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

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SUMMARY

The discussion focuses on proving that the equation 2x - 1 - sin(x) = 0 has exactly one real root using Rolle's Theorem. The Intermediate Value Theorem (IVT) is first applied to establish the existence of at least one root. The key point is that the derivative f'(x) is never zero, which is crucial for applying Rolle's Theorem to conclude that there can only be one real root for the function.

PREREQUISITES
  • Understanding of the Intermediate Value Theorem (IVT)
  • Knowledge of Rolle's Theorem
  • Basic calculus concepts, including derivatives
  • Familiarity with trigonometric functions, specifically sin(x)
NEXT STEPS
  • Study the application of Rolle's Theorem in various functions
  • Explore the Intermediate Value Theorem in more complex scenarios
  • Investigate the properties of derivatives and their implications on function behavior
  • Review examples of proving the existence of roots in equations
USEFUL FOR

Students studying calculus, particularly those focusing on real analysis and the properties of continuous functions. This discussion is beneficial for anyone looking to deepen their understanding of root-finding techniques in mathematical functions.

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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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.
 

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