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If $f(x)=x \ell \operatorname{n} x$, then $f(x)$ attains minimum value at which one of the following points?
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Verified Answer
The correct answer is:
$x=e^{-1}$
Let $f(x)=x \ln x$
$f^{\prime}(x)=\frac{x}{x}+\ln x=1+\ln x$
Put $f(x)=0 \Rightarrow 1+\ln x=0$
$\Rightarrow \ln x=-1 \Rightarrow x=\mathrm{e}^{-1}$
Now, $f^{\prime \prime}(x)=\frac{1}{x}$
$\left.f^{\prime \prime}(x)\right|_{x=e^{-1}}=\frac{1}{e^{-1}}=e>0$
Hence, $f(x)$ attains minimum value at $x=e^{-1}$.
$f^{\prime}(x)=\frac{x}{x}+\ln x=1+\ln x$
Put $f(x)=0 \Rightarrow 1+\ln x=0$
$\Rightarrow \ln x=-1 \Rightarrow x=\mathrm{e}^{-1}$
Now, $f^{\prime \prime}(x)=\frac{1}{x}$
$\left.f^{\prime \prime}(x)\right|_{x=e^{-1}}=\frac{1}{e^{-1}}=e>0$
Hence, $f(x)$ attains minimum value at $x=e^{-1}$.
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