Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 56

Determine the absolute maximum and absolute minimum values of $\displaystyle f(x) = x + \cot \left( \frac{x}{2} \right)$ on the interval $\displaystyle \left[ \frac{\pi}{4}, \frac{7\pi}{4}\right]$.

Taking the derivative of $f(x)$ we have,

$
\begin{equation}
\begin{aligned}
f'(x) &= 1 + \left( -\csc^2 \left( \frac{x}{2}\right) \cdot \left( \frac{1}{2} \right) \right)\\
\\
f'(x) &= 1 - \frac{1}{2} \csc^2 \left( \frac{x}{2} \right)
\end{aligned}
\end{equation}
$

Solving for critical numbers, when $f'(x) = 0$
$0 = \frac{1}{2} \csc^2 \left( \frac{x}{2} \right)$


$
\begin{equation}
\begin{aligned}
\frac{1}{2} \csc^2 \left( \frac{x}{2} \right) &= 1 &&;\text{ recall that } \csc x = \frac{1}{\sin x}\\
\\
2 \sin^2 \left( \frac{x}{2} \right) &= 1\\
\\
\sin^2 \left( \frac{x}{2} \right) &= \frac{1}{2}\\
\\
\sin \left( \frac{x}{2} \right) &= \sqrt{\frac{1}{2}}\\
\\
\frac{x}{2} &= \frac{\pi}{4} \text{ and } \frac{3\pi}{4}\\
\\
x &= \frac{\pi}{4} (2) \text{ and } x = \frac{3\pi(2)}{4}\\
\\
x &= \frac{\pi}{2} \text{ and } x = \frac{3 \pi}{2}

\end{aligned}
\end{equation}
$


It shows that we have either absolute maximum and minimum values at $\displaystyle x = \frac{\pi}{2}$ and $\displaystyle x = \frac{3\pi}{2}$
So,

$
\begin{equation}
\begin{aligned}
\text{when } &= \frac{\pi}{2}\\
\\
f \left( \frac{\pi}{2} \right) &= \frac{\pi}{2} + \cot \left( \frac{\frac{\pi}{2}}{2} \right)\\
\\
f\left( \frac{\pi}{2} \right) &= 2.5708\\
\\
\\
\\
\text{when } &= \frac{3 \pi}{2}\\
\\
f\left( \frac{3\pi}{2} \right) &= \frac{3\pi}{2} + \cot \left(\frac{\frac{3\pi}{2}}{2} \right)\\
\\
f \left( \frac{3\pi}{2} \right) &= 3.7124
\end{aligned}
\end{equation}
$

Evaluating $f(x)$ at end points $\displaystyle x = \frac{\pi}{2}$ and $\displaystyle x = \frac{7\pi}{2}$

$
\begin{equation}
\begin{aligned}
\text{when } x &= \frac{\pi}{4},\\
\\
f \left( \frac{\pi}{4} \right) &= \frac{\pi}{4} + \cot \left( \frac{\frac{\pi}{4}}{2} \right)\\
\\
f \left( \frac{\pi}{4} \right) &= 3.1996
\\
\\
\\
\text{when } x = \frac{7\pi}{4}\\
\\
f \left( \frac{7\pi}{4} \right) &= \frac{7\pi}{4} + \cot \left( \frac{\frac{7\pi}{4}}{4} \right)\\
\\
f \left( \frac{7\pi}{4} \right) &= 3.0836
\end{aligned}
\end{equation}
$

Therefore, we have absolute maximum value at $\displaystyle f \left( \frac{3\pi}{4} \right) = 3.7124 $ and the absolute minimum value at $\displaystyle f \left( \frac{\pi}{2} \right) = 2.5708$ on the interval $\displaystyle \left[ \frac{\pi}{4}, \frac{7\pi}{4}\right]$