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April 16, 2014

April 16, 2014

Posted by **Benjamin** on Monday, November 12, 2012 at 6:05pm.

- Calculus -
**PayDay**, Monday, November 12, 2012 at 6:15pmAt 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.

- Calculus -
**Reiny**, Monday, November 12, 2012 at 7:00pmf ' (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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