What Does Rolle's Theorem Mean for Differentiable Functions on Closed Intervals?

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Rolle's Theorem requires three hypotheses to be verified: the function must be continuous on a closed interval, differentiable on the open interval, and equal at the endpoints. For the function f(x) = sin(2πx) on the interval [-1, 1], it is continuous and differentiable. The derivative is correctly calculated as f'(x) = 2πcos(2πx). To find the numbers c that satisfy the conclusion of Rolle's Theorem, one must set the derivative equal to zero, leading to the equation 2πcos(2πc) = 0. The discussion emphasizes understanding the theorem's requirements and correctly applying the derivative.
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verify that the 3 hypothesis of rolle's theorem on the given interval . then find all the numbers c that satisfy the conclusion of rolle's theorem.

f(x)=sin2piex [-1,1]

i found the derivative cos 2pie x, but what do i do, and what does the theorem mean when f is differentiable on the open interval (a,b)
 
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what do i do
 
First you go back and read the statement of Rolle's theorem? You are asked to "verify that the 3 hypothesis of rolle's theorem [are true] on the given interval". What are the 3 hypotheses? Being differentiable on the interior of the interval is one of them.

You are also asked to "find all the numbers c that satisfy the conclusion of rolle's theorem." Okay, what is the conclusion of Rolle's theorem?
 
well all these are true, so how do u find all numbers c
 
afcwestwarrior said:
well all these are true, so how do u find all numbers c

f'(c)=0

10 chars...
 
Again, what is the conclusion of Rolle's theorem?
 
afcwestwarrior said:
verify that the 3 hypothesis of rolle's theorem on the given interval . then find all the numbers c that satisfy the conclusion of rolle's theorem.

f(x)=sin2piex [-1,1]

i found the derivative cos 2pie x, but what do i do, and what does the theorem mean when f is differentiable on the open interval (a,b)

No The derivative of \sin (2\pi x ) is not that. Using the chain rule it is 2\pi \cos (2\pi x)
 

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