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The minimum value of the function $f(x)=x \log x$ is
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Verified Answer
The correct answer is:
$-\frac{1}{\mathrm{e}}$
$f(x)=x \log x$
$\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}}{\mathrm{x}}+\log \mathrm{x}=1+\log \mathrm{x}$
When $1+\log \mathrm{x}=0 \Rightarrow \mathrm{x}=\frac{1}{\mathrm{e}}$
$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \Rightarrow\left[\mathrm{f}^{\prime \prime}(\mathrm{x})\right]_{\mathrm{x}=\frac{1}{\mathrm{e}}}=\mathrm{e}>0$
Thus $\mathrm{f}(\mathrm{x})$ is minimum at $\mathrm{x}=\frac{1}{\mathrm{e}}$ and
$f(x)=\left(\frac{1}{e}\right) \log \left(\frac{1}{e}\right)=\frac{-1}{e}$
$\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}}{\mathrm{x}}+\log \mathrm{x}=1+\log \mathrm{x}$
When $1+\log \mathrm{x}=0 \Rightarrow \mathrm{x}=\frac{1}{\mathrm{e}}$
$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \Rightarrow\left[\mathrm{f}^{\prime \prime}(\mathrm{x})\right]_{\mathrm{x}=\frac{1}{\mathrm{e}}}=\mathrm{e}>0$
Thus $\mathrm{f}(\mathrm{x})$ is minimum at $\mathrm{x}=\frac{1}{\mathrm{e}}$ and
$f(x)=\left(\frac{1}{e}\right) \log \left(\frac{1}{e}\right)=\frac{-1}{e}$
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