Proving 2x-1-sinx=0 has exactly one real root

[Sczisnad]

Homework Statement

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

Homework Equations

*N/A

The Attempt at a Solution

*Ok so normally to find the real roots for x I would solve for x when I set the function equals to 0, but with this function there is an x and a sinx.

2x-1-sinx=0
2x-sinx=1 // Here is where I run into problems, I cannot isolate x
sinx=2x-1 // I set the sinx equal to all of the other parts.

So this tells me that the equation equals 0 when sinx is equal to 2x-1? Is that assumption correct? Well either way I graphed the equation and it appears that the function crosses the x-axis only once around .889. Also using the solve function on my calculator I find that the x intercept is about .887862211571.

*How do I prove that this answer is the only answer and how do I go about calculating this without my calculator?

 

[╔(σ_σ)╝]

When x>= 1 what happens ?

What happens when 1/2 <= x <= -1 ?

What about when 1/2 <= x ?

Can you use calculus to find critical points ?

 

[Dick]

Homework Statement

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

Homework Equations

*N/A

The Attempt at a Solution

*Ok so normally to find the real roots for x I would solve for x when I set the function equals to 0, but with this function there is an x and a sinx.

2x-1-sinx=0
2x-sinx=1 // Here is where I run into problems, I cannot isolate x
sinx=2x-1 // I set the sinx equal to all of the other parts.

So this tells me that the equation equals 0 when sinx is equal to 2x-1? Is that assumption correct? Well either way I graphed the equation and it appears that the function crosses the x-axis only once around .889. Also using the solve function on my calculator I find that the x intercept is about .887862211571.

*How do I prove that this answer is the only answer and how do I go about calculating this without my calculator?

No, you can't isolate x. And you can't calculate it exactly even using the calculator. What you can do is use the intermediate value theorem to show there is at least one root. Then use the mean value theorem to show there can't be two.