Question

f(x)=cosx+2cos^2x
defined for pi/2 12 years ago

Answer 1

\[f(x)=\cos(x)+2\cos^2(x)\] ?

Answer 2

is that the function?

Answer 3

yup!

Answer 4

did you find the derivative?

Answer 5

yup

Answer 6

i got
\[f'(x)=\sin(x)(1-4\cos(x))\]

Answer 7

I got (-sinx)(2cosx)

Answer 8

hmm

Answer 9

\[-\sin(x)-4\cos(x)\sin(x)\] is a start right?

Answer 10

maybe the problem is with the derivative of \(2\cos^2(x)\)

Answer 11

by the chain rule it is \(-4\cos(x)\sin(x)\)

Answer 12

hmm

Answer 13

actually yeah

Answer 14

ok then set it equal to zero and solve
it will be goofy

Answer 15

uh... i dont even know where to start

Answer 16

\[f'(x)=\sin(x)(1-4\cos(x))\]\[\sin(x)(1-4\cos(x))=0\]

Answer 17

right. I got up to there lol

Answer 18

so \(\sin(x)=0\) which means in your interval \(x=\frac{\pi}{2}\) which is not in fact in your interval is it?

Answer 19

so the other possibility is that
\[1-4\cos(x)=0\] giving
\[\cos(x)=\frac{1}{4}\]

Answer 20

as for that one, \(x=\cos^{-1}(\frac{1}{4})\) and i have no idea what that is , use a calculator or just leave it at that

Answer 21

oh damn i made an algebra mistake!!

Answer 22

it is \[\sin(x)(-1-4\cos(x))=0\]

Answer 23

so \[\cos(x)=-\frac{1}{4}\] or
\[x=\cos^{-1}(-\frac{1}{4})\]

Answer 24

sorry about that

Answer 25

thanks!