Question

Find the equation for the funciton f that has the given derivative and whose graph passes through the given point.

Answer 1

f'(x) = 2 sin (4x)

Point is \[(\pi/4, -17/2)\]

Answer 2

Hm, integrate the function?

Answer 3

good idea then find C by replacement

Answer 4

Yes, since \(x\) and \(y\) are given.

Answer 5

well integrating would be -2cos(4x) + c right?

Answer 6

f(x)=-0.5cos(4x)+C
and sub that point into it

Answer 7

then u can find C

Answer 8

ah okay

Answer 9

\[\frac{-\cos(4(\pi/4))}{2} \] which would be \[\frac{-\cos(4\pi/16)}{2}\]???

Answer 10

\[\frac{-\cos4(−17/2)}{2} = \frac{-\cos4(−68/8)}{2} \] ???

Answer 11

err forget that 4 before the (-68/8) :(

Answer 12

I think that you are confused; one value is \(x\) and the other is \(y\) you can't sub it into the same thing.

Answer 13

Yeah I'm konfused, Idk what I'm doing :(

Answer 14

so I ijust use the first point then? pi/4

Answer 15

\[-\frac{17}{2} =-0.5\cos\left(4*\frac{\pi}{4}\right) +C\]

Answer 16

Solve for \(C\)

Answer 17

\[2\int sin(4x)=\frac{1}{2}\cos(4x)+C\]
\[f(x)=\frac{1}{2}\cos(4x)+C\]
\[f(\frac{\pi}{4})=\frac{1}{2}\cos(2\pi)+C=-\frac{17}{2}\]

Answer 18

\[\frac{1}{2}+C=-\frac{17}{2}\]
\[C=\frac{-17}{2}-\frac{1}{2}=-9\] if i did not mess up

Answer 19

-0.5cos(4*pi/4)+C=-17/2
that is -0.5cos(pi)+C=-17/2
-0.5cos(pi)=0.5
0.5+C=-17/2
C=-9

Answer 20

ho ho ho i did mess up!

Answer 21

twice

Answer 22

So the equation is 12cos(4x)+C or 12cos(4x)+ (-9)?

Answer 23

f(x) = that :P

Answer 24

-0.5cos(4x)+(-9)=f(x)

Answer 25

\[2\int\sin(4x)=-\frac{1}{2}\cos(4x)+C\] and
\[f(\frac{\pi}{4})=-\frac{1}{2}\cos(\pi)+C=-\frac{17}{2}\]
\[C=-9\]

Answer 26

idk why it said 12cos LOL :( stoopid copy/paste

Answer 27

Thanks all!

Answer 28

i thk it is book's problem -.-