It is given that f'''(x) = cos x, f(0)=8, f'(0)=4 and f''(0)=9. We have to find f(x).

f'''(x) = cos x

Integrating f'''(x) gives f''(x) = sin x + C1

Integrating f''(x) gives f'(x) = -cos x + C1*x + C2

Integrating f'(x) gives f(x) = -sin x + C1*x^2/2 + C2*x + C3

Now f(0)=8, f'(0)=4 and f''(0)=9

=> f''(0) = sin 0 + C1 = 9

=> C1 = 9

f'(0) = -cos 0 + 9*0 + C2

=> -1 + C2 = 4

=> C2 = 5

f(0) = -sin 0 + 9*0^2/2 + 5*0 + C3 = 8

=> C3 = 8

**The required function f(x) = -sin x + 9*x^2/2 + 5*x + 8**