Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a) = f(b).

f(x) = cot (x/2) , [π, 9π]

Question

Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a) = f(b). 

f(x) = cot (x/2) , [π, 9π]

yrelhan4 · Answer

One of the most imp conditions for rolle's th. to be applicable is the function has to continuous and differentiable all throughout. 
is this function continuous?

anonymous · Answer

thats what i dont get how do you know it is continuous or differentiable at the interval?

anonymous · Answer

@yrelhan4

yrelhan4 · Answer

you should know the graph of cotx? knowing that you can draw cot(x/2)..

anonymous · Answer

I still dont understand...

john_es · Answer

The functión Cot[x/2] is not continuos in the points 
$$x=2n\pi, \ \ \text{where} \ \ n=0,1,2,3,...$$
You can check it with a calculator. So the function is not continuos in the domain the problem gives you, so the Rolle's theorem can not be applied. If the inteverval was,
$$[\pi/2,3\pi/2]$$
Then we could apply the Rolle's theorem.