## Calculus: Early Transcendentals 8th Edition

(a) $f$ is increasing on the interval $(\frac{\pi}{2}, \frac{3\pi}{2})$ $f$ is decreasing on the intervals $(0, \frac{\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi)$ (b) The local maximum is $f(\frac{3\pi}{2}) = 2$ The local minimum is $f(\frac{\pi}{2}) = -2$ (c) The graph is concave up on this interval: $(\frac{\pi}{6}, \frac{5\pi}{6})$ The graph is concave down on these intervals: $(0, \frac{\pi}{6})\cup (\frac{5\pi}{6}, 2\pi)$ The points of inflection are $(\frac{\pi}{6},-\frac{1}{4})$ and $(\frac{5\pi}{6},-\frac{1}{4})$
(a) $f(x) = cos^2~x-2~sin~x,~~~~0 \leq x \leq 2\pi$ We can find the points where $f'(x) = 0$: $f'(x) = -2~cos~x~sin~x-2~cos~x = 0$ $(-2~cos~x)~(sin~x+1) = 0$ $cos~x = 0$ or $sin~x = -1$ $x = \frac{\pi}{2}, \frac{3\pi}{2}$ When $\frac{\pi}{2} \lt x \lt \frac{3\pi}{2}~~$ then $f'(x) \gt 0$ $f$ is increasing on the interval $(\frac{\pi}{2}, \frac{3\pi}{2})$ When $0 \lt x \lt \frac{\pi}{2}$ and $\frac{3\pi}{2} \lt x \lt 2\pi~~$ then $f'(x) \lt 0$ $f$ is decreasing on the intervals $(0, \frac{\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi)$ (b) $f(\frac{\pi}{2}) = cos^2~\frac{\pi}{2}-2~sin~\frac{\pi}{2} = -2$ $f(\frac{3\pi}{2}) = cos^2~\frac{3\pi}{2}-2~sin~\frac{3\pi}{2} = 2$ The local maximum is $f(\frac{3\pi}{2}) = 2$ The local minimum is $f(\frac{\pi}{2}) = -2$ (c) We can find the points where $f''(x) = 0$: $f''(x) = 2~sin^2~x-2~cos^2~x+2~sin~x= 0$ $sin^2~x-cos^2~x+sin~x= 0$ $sin^2~x-(1-sin^2~x)+sin~x= 0$ $2~sin^2~x+sin~x-1= 0$ $(2~sin~x-1)(sin~x+1)=0$ $sin~x = \frac{1}{2}$ or $sin~x = -1$ $x = \frac{\pi}{6}, \frac{5\pi}{6}$ or $x = \frac{3\pi}{2}$ Note that $x = \frac{3\pi}{2}$ is not a point of inflection because the concavity does not change at this point. The graph is concave down when $f''(x) \lt 0$ The graph is concave down on these intervals: $(0, \frac{\pi}{6})\cup (\frac{5\pi}{6}, 2\pi)$ The graph is concave up when $f''(x) \gt 0$ The graph is concave up on this interval: $(\frac{\pi}{6}, \frac{5\pi}{6})$ $f(\frac{\pi}{6}) = cos^2~\frac{\pi}{6}-2~sin~\frac{\pi}{6} = -\frac{1}{4}$ $f(\frac{5\pi}{6}) = cos^2~\frac{5\pi}{6}-2~sin~\frac{5\pi}{6} = -\frac{1}{4}$ The points of inflection are $(\frac{\pi}{6},-\frac{1}{4})$ and $(\frac{5\pi}{6},-\frac{1}{4})$