Given f ' ' ( x ) = − 36 sin ( 6 x ) and f ' ( 0 ) = − 5 and f ( 0 ) = − 6 . Find f ( π/ 6 )

Question

So I found f'(x) from f''(x) to be 6cos(6x)+c

since f'(0=-5 I got 6cos(6*0)+c=-5 where c=-11

so for f'(x)=6cos(6x)-11x+c

f(x)=sin(6x)-11x+c

since f(0)=-6

f(x)=sin(6*0)-11*0+c=-6

c=-6

so f(pi/6)= sin(6*(pi/6)-11(pi/6)-6=-11.7048 which is wrong..

Answer 1

f'(x) = -36 ∫sin(6x)dx = 6cos(6x) + C₁

Since f'(0) = -5, 6cos0 + C₁ = -5. So, C₁ = -11

f'(x) = 6cos(6x) - 11

f(x) = ∫f'(x)dx = sin(6x) - 11x + C₂

Since f(0) = -6, sin0 - 0 + C₂ = -6. So, C₂ = -6

f(x) = sin(6x) -11x - 6

f(π/6) = sinπ - 11(π/6) - 6 = -11π/6 - 6 ≈ -11.76