Calculus: Early Transcendentals 8th Edition

$f(\frac{\pi}{6})=\frac{3\sqrt 3}{2}$ is the absolute maximum value. $f(\frac{\pi}{2})=0$ is the absolute minimum value.
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.