The minimum value

Solution:

(c) $\frac{-1}{e}$

Here,

$f(x)=x \log _{e} x$

$\Rightarrow f^{\prime}(x)=\log _{e} x+1$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow \log _{e} x+1=0$

$\Rightarrow \log _{e} x=-1$

$\Rightarrow x=e^{-1}$

Now,

$f^{\prime \prime}(x)=\frac{1}{x}$

$\Rightarrow f^{\prime \prime}\left(e^{-1}\right)=e>0$

So, $x=e^{-1}$ is a local minima.

Hence, the minimum value of $f(x)=f\left(e^{-1}\right)$.

$\Rightarrow e^{-1} \log _{e}\left(e^{-1}\right)=-e^{-1}=\frac{-1}{e}$