Explain why rolle's theorem does not apply to the function even though there exist a and b such that f(a) = f(b). (select all that apply.) f(x) = cot x 2 , [π, 9π] there are points on the interval [a, b] where f is not continuous. none of these. there are points on the interval (a,
b.where f is not differentiable. f(a) does not equal f(b) for all possible values of a and b in the interval [π, 9π]. f '(a) does not equal f '(b) for any values in the interval [π, 9π].

The reason Rolle's theorem does not apply to the function cot x^2 on the given interval is that on the interval there are points where f is not continuous, and as such not differentiable either.

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MrRoyal MrRoyal · Answer 2 · 2021-09-25T11:00:17+02:00

Rolle's theorem is used to determine if a function is continuous and also differentiable.

The reasons Rolle's theorem does not apply to [tex]f(x) = \cot x^2\ [\pi, 9\pi][/tex] are:

There are points on the interval [a, b] where f is not continuous.
There are points on the interval (a, b) where f is not differentiable.

The conditions for Rolle's theorem to be true are:

[tex]f(a) = f(b)[/tex].
[tex]f(x)[/tex] must be continuous in [tex][a,b][/tex].
[tex]f(x)[/tex] must be differentiable in [tex](a,b)[/tex].

The given function is a cotangent function.

Cotangent functions are not continuous at [tex]x = n\pi[/tex], where [tex]n \ne 0[/tex].

This means that [tex]f(x) = \cot x^2\ [\pi, 9\pi][/tex] will not be continuous at [tex]\pi, 2\pi, 3\pi.....,9\pi[/tex]

Also;

Cotangent functions are not differentiable at [tex]x = n\pi[/tex], where [tex]n \ne 0[/tex].

This means that [tex]f(x) = \cot x^2\ [\pi, 9\pi][/tex] will not be differentiable at [tex]2\pi, 3\pi.....,8\pi[/tex]

Hence, Rolle's theorem does not apply for the given function because:

There are points on the interval [a, b] where f is not continuous.
There are points on the interval (a, b) where f is not differentiable.