Calc 1 Question - Rolle?

Question

Hi im stuck on this one. Anyone got a step by step. Thanks
It has to do with Rolles Theorem, and mean value theorems.

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

Answer 1

Rolle's Theorem says that if a function f has two roots, a and b, where
f(a) = f(b) = 0,
then somewhere between a and b there exists a number c such that f'(c) = 0.

So if there is no point at which the derivative f'(x) is equal to 0, then f cannot have two roots a and b.

Given the function
f(x) = 2x - 1 - sin x
we differentiate to get
f'(x) = 2 - cos x
Obviously f'(x) can never be zero, since cos x is always between (or equal to) 1 and -1.
So, by Rolle's theorem, f(x) cannot have two roots.

But f(-1) < 0 and f(1) > 0, so by the mean value theorem,
there must be a root r between -1 and 1 where
f(r) = 0

So, there is at least one root, but there cannot be two roots.
Therefore the number of roots must be exactly one.

Answer 2

Notice that the derivative is 2 + cos x. Since -1 <= cos x <= 1, this function is always increasing. At x = -1 the function is < 0, and at x = 3 the function > 0. Since it is continuous, it takes on the value 0 (between the negative value and the positive value). Since it is always increasing, it takes on the value 0 only once.