Answer
$f(\frac{\pi}{6})=\frac{3\sqrt 3}{2}$ is the absolute maximum value.
$f(\frac{\pi}{2})=0$ is the absolute minimum value.
Work Step by Step
The Closed Interval Method
1) Function $f$ must be continuous on the closed interval $[a, b]$
2) Find the values of $f$ at the critical points in $(a,b)$ and the values of $f$ at the endpoints.
3) Compare the values.
- The largest is the absolute maximum value.
- The smallest is the absolute minimum value.
$$f(t)=2\cos t+\sin 2t,\hspace{1.5cm}[0, \pi/2]$$
First, $f(t)$ is continuous on the interval $[0,\pi/2]$, so the Closed Interval Method can be applied here.
1) Find the critical points of $f$
$$f'(t)=-2\sin t+2\cos 2t$$
$$f'(t)=-2\sin t+2(1-2\sin^2 t)$$
$$f'(t)=-2\sin t+2-4\sin^2 t$$
$$f'(t)=-2(2\sin^2 t+\sin t-1)$$
$$f'(t)=-2(\sin t+1)(2\sin t-1)$$
For $$f'(t)=0$$
$$\sin t=-1\hspace{.5cm}or\hspace{.5cm}\sin t=\frac{1}{2}$$
- For $\sin t=-1$, there is no value of $t$ in the interval $(0,\pi/2)$ that can satisfy the equation.
- For $\sin t=\frac{1}{2}$, where t is in the interval $(0,\pi/2)$, $t=\frac{\pi}{6}.$
$t=\frac{\pi}{6}$ would be evaluated next.
2) Calculate the values of $f$ at critical points and endpoints.
- Critical points: $t=\frac{\pi}{6}$
$$f(\frac{\pi}{6})=2\cos\frac{\pi}{6}+\sin\frac{2\pi}{6}=\frac{2\sqrt 3}{2}+\frac{\sqrt 3}{2}=\frac{3\sqrt 3}{2}$$
- End points: $t=0$ and $t=\frac{\pi}{2}$
$$f(0)=2\cos0+\sin(2\times0)=2\times1+0=2$$
$$f(\frac{\pi}{2})=2\cos\frac{\pi}{2}+\sin(2\times\frac{\pi}{2})=2\times0+\sin\pi=0+0=0$$
3) Comparison
Among the results:
- $f(\frac{\pi}{6})=\frac{3\sqrt 3}{2}$ is the largest, so $f(\frac{\pi}{6})=\frac{3\sqrt 3}{2}$ is the absolute maximum value.
- $f(\frac{\pi}{2})=0$ is the smallest, so $f(\frac{\pi}{2})=0$ is the absolute minimum value.
