find the absolute maximum and absolute minimum values of ( f ) on the given interval.\n\n( f(t)=5 t+5 cot…

Explanation:

Step1: Find the derivative of (f(t))

Use the sum rule ((u + v)^\prime=u^\prime+v^\prime) and the chain - rule ((\cot u)^\prime=-\csc^{2}u\cdot u^\prime). If (f(t)=5t + 5\cot(\frac{t}{2})), then (f^\prime(t)=5+5\times(-\csc^{2}(\frac{t}{2}))\times\frac{1}{2}=5-\frac{5}{2}\csc^{2}(\frac{t}{2})). Set (f^\prime(t) = 0), so (5-\frac{5}{2}\csc^{2}(\frac{t}{2})=0). [ \begin{align*} \frac{5}{2}\csc^{2}(\frac{t}{2})&=5\ \csc^{2}(\frac{t}{2})&=2\ \csc(\frac{t}{2})&=\pm\sqrt{2} \end{align*} ] Since (t\in[\frac{\pi}{4},\frac{7\pi}{4}]), then (\frac{t}{2}\in[\frac{\pi}{8},\frac{7\pi}{8}]) and (\csc(\frac{t}{2})=\sqrt{2}) (because (\csc x=\frac{1}{\sin x}\gt0) for (x\in(0,\pi)) in the given interval of (\frac{t}{2})). (\sin(\frac{t}{2})=\frac{1}{\sqrt{2}}), so (\frac{t}{2}=\frac{\pi}{4}) or (\frac{3\pi}{4}), and (t = \frac{\pi}{2}) or (t=\frac{3\pi}{2}).

Step2: Evaluate (f(t)) at critical points and endpoints

• When (t=\frac{\pi}{4}): (f(\frac{\pi}{4})=5\times\frac{\pi}{4}+5\cot(\frac{\pi}{8})). Since (\cot(\frac{\pi}{8})=\sqrt{2}+1), then (f(\frac{\pi}{4})=\frac{5\pi}{4}+5(\sqrt{2} + 1)\approx\frac{5\times3.14}{4}+5\times(1.414 + 1)\approx3.925+12.07\approx15.995).
• When (t=\frac{\pi}{2}): (f(\frac{\pi}{2})=5\times\frac{\pi}{2}+5\cot(\frac{\pi}{4})=\frac{5\pi}{2}+5\times1\approx\frac{5\times3.14}{2}+5=7.85 + 5=12.85).
• When (t=\frac{3\pi}{2}): (f(\frac{3\pi}{2})=5\times\frac{3\pi}{2}+5\cot(\frac{3\pi}{4})=\frac{15\pi}{2}+5\times(-1)\approx\frac{15\times3.14}{2}-5=23.55-5 = 18.55).
• When (t=\frac{7\pi}{4}): (f(\frac{7\pi}{4})=5\times\frac{7\pi}{4}+5\cot(\frac{7\pi}{8})). Since (\cot(\frac{7\pi}{8})=-\cot(\frac{\pi}{8})=-(\sqrt{2}+1)), then (f(\frac{7\pi}{4})=\frac{35\pi}{4}-5(\sqrt{2}+1)\approx\frac{35\times3.14}{4}-12.07\approx27.475-12.07\approx15.405).

Answer:

absolute minimum value: (12.85) (when (t = \frac{\pi}{2})) absolute maximum value: (18.55) (when (t=\frac{3\pi}{2}))