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October 1, 2014

October 1, 2014

Posted by **Anastasia** on Saturday, February 26, 2011 at 10:31pm.

- Calculus-IVP -
**MathMate**, Saturday, February 26, 2011 at 10:47pmInitial value problem.

f"(x)=6x

f'(x)=3x² + C

f(x)=x³ + Cx + D

(each integration introduces one integration constant: C and D)

Given f(0)=1, and f'(0)=0

f(0) = 1 = 0³+C(0)+D => D=1

f'(0) = 0 = 3(0)² + C => C=0

Therefore

f(x)=x³+1

Since the function is a polynomial, its domain is (-∞,∞).

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