Question

find the derivative of f(x) = cos(x) − sin^2(x) on [0,2pi]

Answer 1

i now know that its -sin(x)(2cos(x)+1)

Answer 2

can yu guys help me find the critical points

Answer 3

Are you sure it ask you to find f'(x) on [a,b]?

Answer 4

If so, you did that. You are done.

Answer 5

yes. may you help me find the critical points of this now

Answer 6

Set f'(x)=0 and solve for x to find the critical numbers.

Answer 7

You only need to solve this on the interval [0,2 pi]

Answer 8

can we do it together please

Answer 9

You are already factored f'(x)
Now set both factors equal to 0.

Answer 10

how?
can we work out the problem

Answer 11

step by step please

Answer 12

Yes. I'm walking you through it.
You have -sin(x)*(1+2cos(x))=0
Set both factors equal to 0.

Answer 13

ok

Answer 14

Just like if you have a*b=0 then either a=0 or b=0 or both=0

Answer 15

http://www.wolframalpha.com/input/?i=-sin%28x%29%282cos%28x%29%2B1%29%3D0

Answer 16

this is all i have

Answer 17

im confused about the results

Answer 18

Use your unit circle.
Keep in mind you only have to look between 0 and 2pi

Answer 19

You have -sin(x)=0 or 1+2cos(x)=0
Solve both of these for the trig function.
Solve the first one for sin(x) and solve the second one for cos(x).
Tell me what you get after doing this.

Answer 20

-sinx=pin

Answer 21

\[\pi n\]

Answer 22

Well you are trying to solve -sin(x)=0 for x.
You could divide both sides by -1 since -1 will never be 0.
Try solving this: sin(x)=0
Then we will come back to 1+2cos(x)=0