Calculus: Early Transcendentals 8th Edition

The absolute maximum is $f(\frac{3\pi}{2})= 3.71$ The absolute minimum is $f(\frac{\pi}{2})= 2.57$
$f(t) =t + cot(\frac{t}{2})$ We can find the points where $f'(t) = 0$: $f'(t) = 1 - \frac{1}{2~sin^2~(t/2)} = 0$ $1 - \frac{1}{2~sin^2~(t/2)} = 0$ $\frac{1}{2~sin^2~(t/2)} = 1$ $2~sin^2~(\frac{t}{2}) = 1$ $sin^2~(\frac{t}{2}) = \frac{1}{2}$ $sin~(\frac{t}{2}) = \pm \frac{1}{\sqrt{2}}$ $\frac{t}{2} = \frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4},...$ $t = \frac{\pi}{2},\frac{3\pi}{2}$ We can verify the values of these two points and the endpoints of the interval: $f(\frac{\pi}{4}) = (\frac{\pi}{4})+cot(\frac{\pi}{8}) = 3.20$ $f(\frac{\pi}{2}) = (\frac{\pi}{2})+cot(\frac{\pi}{4}) = 2.57$ $f(\frac{3\pi}{2}) = (\frac{3\pi}{2})+cot(\frac{3\pi}{4}) = 3.71$ $f(\frac{7\pi}{4}) = (\frac{7\pi}{4})+cot(\frac{7\pi}{8}) = 3.08$ The absolute maximum is $f(\frac{3\pi}{2})= 3.71$ The absolute minimum is $f(\frac{\pi}{2})= 2.57$