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

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

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

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(B) Given $f(x) = x \cdot e^{x-x^2}$.To find the intervals of increase or decrease,we compute the derivative $f'(x)$.Using the product rule and chain rule: $f'(x) = 1 \cdot e^{x-x^2} + x \cdot e^{x-x^2} \cdot (1-2x)$.$f'(x) = e^{x-x^2} [1 + x(1-2x)] = e^{x-x^2} [1 + x - 2x^2]$.Factoring the quadratic expression: $1 + x - 2x^2 = -(2x^2 - x - 1) = -(2x+1)(x-1) = (2x+1)(1-x)$.So,$f'(x) = e^{x-x^2} (2x+1)(1-x)$.Since $e^{x-x^2} > 0$ for all $x \in R$,the sign of $f'(x)$ depends on $(2x+1)(1-x)$.$f'(x) > 0$ when $(2x+1)(1-x) > 0$,which implies $-\frac{1}{2} Thus,$f(x)$ is increasing in the interval $\left(-\frac{1}{2}, 1\right)$.

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

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