Find f. f '''(x) = cos(x), f(0) = 4, f '(0) = 2, f ''(0) = 6

Given that every integration has constant C, and is solved by the 0 conditions stated in the problem;
f'''(x) = Cos(x)
f''(x) = sin(x) + 6 after f''(0) = 6
f'(x) = -cos(x) + 6x + 2 after f'(0) = 2
f(x) = -sin(x) + 3x^2 + 2x + 4 after f(0) = 4