Given f"(x)=-8sin3x and f'(0)=-1 and f(0)=-6. Find f(pi/5).
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?
i mean i got stuck with f(pi/5)
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
You're on the right track, but let's go through the process step by step to make sure we get the correct answer.
Given that f''(x) = -8sin(3x) and we want to find f(pi/5), we need to integrate f''(x) twice to find f(x).
First, we integrate f''(x) once to find f'(x):
∫(-8sin(3x)) dx = -8∫sin(3x) dx
Using the integration formula for sin(ax), where a is a constant:
= -8(-1/3cos(3x))
= 8/3cos(3x) + C1
Now, we are given that f'(0) = -1. So, let's substitute x=0 into f'(x) and solve for the constant C1:
f'(0) = 8/3cos(3(0)) + C1
-1 = 8/3cos(0) + C1
-1 = 8/3 + C1
C1 = -1 - 8/3
C1 = -3/3 - 8/3
C1 = -11/3
Now we have f'(x) = 8/3cos(3x) - 11/3.
Next, we integrate f'(x) to find f(x):
∫(8/3cos(3x) - 11/3) dx
= 8/3∫cos(3x) dx - 11/3∫1 dx
Using the integration formulas for cos(ax) and 1:
= 8/3(1/3sin(3x)) - 11/3x + C2
= 8/9sin(3x) - 11/3x + C2
Now, we're given f(0) = -6. So let's substitute x=0 into f(x) and solve for the constant C2:
f(0) = 8/9sin(3(0)) - 11/3(0) + C2
-6 = 0 - 0 + C2
C2 = -6
Finally, we have f(x) = 8/9sin(3x) - 11/3x - 6.
To find f(pi/5), substitute x = pi/5 into f(x):
f(pi/5) = 8/9sin(3(pi/5)) - 11/3(pi/5) - 6
Now you can solve this equation to find the numerical value of f(pi/5).