Find a cubic function f(x) = ax3 + bx2 + cx + d that has a local maximum value of 4 at x = −3 and a local minimum value of 0 at x = 1.find a cubic function f(x) = ax3 + bx2 + cx + d that has a local maximum value of 4 at x = −3 and a local minimum value of 0 at x = 1.

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).