Expert: Paul Klarreich - 3/15/2006

Question
Im really not sure at all what to do here. PLease help me.

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

Thanks in advance for your help.

Answer
Hi, Chevonne,
You wrote:
Subject:  IVT/Rolle's Theorm
Question:  Im really not sure at all what to do here. PLease help me.

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

Thanks in advance for your help.
--------------------------------
Here's what you have to do:

1. Show that the equation has AT LEAST ONE real root.
2. Show that it can't have more than one.

For part 1, you want to use the IVT, as you said.  That means, you must show:

f(x) = 2x - 1 - sin x

(1) is continuous everywhere.
(2) is negative somewhere.
(3) is positive somewhere.

Then, since zero is intermediate between negative and positive numbers, there must be an x such that f(x) = 0.

For part 2, you want to use Rolle's theorem, as you said.  That means:

First show f(x) is continuous and differentiable everywhere, therefore on any closed interval.

Then show that f'(x) is never zero.  (Use your knowledge of the sine and cosine functions.)

Now if there were TWO zeros, call them a and b, then there would exist c, in (a,b) where f'(c) = 0.  

Hopefully, you already proved that was impossible.

If you can't fill in the details, let me know.