Answer
inflection point $(0,0)$
$y(-\frac{\pi}{3})=\frac{4\pi}{3}-\sqrt 3$ as a local maximum,
$y(\frac{\pi}{3})=\sqrt 3-\frac{4\pi}{3}$ as a local minimum.
Concave up on $(0,\frac{\pi}{2})$ and concave down on $(-\frac{\pi}{2},0)$
Work Step by Step
Step 1. Given the function $y=tan(x)-4x, -\frac{\pi}{2}\lt x \lt\frac{\pi}{2}$, we have $y'=sec^2x-4$ and $y''=2sec^2(x)tan(x)$
Step 2. 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 3. The extrema happen when $y'=0$, undefined, or at endpoints; thus we have $sec(x)=\pm2$ and the critical points within the domain are $x=-\frac{\pi}{3}, \frac{\pi}{3}$.
Step 4. We have $y(-\frac{\pi}{3})=\frac{4\pi}{3}-\sqrt 3$ and $y(\frac{\pi}{3})=\sqrt 3-\frac{4\pi}{3}$.
Step 5. Test signs of $y'$ across critical points $(-\frac{\pi}{2})..(+)..(-\frac{\pi}{3})..(-)..(\frac{\pi}{3})..(+)..(\frac{\pi}{2})$; thus we have $y(-\frac{\pi}{3})=\frac{4\pi}{3}-\sqrt 3$ as a local maximum, $y(\frac{\pi}{3})=\sqrt 3-\frac{4\pi}{3}$ as a local minimum.
Step 6. There is no absolute maximum or absolute minimum in the interval.
Step 7. To identify the intervals on which the function is 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 $(0,\frac{\pi}{2})$ and concave down on $(-\frac{\pi}{2},0)$
