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 $f'(x)$ using the product rule:$f'(x) = (1)e^{-

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(A) Given the function $f(x) = x e^{-x}$.To find the critical points,we calculate the first derivative $f'(x)$ using the product rule:$f'(x) = (1)e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$.Setting $f'(x) = 0$ for maxima or minima:$e^{-x}(1 - x) = 0$.Since $e^{-x} 
eq 0$ for any $x \in R$,we have $1 - x = 0$,which gives $x = 1$.Now,we find the second derivative $f''(x)$ to check for maxima:$f''(x) = \frac{d}{dx}[e^{-x} - x e^{-x}] = -e^{-x} - [e^{-x} - x e^{-x}] = e^{-x}(x - 2)$.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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