Answer:
Option A
Explanation:
We have ,
f(x)= x log x
f'(x)= 1+log x
For maxima or minima , put f '(x)= 0
$\therefore$ 1+ log x=0
$\Rightarrow$ logx =-1
$\Rightarrow$ x = e-1
Now, f "(x) = $\frac{1}{x}$
$\Rightarrow$ $ f^{''}(e^{-1})=\frac{1}{e^{-1}}, e >0$
$\therefore$ f(x) is decreasing
Hence, minimum value of f(x) at x= e-1 is
$\Rightarrow $ $f^{}(e^{-1})=e^{-1}\log e^{-1}=\frac{1}{e}(-log e)=-\frac{1}{e}$
