The function f(x) = x e ^ x(1 - x) , x ∈ R , is

Let $f\left(x\right) = x{e}^{x\left(1 - x\right)}$ then $f\left(x\right)$ is:

Now $f\left(x\right) = x{e}^{x\left(1 - x\right)}$

Differentiating both side w.r.t $x$ we get,

${f}^{'}\left(x\right) = x{e}^{x\left(1 - x\right)}\left(1 - 2x\right) + {e}^{x\left(1 - x\right)}$

$= - {e}^{x\left(1 - x\right)}\left(2{x}^{2} - x - 1\right)$

$= - {e}^{x\left(1 - x\right)}\left(2x + 1\right)\left(x - 1\right)$

Now finding critical point by equating ${f}^{'}\left(x\right) = 0$

We get, $x = \dfrac{- 1}{2} \text{ and } x = 1$

$∴ f\left(x\right)$ is decreasing in \left(- \infty , - \dfrac{1}{2}\right) \cup \left(1, \infty\right)$</p><p>And increasing in \left(- \dfrac{1}{2}, 1\right)$