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.) (1), There are points on the interval (a, b) where f is not differentiable. There are points on the interval [a, b] where f is not continuous. Of(a) does not equal f(b) for all possible values of a and b in the interval [π, 3π]. f'(a) does not equal f'(b) for any values in the interval [z, 3π]. None of these. DOO f(x) = cot [П, ЗП]

College Algebra (MindTap Course List)
12th Edition
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Author:R. David Gustafson, Jeff Hughes
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Chapter3: Functions
Section3.3: More On Functions; Piecewise-defined Functions
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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
[π, 3π]
There are points on the interval (a, b) where f is not differentiable.
There are points on the interval [a, b] where f is not continuous.
Of(a) does not equal f(b) for all possible values of a and b in the interval [π, 3π].
O f'(a) does not equal f'(b) for any values in the interval [z, 3π].
None of these.
Transcribed Image Text: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 [π, 3π] There are points on the interval (a, b) where f is not differentiable. There are points on the interval [a, b] where f is not continuous. Of(a) does not equal f(b) for all possible values of a and b in the interval [π, 3π]. O f'(a) does not equal f'(b) for any values in the interval [z, 3π]. None of these.
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