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 on both f(x) and f'(x). The following can be deduced:
f(0.5) = 1 --> a/8 + b/4 + c/2 + d = 1 --> a + 2b + 4c + 8d = 8 ...(1)
f(1.5) = -1 --> 27a/8 + 9b/4 + 3c/2 + d = -1 --> 27a + 18b + 12c + 8d = -8 ...(2)
f'(0.5) = 0 --> 3a/4 + b + c = 0 --> 3a + 4b + 4c = 0 ...(3)
f'(1.5) = 0 --> 27a/4 + 3b + c = 0 --> 27a + 12b + 4c = 0 ...(4)
From here it's just a matter of solving these simultaneous equations. Time for an algebraic crunch:
(4) - (3):
24a + 8b = 0
i.e. 3a + b = 0 ...(5)
(2) - (1):
26a + 16b + 8c = -16
but from (3): 4c = -3a - 4b --> 8c = - 6a - 8b
then 26a + 16b - 6a - 8b = -16
20a + 8b = -16
i.e. 5a + 2b = - 4 ...(6)
2*(5) - (6):
6a + 2b - 5a - 2b = 4
.: a = 4
put a = 4 in (5):
12 + b = 0
.: b = -12
put a = 4 and b = -12 in (1) and (2):
4 - 24 + 4c + 8d = 8
i.e. c + 2d = 7 ...(7)
& 108 -216 + 12c + 8d = -8
i.e. 3c + 2d = 25 ...(8)
(8) - (7):
2c = 18
.: c = 9
put c = 9 in (7):
9 + 2d = 7
.: d = -1
Hence the required equation is y = 4x³ - 12x² + 9x - 1.
