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

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(A) Given the function $f(x) = x e^{-x}$.To find the critical points, we calculate the first derivative:$f'(x) = (1)e^{-x} + x(-e^{-x}) = e^{-x}(1 - x

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(A) Given the function $f(x) = x e^{-x}$.To find the critical points, we calculate the first derivative:$f'(x) = (1)e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$.Setting $f'(x) = 0$ for extrema:$e^{-x}(1 - x) = 0$.Since $e^{-x} 
eq 0$ for any real $x$, we have $1 - x = 0$, which gives $x = 1$.Now, we check the second derivative to confirm the nature of the extremum:$f''(x) = \frac{d}{dx}[e^{-x} - x e^{-x}] = -e^{-x} - (e^{-x} - x e^{-x}) = e^{-x}(x - 2)$.Evaluating at $x = 1$:$f''(1) = e^{-1}(1 - 2) = -e^{-1} Since the second derivative is negative at $x = 1$, the function attains a maximum value at $x = 1$.

The function $f(x) = x e^{-x}, \forall x \in R$ attains a maximum value at $x$ equal to

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