The minimum value of the function $f(x)=x \log x$ is

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Aicharya · Accepted Answer

$f(x)=x \log x$ $	herefore \quad \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}}{\mathrm{x}}+\log \mathrm{x}=1+\log \mathrm{x}$ When $1+\log \mathrm{x}=0 \Rightarrow \mathrm{x}=\frac{1}{\mathrm{e}}$ $\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \Rightarrow\left[\mathrm{f}^{\prime \prime}(\mathrm{x})ight]_{\mathrm{x}=\frac{1}{\mathrm{e}}}=\mathrm{e}>0$ Thus $\mathrm{f}(\mathrm{x})$ is minimum at $\mathrm{x}=\frac{1}{\mathrm{e}}$ and $f(x)=\left(\frac{1}{e}ight) \log \left(\frac{1}{e}ight)=\frac{-1}{e}$

The minimum value of the function $f(x)=x \log x$ is

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