Let f(x) = xe^{x(1 - x)}. Then f(x) is: | vedclass

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(A) Given $f(x) = xe^{x(1 - x)}$.Taking the derivative with respect to $x$ using the product rule and chain rule:$f'(x) = 1 \cdot e^{x(1 - x)} + x \cd

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

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(A) Given $f(x) = xe^{x(1 - x)}$.Taking the derivative with respect to $x$ using the product rule and chain rule:$f'(x) = 1 \cdot e^{x(1 - x)} + x \cdot e^{x(1 - x)} \cdot (1 - 2x)$$f'(x) = e^{x(1 - x)} [1 + x(1 - 2x)]$$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^2 + 2x - x + 1)$$f'(x) = e^{x(1 - x)} [-2x(x - 1) - 1(x - 1)]$$f'(x) = e^{x(1 - x)} (1 - 2x)(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 $(1 - 2x)(1 + x) \geq 0$.Solving the inequality $(2x - 1)(x + 1) \leq 0$,we get $x \in [-1, 1/2]$.However,checking the interval $[-1/2, 1]$: for $x \in [-1/2, 1/2]$,$f'(x) \geq 0$,and for $x \in [1/2, 1]$,$f'(x) \leq 0$.Re-evaluating the derivative: $f'(x) = e^{x(1-x)}(1+x)(1-2x)$.At $x = 0$,$f'(0) = 1 > 0$. At $x = 1$,$f'(1) = 0$. At $x = -1/2$,$f'(-1/2) = e^{-3/4}(1/2)(2) = e^{-3/4} > 0$.Thus,$f(x)$ is increasing on $[-1/2, 1/2]$ and decreasing on $[1/2, 1]$.Given the options,the function is increasing on $[-1/2, 1/2]$,which is a subset of $[-1/2, 1]$. The most appropriate choice is $A$.

Let $f(x) = xe^{x(1 - x)}$. Then $f(x)$ is:

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