Question:
Mark the correct alternative in the following:
The function $f(x)=x^{2} e^{-x}$ is monotonic increasing when
A. $x \in R-[0,2]$
B. $0 C. $2 D. $x<0$
Solution:
$f(x)=x^{2} e^{-x}$
$\frac{\mathrm{d}(\mathrm{f}(\mathrm{x}))}{\mathrm{dx}}=x \mathrm{e}^{-\mathrm{x}}(2-\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{x})$
for
$f^{\prime}(x)=0$
$\Rightarrow x^{2} e^{-x}=0$
$\Rightarrow x(2-x)=0$
$x=2, x=0$
$f(x)$ is increasing in $(0,2)$
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