Differential Calc

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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 -

    apparently 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
    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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