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

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Aicharya · Accepted Answer

$\begin{aligned} \mathrm{f}(x) & =x \mathrm{e}^{x(1-x)} \ 	herefore \quad \mathrm{f}^{\prime}(x) & =x \mathrm{e}^{x(1-x)}[x(-1)+(1-x)]+\mathrm{e}^{x(1-x)} \ & =\mathrm{e}^{x(1-x)}\left(x-2 x^2+1ight) \end{aligned}$ For $\mathrm{f}(x)$ to be increasing, $\mathrm{f}^{\prime}(x) \geq 0$ $\begin{aligned} & \Rightarrow \mathrm{e}^{x(1-x)}\left(x-2 x^2+1ight) \geq 0 \ & \Rightarrow x-2 x^2+1 \geq 0 \ & \Rightarrow 2 x^2-x-1 \leq 0 \ & \Rightarrow(2 x+1)(x-1) \leq 0 \ & \Rightarrow x \in\left[-\frac{1}{2}, 1ight] \end{aligned}$ For $\mathrm{f}(x)$ to be decreasing, $\mathrm{f}^{\prime}(x) \leq 0$ $\begin{aligned} & \Rightarrow(2 x+1)(x-1) \geq 0 \ & \Rightarrow x \in\left(-\infty,-\frac{1}{2}ight] \cup[1, \infty) \end{aligned}$

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

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