What is the Proof for Rolle's Theorem Using the Intermediate Value Theorem?

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In summary, the conversation discusses how to prove the existence of a point c within the interval [0,3] where f(c) = c for a given function that is continuous and differentiable on that interval. By using Rolle's theorem and defining a new function g(x), it is shown that there must exist a point c within [1,3] where f(c) = c. The information provided about f(0)=1 is not necessary and the problem only requires the function to be continuous. The Intermediate Value Theorem is also used to support the proof.
  • #1
rogain
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Homework Statement


A function is continuous/differentiable on the interval [0,3]. I have been given that f(0)=1, f(1)=2, f(3)=2.
I need to prove that there exists a c within the interval [0,3] such that f(c) = c.

Homework Equations



The Attempt at a Solution


f(1)=f(3)=2. From Rolle's theorem, there exists a c within [1,3] such that f`(c)=0. From this I am able to show that at some point c within [1,3] the function must be constant i.e f(c)=k.
I am not sure how to show that the constant is equal to the point i.e k=c.
 
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  • #2
The problem does not ask you about the derivative of f so Rolle's theorem is unnecessary. Also it makes no sense to say that a function is constant at a point! Every function has a single value at every point of its domain.

Define g(x)= f(x)- x. g(0)= 1- 0= 0, g(1)= 2- 1= 1, and g(3)= 2- 3= -1. That is, g(1)> 0 and g(3)< 0. What does that tell you?


(Much of the information given in this problem is unnecessary! The function does not have to be differentiable, just continuous. The information about f(0)= 1 is irrelevant. And, in fact, the desired point must be in the interval [1, 3].)
 
  • #3
HallsofIvy said:
Define g(x)= f(x)- x. g(0)= 1- 0= 0, g(1)= 2- 1= 1, and g(3)= 2- 3= -1. That is, g(1)> 0 and g(3)< 0. What does that tell you?

Thanks for the response. This helped a lot.
 
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  • #4

What is Rolle's theorem?

Rolle's theorem is a fundamental theorem in calculus that states that if a function is continuous on a closed interval and differentiable on the open interval, and the function's values are equal at the endpoints of the interval, then there exists at least one point in the interval where the derivative of the function is equal to zero.

What is the significance of Rolle's theorem?

Rolle's theorem is important because it provides a necessary condition for a function to have a critical point or a point of inflection. It also enables us to prove other useful theorems such as the mean value theorem and the fundamental theorem of calculus.

How is Rolle's theorem used in practical applications?

Rolle's theorem has various applications in physics, engineering, and economics. It is used to determine the maximum and minimum values of a function and to find the points of inflection in a curve. It also helps in optimization problems, such as finding the maximum profit or minimum cost in business.

What are the limitations of Rolle's theorem?

One limitation of Rolle's theorem is that it only provides a sufficient condition for the existence of a point where the derivative is equal to zero. It does not guarantee that there will be only one such point or that the function will not have other critical points. Additionally, the function must be continuous and differentiable on the entire interval for the theorem to be applicable.

Can Rolle's theorem be extended to higher dimensions?

Yes, Rolle's theorem can be extended to higher dimensions through the Mean Value Theorem for Multivariable Functions. This theorem states that if a function is continuous and differentiable on a closed and connected set in n-dimensional space, and the function's values are equal at the endpoints of a line segment, then there exists at least one point on the line segment where the gradient of the function is equal to zero.

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