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

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(A) Given function is $f(x) = x e^{x-x^2}$.To find the intervals of increase and decrease,we calculate the derivative $f'(x)$:$f'(x) = 1 \cdot e^{x-x^

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increasing in $\left(-\frac{1}{2}, 1\right)$

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(A) Given function is $f(x) = x e^{x-x^2}$.To find the intervals of increase and decrease,we calculate the derivative $f'(x)$:$f'(x) = 1 \cdot e^{x-x^2} + x \cdot e^{x-x^2} \cdot (1-2x)$$f'(x) = e^{x-x^2} [1 + x - 2x^2]$$f'(x) = e^{x-x^2} [-(2x^2 - x - 1)]$$f'(x) = -e^{x-x^2} (2x+1)(x-1)$.For the function to be increasing,$f'(x) > 0$:$-e^{x-x^2} (2x+1)(x-1) > 0$$(2x+1)(x-1) The roots are $x = -\frac{1}{2}$ and $x = 1$. The inequality holds for $x \in \left(-\frac{1}{2}, 1\right)$.Thus,the function is increasing in $\left(-\frac{1}{2}, 1\right)$.

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

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