Mark M. answered 12/02/22
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
f'(x) = ∫f"(x)dx = -cosx + sinx + C
Since f'(0) = 5, we have -1 + 0 + C = 5. So, C = 6
f'(x) = -cosx + sinx + 6
f(x) = ∫f'(x)dx = -sinx - cosx + 6x + K
Since f(0) = 10, we get 0 - 1 + 0 + K = 10. So, K = 11
Therefore, f(x) = -sinx - cosx + 6x + 11