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(D) Given $f(x) = xe^{x(1-x)}$.Taking the derivative with respect to $x$:$f'(x) = e^{x(1-x)} \cdot (1) + x \cdot e^{x(1-x)} \cdot (1-2x)$$f'(x) = e^{x

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

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(D) Given $f(x) = xe^{x(1-x)}$.Taking the derivative with respect to $x$:$f'(x) = e^{x(1-x)} \cdot (1) + 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)$$f'(x) = -e^{x(1-x)} (2x+1)(x-1)$$f'(x) = -2e^{x(1-x)} (x + \frac{1}{2})(x - 1)$.For $x \in [-\frac{1}{2}, 1]$,the term $(x + \frac{1}{2}) \ge 0$ and $(x - 1) \le 0$.Thus,the product $(x + \frac{1}{2})(x - 1) \le 0$.Since $-2e^{x(1-x)}$ is always negative,$f'(x) \le 0$ for all $x \in [-\frac{1}{2}, 1]$.Therefore,$f(x)$ is a decreasing function on $\left[ -\frac{1}{2}, 1 \right]$.

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

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