Question

Using Rolles Theorem. I have the exact problem in the comments. Pleaseee help me

Answer 1

\[f \left( x \right)=\cos x~\in~[0.2\pi]\]
\[1. \cos~ x~ is ~defined~ \in[0'2\pi]\]
\[2.f'\left( x \right)=-\sin x,\],which exists in \[\left( 0,2\pi \right)\]
3.f(0)=cos 0=1
\[f(2\pi )=\cos 2\pi=1 \]
\[f \left( 0 \right)=f \left( 2 \pi \right)\]
It satisfies all the conditions of Rolle;s theorem.
there exists at least one point in (0,2 pi) such that f'(c)=0
-sin c=0,\[\sin c=0=\sin 0,\sin \pi,\sin 2\pi ,c=0,\pi,2\pi,c=\pi \in \left( 0,2\pi \right)\]
Hence Rolle's theorem is verified.

Answer 2

Okay I understand that part. I'm confused with the second part now

Answer 3

how did you do the rest? The part about the sin c

Answer 4

f'(c)=0
-sinc=0
\[\sin c=0=\sin 0,\sin \pi,\sin 2\pi ,c=0,\pi,2\pi \]
we have to find c in open interval \[\left( 0,2\pi \right) ~that~means~c \neq 0 ~and~c \neq 2\pi \]

Answer 5

So then the only point is c= pi?

Answer 6

yes,we have to find at least one point.
they can be more.