Question

Find the critical numbers of the function on the interval 0 ≤ θ < 2π.
f(θ) = 2cos(θ) + (sin(θ))^2

θ = (smaller value)

θ = (larger value)

Answer 1

Find f'.

Answer 2

Have you done that yet?

Answer 3

I will give you this:
\[(\cos(\theta))'=-\sin(\theta)\]
\[(\sin(\theta))'=\cos(\theta)\]
\[([g(\theta)]^2)'=2[g(\theta)]^{2-1}=2g(\theta) \]

Answer 4

Yes but im having trouble when i make the derivative equal to 0

Answer 5

So you got:
\[f'(\theta)=-2\sin(\theta)+2\sin(\theta)(\cos(\theta)) ?\]

Answer 6

Are you still there @dainel40 ?

Answer 7

yes sorry

Answer 8

yes I got that

Answer 9

@myininaya still there?

Answer 10

Ok set equal to 0
Look and see that both terms have the common factor sin(theta)

Answer 11

then factor the expression using that.

Answer 12

factor out a \(2\sin(\theta)\)

Answer 13

hello! i didn't see you

Answer 14

I was away for a sec.

Answer 15

im getting f'(0) = pi not sure if its right

Answer 16

\[f'(\theta)=0 \]
when theta=?
Did you factor the expression using what satellite or I said? Either way will work but satellite's looks prettier.

Answer 17

You should get two values.

Answer 18

yes, im getting pi

Answer 19

umm

Answer 20

\[2 \sin(\theta)(-1+\cos(\theta))=0\] ?
Setting both equal to 0?
Remember you are looking at the interval [0,2pi)

Answer 21

ok im getting pi, 2pi now

Answer 22

I define a critical number c as a number that satisfies f'(c)=0 or f'(c) dne and also c exists as an element of the domain of f.

Answer 23

Does 2pi exist in the domain of f?

Answer 24

no because it is not included

Answer 25

So our only critical numbers are?

Answer 26

pi

Answer 27

and ?

Answer 28

there is another number you can plug in that will give you 0

Answer 29

what is it

Answer 30

0?

Answer 31

yep.

Answer 32

so are critical numbers are 0 and pi

Answer 33

yay that was correct!!!

Answer 34

thanks alot!!!

Answer 35

lol

Answer 36

np