Question:
The minimum value of $x \log _{e} x$ is equal to
(a) $\mathrm{e}$
(b) $1 / \mathrm{e}$
(c) $-1 / e$
(d) $2 / e$
(e) $-\mathrm{e}$
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 $f(x)=f\left(e^{-1}\right)$.
$\Rightarrow e^{-1} \log _{e}\left(e^{-1}\right)=-e^{-1}=\frac{-1}{e}$