# How do I use the Mean Value Theorem to so 2x-1-sin(x)=0 has exactly one real root?

Aug 27, 2014

This is a proof, so there is no short answer.

To show that there is exactly 1 real root, we start with the IVT to show that roots exist. Then we use a very common math technique which is proof by contradiction.

Let $f \left(x\right) = 2 x - 1 - \sin x$. This is a continuous and differentiable function.

$f \left(0\right) = 0 - 1 - 0 < 0$
$f \left(1\right) = 2 - 1 - \sin \left(1\right) = 1 - \sin \left(1\right) > 0$ because $\sin \left(\frac{\pi}{2}\right) = 1$
Therefore, by IVT, there exists some $c$ such that $f \left(c\right) = 0$.

To prove there are no other real roots, let's suppose that there are two roots: $a$ and $b$, so $f \left(a\right) = 0$ and $f \left(b\right) = 0$.

By Rolle's Theorem, there exists a $c$ such that $f ' \left(c\right) = 0$.

$f ' \left(x\right) = 2 - \cos x$

But $f ' \left(x\right) > 0$ for all $x$ because $\cos x$ is never larger than 1.
So, by contradiction there are not two real roots $a$ and $b$.

You can substitute MVT for Rolle's Theorem if you like. Rolle's Theorem is used to prove MVT.

For proofs, to show something is true, you have to show every case, usually using algebra like you've done with limits. However, to show something is false, you only have to show a single case; this is why proof by contradiction is powerful.