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September 3, 2014

September 3, 2014

Posted by **Joey** on Monday, November 5, 2012 at 7:11pm.

0 = f(1) = a+b+c+d

2 = f(-2) = -8a+4b-2c+d

0 = f'(1) = 3a+2b+c

0 = f'(-2) = 12a-4b+c

"since we now have four equations and four unknowns, we can determine unique solution. Solving for "a" and "b", a=, b="

Please show the steps, I am so confused and no one knows how to do it.

Thank you so much in advance!

- Differential Calc -
**Steve**, Monday, November 5, 2012 at 9:39pmapparently f(x) = ax^3 + bx^2 + cx + d

so f'(x) = 3ax^2 + 2bx + c

and you were given f(1), f(-2) and f'(1), f'(-2)

take the last two equations, equate c:

3a+2b = 12a-4b

6b = 9a

2b = 3a

take the first two equations, equate d:

a+b+c = -8a+4b-2c

so,

2a+2b+2c = -16a+8b-4c

but 2b = 3a, so

2a+3a+c = -16a + 12a - 4c

9a = -5c

start substituting those values back in, and you get

a,b,c,d = 4/27, 2/9, -8/9, 14/27

y = 1/27 (4x^3 + 6x^2 - 24x + 14)

= 2/27(x-1)^2 (2x+7)

f(1) = 0 and f(-2) = 2

y' = 4/9 (x^2 + x - 2) = 4/9 (x+2)(x-1)

it's easy to see that f'(1) = f'(-2) = 0

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