-/2 points details my notes scalc9 3.1.059. find the absolute maximum and absolute minimum values of f on…

Explanation:

Step1: Find the derivative of f(t)

Using the chain - rule, if (y = 2\cos(t)+\sin(2t)), then (f^\prime(t)=- 2\sin(t)+2\cos(2t)). Recall that (\cos(2t)=1 - 2\sin^{2}(t)), so (f^\prime(t)=-2\sin(t)+2(1 - 2\sin^{2}(t))=-4\sin^{2}(t)-2\sin(t)+2).

Step2: Set the derivative equal to zero

Let (x = \sin(t)), then (-4x^{2}-2x + 2=0), or (2x^{2}+x - 1 = 0). Factoring gives ((2x - 1)(x + 1)=0). So (x=\sin(t)=\frac{1}{2}) or (\sin(t)=-1). In the interval ([0,\frac{\pi}{2}]), when (\sin(t)=\frac{1}{2}), (t=\frac{\pi}{6}), and (\sin(t)=-1) has no solutions in ([0,\frac{\pi}{2}]).

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})