Unique fixed point on $[0, 2\pi ]$

Question

In the following question I am looking for an explanation of the answer which is given. How did they find the critical point?

Prove that $f(x) = \pi + \frac{1}{2}\sin \left( \frac{x}{2} \right)$ has a unique fixed point on $[0,2 \pi]$

Checking the range of the function $f$:

we have that $f(0) = \pi $ and $f(2\pi ) = \pi$.

The derivative of $f(x)$ is:

$$f'(x) = \frac{1}{4} \cos \left( \frac{x}{2} \right)$$.

Here is what I am having trouble understanding. The only critical point inside the interval is at $x = \pi$. How did they find this critical point? Do I then take that critical point evaluate the function at it and conldude that $0 \leq f(x) \leq 2\pi$ for all $x \in [0, 2\pi]$ and thus a fixed point exists?

Answer 1

You have that $f(0)=\pi$, $f(\pi)=\pi+0.5$, $f(2\pi)=\pi$. According to intermediate value theorem, $\pi \le f(x) \le \pi+0.5$ when $0 \le x \le \pi$. We cannot have fixed point there. On the other hand, $\pi \le f(x) \le \pi+0.5$ when $\pi \le x \le 2\pi$. Because there are no critical points between $\pi$ and $2\pi$, we know that $x=f(x)$ has only one solution on this interval.

Answer 2

It is not clear why you are checking for critical points.

Note that $\sup_x |f'(x)| = {1 \over 4} < 1 $, and $f([0,2 \pi]) \subset [0, 2 \pi]$.

Hence $f$ is a contraction map and has a unique fixed point.