Find the interval(s) on which $f(x) = x \log x$ is strictly decreasing for $x > 0$.

Explanation

1. 01

   $f'(x) = \log x + 1$, which is 0 at $x = e^{-1}$.

2. 02

   $f'(x) > 0$ on $(e^{-1}, \infty)$ and $f'(x) < 0$ on $(0, e^{-1})$.

3. 03

   So $f$ is strictly decreasing on $(0, e^{-1})$.