Question

Given f′′(x)=−9sin(3x) and f′(0)=−4 and f(0)=0.

Find f(π/5)=

Answer 1

I got f'(x)=-3cos(3x)+c
c=-1
f(x)=sin(3x)-x-D
D=0
F(pi/5)=0.3227

Answer 2

the correct answer is -3.447

Answer 3

I don't know what i did wrong

Answer 4

If f''(x)=-9sin(3x) then f'(x)=acos(3x)+c
Now differentiate acos(3x) to get a: -3asin(3x)=-9sin(3x), so a=3.

Answer 5

what is acos?

Answer 6

acos is a*cos

Answer 7

f''(x)=−9sin(3x)
f'(x) = 3cos(3x)+C
-4=3+C
C=-7

f'(x) = 3cos(3x)-7
f(x) = sin(3x)-7x+D
D=0

f(x) = sin(3x)-7x+0
f(π/5)=sin(3*π/5)-7*π/5

f(π/5)=_______

Answer 8

So your f' is right.
Now f'(x)=3cos(3x)+c, f'(0)=-4, so 3cos(0)+c=-4, so 3+c=-4, so c=-7.

f(x)=sin(3x) -7x + d.

f(0)=0: sin(0)-7*0+d=0, so d=0

f(x)=sin(3x)-7x.

f(pi/5)=sin(3pi/5)-7pi/5 which is about 0.951-
4.398=-3.447

Answer 9

You had -x instead of -7x in f.

Answer 10

oh ok