Initial 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 (-∞,∞).
let d2y/dx2= 6x. Find a solution to the differential equation that is continuous for negative infinity to positive infinity and whose graph passes through the point (0,1) and has a horizontal tangent there.
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