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

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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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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