Answer
$y(0)=1, y(\frac{4\pi}{3})=-2$ as local minima, and $ y(\frac{\pi}{3})=2, y(2\pi)=1$ as local maxima
$y(\frac{\pi}{3})=2$ as an absolute maximum and $y(\frac{4\pi}{3})=-2$ as an absolute minimum in the interval.
inflection points $(\frac{5\pi}{6},0)$ and $(\frac{11\pi}{6}, 0)$
concave up on $(\frac{5\pi}{6},\frac{11\pi}{6})$ and concave down on $(0,\frac{5\pi}{6})$ and $(\frac{11\pi}{6}, 2\pi)$
See graph.
Work Step by Step
Step 1. Given the function $y=cos(x)+\sqrt 3sin(x), 0\leq x\leq2\pi$, we have $y'=-sin(x)+\sqrt 3cos(x)$ and $y''=-cos(x)-\sqrt 3sin(x)$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints; thus we have $tan(x)=\sqrt 3, x=\frac{\pi}{3}, \frac{4\pi}{3}$ and the critical points within the domain are $x=0, \frac{\pi}{3}, \frac{4\pi}{3}, 2\pi$.
Step 3. We have $y(0)=1, y(\frac{\pi}{3})=2, y(\frac{4\pi}{3})=-2, y(2\pi)=1$.
Step 4. Test signs of $y'$ across critical points $(0)..(+)..(\frac{\pi}{3})..(-)..(\frac{4\pi}{3})..(+)..(2\pi)$, we have $y(0)=1, y(\frac{4\pi}{3})=-2$ as local minima, and $ y(\frac{\pi}{3})=2, y(2\pi)=1$ as local maxima
Step 5.We can identify $y(\frac{\pi}{3})=2$ as an absolute maximum and $y(\frac{4\pi}{3})=-2$ as an absolute minimum in the interval.
Step 6. The inflection points can be found when $y''=0$ or it does not exist. We have $tan(x)=-\frac{\sqrt 3}{3}, x=\frac{5\pi}{6}, \frac{11\pi}{6}$ which gives coordinates $(\frac{5\pi}{6},0)$ and $(\frac{11\pi}{6}, 0)$
Step 7. To identify the intervals on which the functions are concave up and concave down, we need to examine the sign of $y''$ on different intervals. We have $(0)..(-)..(\frac{5\pi}{6})..(+)..(\frac{11\pi}{6})..(-)..(2\pi)$,
Step 8. The function is concave up on $(\frac{5\pi}{6},\frac{11\pi}{6})$ and concave down on $(0,\frac{5\pi}{6})$ and $(\frac{11\pi}{6}, 2\pi)$
Step 9. See graph.
