Single Variable Calculus, Chapter 4, 4.2, Section 4.2, Problem 4

Show that the function $f(x) = \cos 2x$ satisfies the three hypothesis of Rolle's Theorem on the interval $\displaystyle \left[ \frac{\pi}{8}, \frac{7\pi}{8}\right]$. Then find all numbers $c$ that satisfy the conclusion of Rolle's Theorem.

We know that $\cos 2x$ is trigonometric function that is continuous everywhere. Hence, $f(x)$ is continuous or the closed interval $\displaystyle \left[ \frac{\pi}{8}, \frac{7\pi}{8}\right]$
Next, if we take the derivative of $f(x)$, we get...

$\begin{equation} \begin{aligned} f'(x) &= -\sin2x(2)\\ \\ f'(x) &= -2 \sin 2x \end{aligned} \end{equation}$


We also know that $f'(x) = - 2 \sin 2x$ is a quadratic function that is differentable everywhere, hence, $f$ is differentiable on the open interval $\displaystyle \left[ \frac{\pi}{8}, \frac{7\pi}{8}\right]$

Lastly, if $\displaystyle f \left( \frac{\pi}{8} \right) = f\left( \frac{7\pi}{8} \right)$


$\begin{equation} \begin{aligned} \cos \left[ 2 \left( \frac{\pi}{8} \right) \right] &= \cos \left[ 2 \left( \frac{7\pi}{8} \right) \right] \\ \\ \frac{\sqrt{2}}{2} &= \frac{\sqrt{2}}{2} \end{aligned} \end{equation}$


Since we satisfy all the hypothesis of Rolle's Theorem, we can now solve for $c$ where $f'(c) = 0$, so,

$\begin{equation} \begin{aligned} f'(c) &= -2 \sin 2x = 0\\ \\ -2 \sin 2x &= 0\\ \\ \sin 2x &= 0\\ \\ 2x &= \sin^{-1} [0] && \text{; where } n \text{ is any integer}\\ \\ x &= \frac{\pi n}{2} \end{aligned} \end{equation}$

If we substitute $n = 0, 1$ and $2$, we get...
$x = 0$, $\displaystyle x = \frac{\pi}{2}$ and $x = \pi$

But, only $\displaystyle x = \frac{\pi}{2}$ is in the interval $\displaystyle \left[ \frac{\pi}{8}, \frac{7\pi}{8} \right]$. Therefore, $\displaystyle c = \frac{\pi}{2}$