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

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(A) Given $f(x) = x e^{x(1-x)}$.Applying the product rule,$f'(x) = x \cdot e^{x(1-x)} \cdot \frac{d}{dx}(x-x^2) + e^{x(1-x)} \cdot 1$.$f'(x) = x e^{x(

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increasing in $\left[-\frac{1}{2}, 1\right]$

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(A) Given $f(x) = x e^{x(1-x)}$.Applying the product rule,$f'(x) = x \cdot e^{x(1-x)} \cdot \frac{d}{dx}(x-x^2) + e^{x(1-x)} \cdot 1$.$f'(x) = x e^{x(1-x)}(1-2x) + e^{x(1-x)}$.$f'(x) = e^{x(1-x)} [x(1-2x) + 1] = e^{x(1-x)} (x - 2x^2 + 1)$.$f'(x) = e^{x(1-x)} (-2x^2 + x + 1) = -e^{x(1-x)} (2x^2 - x - 1)$.$f'(x) = -e^{x(1-x)} (2x+1)(x-1) = e^{x(1-x)} (2x+1)(1-x)$.For $f(x)$ to be increasing,$f'(x) \geq 0$.Since $e^{x(1-x)} > 0$ for all $x \in R$,we need $(2x+1)(1-x) \geq 0$.This inequality holds when $x \in \left[-\frac{1}{2}, 1\right]$.Thus,$f(x)$ is increasing in $\left[-\frac{1}{2}, 1\right]$.

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

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