Question

find f
f''(x)=2+cosx, f(0)=-1, f(pi/2)=0

Answer 1

can you find f' given f''

Answer 2

i got it down to the second anti-derivative x^2-cosx+Cx+D

Answer 3

@freckles

Answer 4

...? not sure how you got that. Can you give your steps

Answer 5

the answer on the back of the book is x^2-cosx-1/2pix

Answer 6

i need help with getting the -1/2pix

Answer 7

first step is
\[f'(x)=2x+\sin(x)+C\] and find \(C\)
but i think there is a typo in your question (maybe not)
are you told a value of the derivative?

Answer 8

ohhh, it's a double prime, it only looked like a single

Answer 9

yeah they are f(0)=-1 and f(pi/2)=0

Answer 10

those are both only for f not the derivative

Answer 11

yeah just f

Answer 12

ok then we have more work to do

Answer 13

\[f(x)=x^2-\cos(x)+Cx+D\] and you need \(C\) and \(D\)

Answer 14

yeah

Answer 15

D is easy enough to find, since you know
\[f(0)=-1\]

Answer 16

yeah i got that one

Answer 17

i get \(D=0\) do you?

Answer 18

no i got d=-1

Answer 19

ok
\[f(0)=0^2-\cos(0)+C\times 0+D=-1\]

Answer 20

since \(\cos(0)=1\) you have
\[-1+D=-1\]

Answer 21

oh no my mistake

Answer 22

so d=0 what about c

Answer 23

now we have
\[f(x)=x^2-\sin(x)+Cx\] and you can find \(D\) by using the fact that
\[f(\frac{\pi}{2})=0\]

Answer 24

i meant of course "find C"

Answer 25

i get c=pi/4

Answer 26

but the answer is x^2-cosx-1/2pix

Answer 27

ok that was wrong

Answer 28

but cos of pi/2 is 0

Answer 29

damn another typo
\[f(\frac{\pi}{2})=\frac{\pi^2}{4}+\frac{\pi}{2}C=0\]

Answer 30

lol

Answer 31

that makes
\[\frac{\pi}{2}C=-\frac{\pi^2}{4}\]

Answer 32

making \(C=-\frac{\pi}{2}\)