Answer: The formula is f(x) = a x ^ 3 + b x ^ 2 + c x + d
f '(x) = 3ax^2 + 2bx + c.
f(- 3) = 3 ==> - 27a + 9b - 3c + d = 3
f '(- 3) = 0 (being a most extreme) ==> 27a - 6b + c = 0.
f(1) = 0 ==> a + b + c + d = 0
f '(1) = 0 (being a base) ==> 3a + 2b + c = 0.
-
Along these lines, we have the four conditions
- 27a + 9b - 3c + d = 3
a + b + c + d = 0
27a - 6b + c = 0
3a + 2b + c = 0
Subtracting the last two conditions yields 24a - 8b = 0 ==> b = 3a.
Along these lines, the last condition yields 3a + 6a + c = 0 ==> c = - 9a.
Consequently, we have from the initial two conditions:
- 27a + 9(3a) - 3(- 9a) + d = 3 ==> 27a + d = 3
a + 3a - 9a + d = 0 ==> d = 5a.
Along these lines, a = 3/32 and d = 15/32.
==> b = 9/32 and c = - 27/32.