Calculus
posted by Benjamin .
Find a cubic function f(x) = ax^3 + bx^2 + cx + d that has a local maximum value of 3 at x = −2 and a local minimum value of 0 at x = 1.

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
in #3
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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