What is the least value of $\sin 2x$ on $[0, \pi]$?

Explanation

1. 01

   $f'(x) = 2 \cos 2x = 0$ inside the interval at $x = \frac{\pi}{4}, \frac{3 \pi}{4}$.

2. 02

   Values: $f(0) = 0$, $f(\pi) = 0$, $f(\frac{\pi}{4}) = 1$, $f(\frac{3 \pi}{4}) = -1$; the least value is $-1$.