The function $f(x)=x \mathrm{e}^{x(1-x)}, x \in \mathbb{R}$, is :
Solution
<p>$f(x) = x{e^{x(1 - x)}},\,x \in R$</p>
<p>$f'(x) = x{e^{x(1 - x)}}\,.\,(1 - 2x) + {e^{x(1 - x)}}$</p>
<p>$= {e^{x(1 - x)}}[x - 2{x^2} + 1]$</p>
<p>$= - {e^{x(1 - x)}}[2{x^2} - x - 1]$</p>
<p>$= - {e^{x(1 - x)}}(2x + 1)(x - 1)$</p>
<p>$\therefore$ $f(x)$ is increasing in $\left( { - {1 \over 2},1} \right)$ and decreasing in $\left( { - \infty ,\, - {1 \over 2}} \right) \cup \left( {1,\infty } \right)$</p>
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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