Thank you for reporting, we will resolve it shortly
Q.
The minimum value of $x \log x$ is equal to
ManipalManipal 2018
Solution:
Let $y=x \log _{e} x$
On differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=x \cdot \frac{1}{x}+\log x=(1+\log x)$
Again, differentiating, we get
$\frac{d^{2} y}{d x^{2}}=\frac{1}{x}$
Put, $\frac{d y}{d x}=0$ for maxima or minima.
$\Rightarrow 1+\log x=0$
$\Rightarrow x=\frac{1}{e}$
$\therefore \left(\frac{d^{2} y}{d x^{2}}\right)_{\left(x=\frac{1}{c}\right)}=e$
$\therefore y$ is minimum at $x=\frac{1}{e}$
$\therefore y_{\min }=\frac{1}{e} \log _{e}\left(\frac{1}{e}\right)=-\frac{1}{e}$