## anonymous 3 years ago find the derivative of f(x) = cos(x) − sin^2(x) on [0,2pi]

1. anonymous

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

2. anonymous

can yu guys help me find the critical points

3. myininaya

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

4. myininaya

If so, you did that. You are done.

5. anonymous

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

6. myininaya

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

7. myininaya

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

8. anonymous

can we do it together please

9. myininaya

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

10. anonymous

how? can we work out the problem

11. anonymous

12. myininaya

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

13. anonymous

ok

14. myininaya

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

15. anonymous
16. anonymous

this is all i have

17. anonymous

18. myininaya

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

19. myininaya

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.

20. anonymous

-sinx=pin

21. anonymous

$\pi n$

22. myininaya

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