## Calculus: Early Transcendentals 8th Edition

(a) $f$ is decreasing on the interval $(\frac{\pi}{4}, \frac{5\pi}{4})$ $f$ is increasing on the intervals $(0, \frac{\pi}{4}) \cup (\frac{5\pi}{4}, 2\pi)$ (b) The local maximum is $f(\frac{\pi}{4}) = \sqrt{2}$ The local minimum is $f(\frac{5\pi}{4}) = -\sqrt{2}$ (c) The graph is concave down on these intervals: $(0, \frac{3\pi}{4})\cup (\frac{7\pi}{4}, 2\pi)$ The graph is concave up on this interval: $(\frac{3\pi}{4}, \frac{7\pi}{4})$ The points of inflection are $(\frac{3\pi}{4},0)$ and $(\frac{7\pi}{4},0)$
(a) $f(x) = sin~x+cos~x,~~~~0 \leq x \leq 2\pi$ We can find the points where $f'(x) = 0$: $f'(x) = cos~x-sin~x = 0$ $cos~x = sin~x$ $tan~x = 1$ $x = \frac{\pi}{4}, \frac{5\pi}{4}$ When $\frac{\pi}{4} \lt x \lt \frac{5\pi}{4}~~$ then $f'(x) \lt 0$ $f$ is decreasing on the interval $(\frac{\pi}{4}, \frac{5\pi}{4})$ When $0 \lt x \lt \frac{\pi}{4}$ and $\frac{5\pi}{4} \lt x \lt 2\pi~~$ then $f'(x) \gt 0$ $f$ is increasing on the intervals $(0, \frac{\pi}{4}) \cup (\frac{5\pi}{4}, 2\pi)$ (b) $f(0) = sin~0+cos~0 = 1$ $f(\frac{\pi}{4}) = sin~\frac{\pi}{4}+cos~\frac{\pi}{4} = \sqrt{2}$ $f(\frac{5\pi}{4}) = sin~\frac{5\pi}{4}+cos~\frac{5\pi}{4} = -\sqrt{2}$ $f(2\pi) = sin~2\pi+cos~2\pi = 1$ The local maximum is $f(\frac{\pi}{4}) = \sqrt{2}$ The local minimum is $f(\frac{5\pi}{4}) = -\sqrt{2}$ (c) We can find the points where $f''(x) = 0$: $f''(x) = -sin~x-cos~x = 0$ $sin~x+cos~x = 0$ $sin~x = -cos~x$ $tan~x = -1$ $x = \frac{3\pi}{4}, \frac{7\pi}{4}$ The graph is concave down when $f''(x) \lt 0$ The graph is concave down on these intervals: $(0, \frac{3\pi}{4})\cup (\frac{7\pi}{4}, 2\pi)$ The graph is concave up when $f''(x) \gt 0$ The graph is concave up on this interval: $(\frac{3\pi}{4}, \frac{7\pi}{4})$ $f(\frac{3\pi}{4}) = sin~\frac{3\pi}{4}+cos~\frac{3\pi}{4} = 0$ $f(\frac{7\pi}{4}) = sin~\frac{7\pi}{4}+cos~\frac{7\pi}{4} = 0$ The points of inflection are $(\frac{3\pi}{4},0)$ and $(\frac{7\pi}{4},0)$