At the local min/max points, the derivative is zero.
f' = 3ax^2 + 2bx + c = 0
Plug in x = -3 and x = 2, and that gives you two equations.
Also, use f = ax^3 + bx^2 + cx + d
Plug in the values (x = -3, f = 3) and (x = 2, f = 0).
That gives you two more equations.
Now solve the four simultaneous linear equations in a, b, c, and d.
f ' (x) = 3ax^2 + 2bx + c
f '(-2) = 0
12a -4b + c = 0 (#!1
f ' (1) = 0
3a + 2b + c = 0 (#2)
#1 - #2 ----> 9a - 6b = 0
or 3a = 2b
also (-2,3) lies on original
-8a + 4b - 2c + d = 3
and (1,0) lies on it
a + b + c + d = 0
subtract those two equations
9a -3b+3c = -3 ,or
3a - b + c = -1 , (#3)
#2 - #3 :
3b = 1
b = 1/3 , but a = 2b/3 = (2/3)(1/3) = 2/9
3(2/9) -1/3 + c = -1
c = -4/3
in a+b+c+d = 0
2/9 + 1/3 - 4/3 + d = 0
d = 7/9
so f(x) = (2/9)x^3 + (1/3)x^2 - (4/3)x + 7/9
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