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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.)

Explain why Rolle's Theorem does not apply to the function even though there exist-example-1
User FBidu
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\begin{gathered} f(x)=\cot ((x)/(2)) \\ \lbrack\pi,9\pi\rbrack \end{gathered}

Let see what Rolle's theorem mean:

Let f(x) be a function that satisfies the following 3 hypotheses:

1. f(x) is continuous on the closed interval [a,b]

2. f(x) is differentiable on the open interval (a,b)

3. f(a) = f(b)

Then there is a number c in (a,b) such that f'(c) = 0

Let's check every condition:

1.

As we can see, the function is not continuous for:


2\pi,4\pi,6\pi,8\pi

2.

Since the function is not continuous for some points over the interval, we can also conclude the function is not differentiable for (a,b).

3.

We can see that f(a) = f(b).

Therefore, the answers are:

There are points on the interval (a,b) where f is not diffenrentiable

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

Explain why Rolle's Theorem does not apply to the function even though there exist-example-1
User Jerry Horton
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