We solve the equation $ f(x) = 0 $ on the domains $ x \le 3 $ and $ x > 3 $. If $ x \le 3, $ then $ f(x) = 2x + 1, $ so we want to solve $ 2x + 1 = 0 $. The solution is $ x = -1/2, $ which satisfies $ x \le 3 $. If $ x > 3, $ then $ f(x) = 8 - 4x, $ so we want to solve $ 8 - 4x = 0 $. The solution is $ x = 2, $ but this value does not satisfy $ x > 3 $. Therefore, the only solution is $ x = \boxed{-\frac{1}{2}} $.