Answer
$y(-\frac{\pi}{6})=-\frac{2\pi}{9}+\frac{\sqrt 3}{3}$ as local minimum,
$y(\frac{\pi}{6})=\frac{2\pi}{9}-\frac{\sqrt 3}{3}$ as local maximum,
- no absolute maximum or absolute minimum in the interval.
- inflection point: $(0,0)$
- concave up on $(-\frac{\pi}{2},0)$ and concave down on $(0,\frac{\pi}{2})$
See graph.
Work Step by Step
Step 1. Given the function $y=\frac{4}{3}x-tan(x), -\frac{\pi}{2}\lt x \lt \frac{\pi}{2}$, we have $y'=\frac{4}{3}-sec^2(x)$ and $y''=2sec^2(x)tan(x)$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints; thus we have $sec(x)=\pm\frac{2}{\sqrt 3}$ and the critical points within the domain are $x= -\frac{\pi}{6}, \frac{\pi}{6}$.
Step 3. We have $y(-\frac{\pi}{6})=\frac{4}{3}(-\frac{\pi}{6})-tan(-\frac{\pi}{6})=-\frac{2\pi}{9}+\frac{\sqrt 3}{3}$, $y(\frac{\pi}{6})=\frac{4}{3}(\frac{\pi}{6})-tan(\frac{\pi}{6})=\frac{2\pi}{9}-\frac{\sqrt 3}{3}$.
Step 4. Test signs of $y'$ across critical points $..(-)..(-\frac{\pi}{6})..(+)..(\frac{\pi}{6})..(-)..$; thus we have $y(-\frac{\pi}{6})=-\frac{2\pi}{9}+\frac{\sqrt 3}{3}$ as local minimum, and $y(\frac{\pi}{6})=\frac{2\pi}{9}-\frac{\sqrt 3}{3}$ as local maximum,
Step 5. There is no absolute maximum or absolute minimum in the interval.
Step 6. The inflection points can be found when $y''=0$ or it does not exist. We have $tan(x)=0, x=0$ which gives coordinates $(0,0)$
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 $(-\frac{\pi}{2})..(+)..(0)..(-)..(\frac{\pi}{2})$,
Step 8. The function is concave up on $(-\frac{\pi}{2},0)$ and concave down on $(0,\frac{\pi}{2})$
Step 9. See graph.
