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

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(A) Given $f(x) = x e^{x-x^2}$.To determine the intervals of increase or decrease,we find the derivative $f'(x)$.Using the product rule and chain rule

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

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(A) Given $f(x) = x e^{x-x^2}$.To determine the intervals of increase or decrease,we find the derivative $f'(x)$.Using the product rule and chain rule:$f'(x) = 1 \cdot e^{x-x^2} + x \cdot e^{x-x^2} \cdot (1-2x)$$f'(x) = e^{x-x^2} [1 + x(1-2x)]$$f'(x) = e^{x-x^2} [1 + x - 2x^2]$$f'(x) = e^{x-x^2} [-(2x^2 - x - 1)]$$f'(x) = -e^{x-x^2} (2x+1)(x-1)$For $f(x)$ to be increasing,$f'(x) \ge 0$.Since $e^{x-x^2} > 0$ for all $x \in R$,we need $-(2x+1)(x-1) \ge 0$,which implies $(2x+1)(x-1) \le 0$.The roots are $x = -\frac{1}{2}$ and $x = 1$.The inequality holds for $x \in \left[-\frac{1}{2}, 1\right]$.Thus,$f(x)$ is increasing on $\left[-\frac{1}{2}, 1\right]$.

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

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