Answer
A. The domain is $[-2\pi, 2\pi]$
B. The y-intercept is $\sqrt{3}$
The x-intercepts are $-\frac{4\pi}{3}, -\frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{3}$
C. The function is not an odd function or an even function.
D. There are no asymptotes.
E. The function is decreasing on the intervals $(-\frac{11\pi}{6}, -\frac{5\pi}{6})\cup (\frac{\pi}{6}, \frac{7\pi}{6})$
The function is increasing on the intervals $(-2\pi, -\frac{11\pi}{6})\cup (-\frac{5\pi}{6}, \frac{\pi}{6})\cup (\frac{7\pi}{6}, 2\pi)$
F. The local minima are $(-\frac{5\pi}{6},-2)$ and $(\frac{7\pi}{6},-2)$
The local maxima are $(-\frac{11\pi}{6},2)$ and $(\frac{\pi}{6},2)$
G. The graph is concave up on the intervals $(-\frac{4\pi}{3}, -\frac{\pi}{3})\cup (\frac{2\pi}{3}, \frac{5\pi}{3})$
The graph is concave down on the intervals $(-2\pi, -\frac{4\pi}{3})\cup (-\frac{\pi}{3}, \frac{2\pi}{3})\cup (\frac{5\pi}{3}, 2\pi)$
The points of inflection are $(-\frac{4\pi}{3},0), (-\frac{\pi}{3},0), (\frac{2\pi}{3},0),$ and $(\frac{5\pi}{3},0)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = sin~x + \sqrt{3}~cos~x,~~~~-2\pi\leq x \leq 2\pi$
A. The domain is given in the question as $[-2\pi, 2\pi]$
B. When $x=0$, then $y = sin~0 + \sqrt{3}~cos~0 = \sqrt{3}$
The y-intercept is $\sqrt{3}$
When $y = 0$:
$sin~x + \sqrt{3}~cos~x = 0$
$sin~x = -\sqrt{3}~cos~x$
$tan~x = -\frac{\sqrt{3}}{1}$
$x = -\frac{4\pi}{3}, -\frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{3}$
The x-intercepts are $-\frac{4\pi}{3}, -\frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{3}$
C. The function is not an odd function or an even function.
D. There are no asymptotes.
E. We can find values of $x$ such that $y' = 0$:
$y' = cos~x-\sqrt{3}~sin~x = 0$
$cos~x = \sqrt{3}~sin~x$
$tan~x = \frac{1}{\sqrt{3}}$
$x = -\frac{11\pi}{6}, -\frac{5\pi}{6}, \frac{\pi}{6}, \frac{7\pi}{6}$
When $-\frac{11\pi}{6} \lt x \lt -\frac{5\pi}{6}$ or $\frac{\pi}{6} \lt x \lt \frac{7\pi}{6}$, then $y' \lt 0$
The function is decreasing on the intervals $(-\frac{11\pi}{6}, -\frac{5\pi}{6})\cup (\frac{\pi}{6}, \frac{7\pi}{6})$
When $-2\pi \lt x \lt -\frac{11\pi}{6}$ or $-\frac{5\pi}{6} \lt x \lt \frac{\pi}{6}$ or $-\frac{7\pi}{6} \lt x \lt 2\pi$, then $y' \gt 0$
The function is increasing on the intervals $(-2\pi, -\frac{11\pi}{6})\cup (-\frac{5\pi}{6}, \frac{\pi}{6})\cup (\frac{7\pi}{6}, 2\pi)$
F. When $x=-\frac{5\pi}{6}$, then $y = sin~(-\frac{5\pi}{6}) + \sqrt{3}~cos~(-\frac{5\pi}{6}) = -2$
When $x=\frac{7\pi}{6}$, then $y = sin~(\frac{7\pi}{6}) + \sqrt{3}~cos~(\frac{7\pi}{6}) = -2$
The local minima are $(-\frac{5\pi}{6},-2)$ and $(\frac{7\pi}{6},-2)$
When $x=-\frac{11\pi}{6}$, then $y = sin~(-\frac{11\pi}{6}) + \sqrt{3}~cos~(-\frac{11\pi}{6}) = 2$
When $x=\frac{\pi}{6}$, then $y = sin~(\frac{\pi}{6}) + \sqrt{3}~cos~(\frac{\pi}{6}) = 2$
The local maxima are $(-\frac{11\pi}{6},2)$ and $(\frac{\pi}{6},2)$
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = -sin~x-\sqrt{3}~cos~x = 0$
$sin~x = -\sqrt{3}~cos~x$
$tan~x = -\frac{\sqrt{3}}{1}$
$x = -\frac{4\pi}{3}, -\frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{3}$
When $-\frac{4\pi}{3}\lt x \lt -\frac{\pi}{3}$ or $\frac{2\pi}{3}\lt x \lt \frac{5\pi}{3}$, then $y'' \gt 0$
The graph is concave up on the intervals $(-\frac{4\pi}{3}, -\frac{\pi}{3})\cup (\frac{2\pi}{3}, \frac{5\pi}{3})$
When $-2\pi \lt x \lt -\frac{4\pi}{3}$ or $-\frac{\pi}{3}\lt x \lt \frac{2\pi}{3}$ or $\frac{5\pi}{3}\lt x \lt 2\pi$, then $y'' \lt 0$
The graph is concave down on the intervals $(-2\pi, -\frac{4\pi}{3})\cup (-\frac{\pi}{3}, \frac{2\pi}{3})\cup (\frac{5\pi}{3}, 2\pi)$
When $x=-\frac{4\pi}{3}$, then $y = sin~(-\frac{4\pi}{3}) + \sqrt{3}~cos~(-\frac{4\pi}{3}) = 0$
When $x=-\frac{\pi}{3}$, then $y = sin~(-\frac{\pi}{3}) + \sqrt{3}~cos~(-\frac{\pi}{3}) = 0$
When $x=\frac{2\pi}{3}$, then $y = sin~(\frac{2\pi}{3}) + \sqrt{3}~cos~(\frac{2\pi}{3}) = 0$
When $x=\frac{5\pi}{3}$, then $y = sin~(\frac{5\pi}{3}) + \sqrt{3}~cos~(\frac{5\pi}{3}) = 0$
The points of inflection are $(-\frac{4\pi}{3},0), (-\frac{\pi}{3},0), (\frac{2\pi}{3},0),$ and $(\frac{5\pi}{3},0)$
H. We can see a sketch of the curve below.
