Find f. f ''(𝜃) = sin(𝜃) + cos(𝜃), f(0) = 4, f '(0) = 1

Answer 1

f(x) = integral of f'(x)

= -cosx + sinx + c, where c is an unknown constant

f''(0) = -1 + 0 + c = 1 , so c = 2

f'(x) = -cosx + sinx + 2

take the integral to get

f(x) = -sinx -cosx + 2x + k, where k is another unknown constant.

f(0) = 4 = 0 -1 + 0 + k

k =5

f(x) = -sinx - cosx + 2x + 5

hopefully. this is a tedious problem, where mistakes easily creep in