f(x) = y = ax³ + bx² + cx + d
f'(x) = 3ax² + 2bx + c
For stat pts, solve 3ax² + 2bx + c = 0
.: delta = (2b)² - 4(3a)(c) = 4b² - 12ac
Require delta > 0 for 2 distinct solutions.
i.e. 4b² - 12ac > 0
4b² > 12ac
.: b² > 3ac
Now, having turning pts at (0.5, 1) and (1.5, -1) has implications...