Question
Write the point where $\text{f}(\text{x})=\text{x}\log_{\text{e}}\text{x}$ attains minimum value.

Answer

Given, $\text{f}(\text{x})=\text{x}\log_{\text{e}}\text{x}$
$\Rightarrow\text{f}'(\text{x})=\log_{\text{e}}\text{x}+1$
For a local maxima or a minima, we must have $\text{f}(\text{x})=0$
$\Rightarrow\log_{\text{e}}\text{x}+1=0$
$\Rightarrow\log_{\text{e}}\text{x}=-1$
$\Rightarrow\text{x}=\frac{1}{\text{e}}$
$\Rightarrow\text{f}\Big(\frac{1}{\text{e}}\Big)=\frac{1}{\text{e}}\ \log_{\text{e}}\Big(\frac{1}{\text{e}}\Big)=-\frac{1}{\text{e}}$
Now, $\text{f}''(\text{x})=\frac{1}{\text{x}}$
At, $\text{x}=\frac{1}{\text{e}}$
$\text{f}''\Big(\frac{1}{\text{e}}\Big)=\frac{1}{\frac{1}{\text{e}}}=\text{e}>0$
So, $\Big(\frac{1}{\text{e}},-\frac{1}{\text{e}}\Big)$ is a point of local minimum.

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