How to use Rolle's Theorem to prove exactly ONE REAL ROOT

In summary, the conversation discusses how to show that the equation 2x-1-sin(x) = 0 has exactly one real root. The first step involves using the Intermediate Value Theorem to prove the existence of at least one value of c where f(c) = 0. The next step requires using Rolle's Theorem to show that there is only one real root, but the speaker is unsure of how to proceed at this point. Clarification is also given regarding the use of f'(x) and the implications of multiple roots for f(x) = 0.
  • #1
Khayyam89
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Homework Statement


Show that the equation 2x-1-sin(x) = 0 has exactly one real root.


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The Attempt at a Solution


I first used the Intermediate Value Theorem to prove that there exists at least one c such that f '(c)=0. The next step requires Rolle's Theorem to prove that there is EXACTLY ONE REAL ROOT, but I have no idea how to proceed at this point.
 
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  • #2
I think you mean that you used the IVT to show there is a least one value of c such that f(c)=0. Not f'(c). f'(x) is never zero. Can you show that? If so what does Rolles theorem tell you if f(x)=0 had multiple roots.
 

FAQ: How to use Rolle's Theorem to prove exactly ONE REAL ROOT

What is Rolle's Theorem?

Rolle's Theorem is a mathematical theorem that states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), and if the function's value at the endpoints of the interval is equal (f(a) = f(b)), then there exists at least one point c in the open interval (a, b) where the derivative of the function is equal to 0 (f'(c) = 0).

How is Rolle's Theorem used to prove exactly one real root?

Rolle's Theorem is used to prove exactly one real root by showing that there is only one point in the interval (a, b) where the derivative of the function is equal to 0. This means that the function has a critical point at that point, and since the function is continuous, it must either have a local maximum or a local minimum at that point. This local extremum can only occur at one point, thus proving that there is only one real root.

What are the conditions for using Rolle's Theorem?

The conditions for using Rolle's Theorem are that the function must be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). Additionally, the function's value at the endpoints of the interval must be equal (f(a) = f(b)).

Can Rolle's Theorem be used to prove more than one real root?

No, Rolle's Theorem can only be used to prove exactly one real root. This is because the theorem states that there is at least one point in the interval (a, b) where the derivative of the function is equal to 0. This means that there can be more than one point where the derivative is equal to 0, but there can only be one local extremum for the function, and therefore only one real root.

Are there any limitations to using Rolle's Theorem to prove exactly one real root?

Yes, there are limitations to using Rolle's Theorem to prove exactly one real root. The function must satisfy the conditions of the theorem, which means that it must be continuous and differentiable on the specified interval. If the function does not meet these conditions, then Rolle's Theorem cannot be applied. Additionally, the theorem can only be used for univariate functions (functions with one independent variable). It cannot be applied to multivariate functions.

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