Determine os extremos absolutos, caso existam, da função $f(t) = t + \cot \left(\dfrac{t}{2}\right)$ no intervalo $\left[\dfrac{\pi}{4}, \dfrac{7\pi}{4}\right]$.

$1 = \dfrac{2\tan \left(\dfrac{\pi}{8}\right)}{1 - \tan^2 \left(\dfrac{\pi}{8}\right)}\ \Rightarrow\ \tan \left(\dfrac{\pi}{8}\right) = \sqrt{2} - 1\ \wedge\ \tan \left(\dfrac{7\pi}{8}\right) = 1 - \sqrt{2}$

$f\left(\dfrac{\pi}{4}\right) = \dfrac{\pi}{4} + \dfrac{1}{\sqrt{2} - 1}$

$f\left(\dfrac{7\pi}{4}\right) = \dfrac{7\pi}{4} + \dfrac{1}{1 - \sqrt{2}}$

$f'(t) = 1 - \dfrac{1}{2} \cdot \csc^2 \left(\dfrac{t}{2}\right)$

$\mathbb{U} = \left[\dfrac{\pi}{4}, \dfrac{7\pi}{4}\right]\ \wedge\ f'(t) = 0\ \Rightarrow\ t = \dfrac{\pi}{2}\ \vee\ t = \dfrac{3\pi}{2}$

$f\left(\dfrac{\pi}{2}\right) = \dfrac{\pi}{2} + 1$

$f\left(\dfrac{3\pi}{2}\right) = \dfrac{3\pi}{2} - 1$

$\dfrac{3\pi}{2} - 1 > \dfrac{\pi}{4} + \dfrac{1}{\sqrt{2} - 1} > \dfrac{7\pi}{4} + \dfrac{1}{1 - \sqrt{2}} > \dfrac{\pi}{2} + 1$

Logo os pontos de extremos absolutos são $t = \dfrac{3\pi}{2} - 1$ e $t = \dfrac{\pi}{2} + 1$.