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

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

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

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

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

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