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. That is, f(x) = (1/32)(3x^3 + 9x^2 - 27x + 15).
