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.

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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).