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(A) Given $f(x) = x e^{x(1 - x)}$.Applying the product rule for differentiation,we get:$f'(x) = 1 \cdot e^{x(1 - x)} + x \cdot e^{x(1 - x)} \cdot \fra

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Increasing on $\left[ -\frac{1}{2}, 1 \right]$

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(A) Given $f(x) = x e^{x(1 - x)}$.Applying the product rule for differentiation,we get:$f'(x) = 1 \cdot e^{x(1 - x)} + x \cdot e^{x(1 - x)} \cdot \frac{d}{dx}(x - x^2)$$f'(x) = e^{x(1 - x)} + x \cdot e^{x(1 - x)} \cdot (1 - 2x)$$f'(x) = e^{x(1 - x)} \{1 + x(1 - 2x)\}$$f'(x) = e^{x(1 - x)} \cdot (1 + x - 2x^2)$$f'(x) = e^{x(1 - x)} \cdot (1 - x)(1 + 2x)$Since $e^{x(1 - x)}$ is always positive for all real $x$,the sign of $f'(x)$ depends on the quadratic expression $(1 - x)(1 + 2x)$.The roots of the quadratic are $x = 1$ and $x = -\frac{1}{2}$.Testing the intervals:For $x \in \left[ -\frac{1}{2}, 1 \right]$,$f'(x) \ge 0$.Thus,$f(x)$ is increasing on the interval $\left[ -\frac{1}{2}, 1 \right]$.

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

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