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July 28, 2014

July 28, 2014

Posted by **Daniel** on Tuesday, December 6, 2011 at 2:07am.

F''(x) = 4 + 6x + 24x^2, f(0) = 3, f(1)=10

Not sure how to find the value of the constant for F'(x).

- calculus -
**drwls**, Tuesday, December 6, 2011 at 2:42amIt is not clear if you want F'(x) or whatever you call f(x).

F'(x) = integral of F"(x)

= 4x + 3x^2 + 8x^3 + C

where C is any constant

F(x) = 2x^2 + x^3 + 2x^4 + Cx + C'

where C' is a different constant.

You could choose C and C' to make F(0) = 3 and F(1) = 10, but the problem does not ask for F(x)

F(0) = 3 = C

F(1) = 2 + 1 + 2 + 3 + C' = 10

C' = 2

F(x) = 2x^2 + x^3 + 2x^4 + 3x + 2

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