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

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})