If f(x) = x e^{x(1-x)}, x in R,then f(x) is | vedclass

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(C) Given $f(x) = x e^{x(1-x)}$.Applying the product rule and chain rule,we find the derivative:$f'(x) = 1 \cdot e^{x(1-x)} + x \cdot e^{x(1-x)} \cdot

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increasing on $[-1/2, 1]$

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(C) Given $f(x) = x e^{x(1-x)}$.Applying the product rule and chain rule,we find the derivative:$f'(x) = 1 \cdot e^{x(1-x)} + x \cdot e^{x(1-x)} \cdot (1-2x)$$f'(x) = e^{x(1-x)} [1 + x - 2x^2]$$f'(x) = -e^{x(1-x)} [2x^2 - x - 1]$Factoring the quadratic expression:$f'(x) = -e^{x(1-x)} (2x + 1)(x - 1)$$f'(x) = -2 e^{x(1-x)} (x + 1/2)(x - 1)$Since $e^{x(1-x)} > 0$ for all $x \in R$,the sign of $f'(x)$ depends on the expression $-(x + 1/2)(x - 1)$.For $x \in [-1/2, 1]$,the product $(x + 1/2)(x - 1) \leq 0$.Therefore,$-(x + 1/2)(x - 1) \geq 0$.Thus,$f'(x) \geq 0$ for $x \in [-1/2, 1]$.Hence,$f(x)$ is increasing on $[-1/2, 1]$.

If $f(x) = x e^{x(1-x)}, x \in R$,then $f(x)$ is

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