpha, \beta) = \left(1, \frac{1}{e}ight)$.

If the function $f(x)=xe^{-x}, x \in R$ attains its maximum value $\beta$ at $x=\alpha$, then $(\alpha, \beta)=$

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