The function f(x) = x e^{-x} for all x in R a | vedclass

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(A) To find the maximum value of the function $f(x) = x e^{-x}$, we first find its derivative with respect to $x$.
Using the product rule, $f'(x)

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$1$

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(A) To find the maximum value of the function $f(x) = x e^{-x}$, we first find its derivative with respect to $x$.
Using the product rule, $f'(x) = (1)e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$.
For critical points, set $f'(x) = 0$.
$e^{-x}(1 - x) = 0$.
Since $e^{-x} 
eq 0$ for any $x$, we have $1 - x = 0$, which gives $x = 1$.
To confirm this is a maximum, we use the second derivative test.
$f''(x) = -e^{-x}(1 - x) + e^{-x}(-1) = e^{-x}(x - 1 - 1) = e^{-x}(x - 2)$.
At $x = 1$, $f''(1) = e^{-1}(1 - 2) = -e^{-1} < 0$.
Since the second derivative is negative at $x = 1$, the function attains a local maximum at $x = 1$.
Thus, $k = 1$.

The function $f(x) = x e^{-x}$ for all $x \in R$ attains a maximum value at $x = k$, then $k = $

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