find the absolute maximum and absolute minimum values of ( f ) on the given interval.\n\n( f(t)=2 cos…

Explanation:

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

Use the sum rule and chain rule. The derivative of (\cos(t)) is (-\sin(t)) and the derivative of (\sin(2t)) is (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 get (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). Factoring gives (f^\prime(t)=-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). Since (\sin(t)+1\geq0) for all (t), we set (2\sin(t)-1 = 0), so (\sin(t)=\frac{1}{2}). In the interval (\left[0,\frac{\pi}{2}\right]), (t=\frac{\pi}{6}).

Step3: Evaluate the function at the critical points and endpoints

Evaluate (f(t)) at (t = 0): (f(0)=2\cos(0)+\sin(0)=2). Evaluate (f(t)) at (t=\frac{\pi}{6}): (f\left(\frac{\pi}{6}\right)=2\cos\left(\frac{\pi}{6}\right)+\sin\left(\frac{\pi}{3}\right)=2\times\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}\approx2.598). Evaluate (f(t)) at (t=\frac{\pi}{2}): (f\left(\frac{\pi}{2}\right)=2\cos\left(\frac{\pi}{2}\right)+\sin(\pi)=0).

Answer:

absolute minimum value: (0) absolute maximum value: (\frac{3\sqrt{3}}{2})