## anonymous 3 years ago Consider f(x)=sin^2(x)-cos(x) on the interval 0 is less than or equal to x which is less than or equal to pi. Determine the local extrema of f(x).

1. anonymous

$f(\theta)=\sin^{2}(\theta)-\cos(\theta)$ $0\le \theta \le \pi$

2. anonymous

I know one of the local extrema is at x=0. There are two more.

3. anonymous

in order to find the local extrema first differentiate f(x) with respect to x and then put f '(x) = 0 this will help u in getting local extrema

4. anonymous

I get:$\sin(\theta)(2\cos(\theta)+1)=0$

5. anonymous

I know the sin (theta) piece is what give the local extrema at x=0. But I can't figure out the cos part.

6. anonymous

ok 2{cos (theta) + 1} = 0 now can we write cos(theta) + 1 = 0?

7. anonymous

cos(theta) = -1 cos(theta) = cos(n* pi) where n is an odd integer is this step clear to u ?

8. anonymous

i m sorry its wrong

9. anonymous

Yes, I am trying to find the angle that gives the odd integer?

10. anonymous

let me correct its 2cos(theta) +1 = 0 2cos(theta) = -1 cos(theta) = -1/2 is this step clear to u?

11. anonymous

as we know cosine function is negative in second and third quadrant therefore , value of theta must lie in pi/2 < theta < 3*pi/2

12. anonymous

2pi/3 or 4 pi/3

13. anonymous

yup

14. anonymous

Do I then restrict those answers based on the domain?

15. anonymous

yeah

16. anonymous

So then 2pi/3 is the only one that works?