Given f"(x)=-8sin3x and f'(0)=-1 and f(0)=-6. Find f(pi/5).

Question

i got stuck with f(pi/f)
f"(x)=-8sin3x

f'(x)=8cos3x+c
f'(0)=8cos3(0)+c=-1
c=-9
f'(x)=8cos3x-9

f(x)=-8sin3x-9x+c
f(0)=-8sin3(0)-9(0)+c=-6
c=-6
f(pi/5)-8sin3(pi/5)-9(pi/5)-6. so am i right?

Answer 1

i mean i got stuck with f(pi/5)

Answer 2

no, if f' = sin(nx)
f = -1/n cos(nx)

remember the chain rule: d/dx cos u = -sinu du/dx

so, ∫sin(nx) = -1/n cos(nx)

So. Let's see what we have:

f''(x) = -8sin3x
f'(x) = 8/3 cos3x + c
-1 = 8/3 + c
c = -11/3

f'(x) = 8/3 cos3x - 11/3
f(x) = 8/9 sin3x - 11/3 x + c
-6 = 0 - 0 + c
c = -6

f(x) = 8/9 sin3x - 11/3 x - 6

f(pi/5) = 8/9 sin 3pi/5 - 11/3 * pi/5 - 6
= 0.8454 - 2.3038 - 6
= -7.4584