find the absolute maximum and absolute minimu...

# Explanation: ## Step1: Find the derivative of \(f(t)\) Using the chain - rule and basic derivative formulas. The derivative of \(y = 2\cos(t)\) is \(y^\prime=- 2\sin(t)\), and the derivative of \(y=\sin(2t)\) is \(y^\prime = 2\cos(2t)\). So \(f^\prime(t)=-2\sin(t)+2\cos(2t)\). Using the double - angle formula \(\cos(2t)=1 - 2\sin^{2}(t)\), we rewrite \(f^\prime(t)\) as \(f^\prime(t)=-2\sin(t)+2(1 - 2\sin^{2}(t))=-4\sin^{2}(t)-2\sin(t)+2\). Let \(x = \sin(t)\), then \(f^\prime(t)=-4x^{2}-2x + 2=-2(2x^{2}+x - 1)=-2(2x - 1)(x + 1)\). Substituting back \(x=\sin(t)\), we have \(f^\prime(t)=-2(2\sin(t)-1)(\sin(t)+1)\). ## Step2: Find the critical points Set \(f^\prime(t)=0\). Then \(-2(2\sin(t)-1)(\sin(t)+1)=0\). \(\sin(t)+1 = 0\) gives \(\sin(t)=-1\), but \(t\in[0,\frac{\pi}{2}]\), so we ignore this solution. \(2\sin(t)-1 = 0\) gives \(\sin(t)=\frac{1}{2}\), and since \(t\in[0,\frac{\pi}{2}]\), then \(t=\frac{\pi}{6}\). ## Step3: Evaluate the function at critical points and endpoints Evaluate \(f(t)\) at \(t = 0\), \(t=\frac{\pi}{6}\), and \(t=\frac{\pi}{2}\). When \(t = 0\), \(f(0)=2\cos(0)+\sin(0)=2\times1 + 0=2\). When \(t=\frac{\pi}{6}\), \(f(\frac{\pi}{6})=2\cos(\frac{\pi}{6})+\sin(\frac{\pi}{3})=2\times\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}\approx2.6\). When \(t=\frac{\pi}{2}\), \(f(\frac{\pi}{2})=2\cos(\frac{\pi}{2})+\sin(\pi)=2\times0 + 0=0\). # Answer: absolute minimum value: \(0\) absolute maximum value: \(\frac{3\sqrt{3}}{2}\)