The minimum value of $f(x)=x \log x$ is :
Solution
$f(x) = x \log x$
$f'(x) = 1 + \log x$
For min$f'(x) = 0$
$= 1 + \log x = 0$
$\log x = -1$
$x = e^{-1} = x = \dfrac{1}{e}$
Now $f'(x) = \dfrac{1}{x}$
$f"(x = e^{-1}) = \dfrac{1}{e^{-1}} = e > 0$
Condition for minimum
Hence $f(x)$ has minimum value
at $x = \dfrac{1}{e}$
Minimum value $= \dfrac{1}{e} \log \dfrac{1}{e}$
$= \dfrac{1}{e}(-1) = -\dfrac{1}{e}$