The maximum value of x e^{-x} is | vedclass

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(B) Let $y = x e^{-x}$.On differentiating with respect to $x$,we get:$\frac{dy}{dx} = x \cdot (-e^{-x}) + e^{-x} \cdot (1) = e^{-x}(1 - x)$.For maximu

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$\frac{1}{e}$

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(B) Let $y = x e^{-x}$.On differentiating with respect to $x$,we get:$\frac{dy}{dx} = x \cdot (-e^{-x}) + e^{-x} \cdot (1) = e^{-x}(1 - x)$.For maximum or minimum values,we set $\frac{dy}{dx} = 0$:$e^{-x}(1 - x) = 0$.Since $e^{-x} 
eq 0$ for any real $x$,we have $1 - x = 0$,which implies $x = 1$.Now,we find the second derivative to check for maxima:$\frac{d^2y}{dx^2} = e^{-x}(-1) + (1 - x)(-e^{-x}) = -e^{-x} - e^{-x} + x e^{-x} = e^{-x}(x - 2)$.At $x = 1$:$\frac{d^2y}{dx^2} = e^{-1}(1 - 2) = -\frac{1}{e} Since the second derivative is negative at $x = 1$,the function has a local maximum at $x = 1$.The maximum value is $y(1) = 1 \cdot e^{-1} = \frac{1}{e}$.

The maximum value of $x e^{-x}$ is

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