Question

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).

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

i m sorry its wrong

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

2pi/3 or 4 pi/3

Answer 13

yup

Answer 14

Do I then restrict those answers based on the domain?

Answer 15

yeah

Answer 16

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