Answer
$y(0)=0$ is a local minimum and there are no local maxima;
absolute minimum is at $y(0)=0$ and there is no absolute maximum;
concave down on $(-\infty,0)$ and $(0,\infty)$.
See graph.
Work Step by Step
Step 1. Given the function $y=\sqrt {|x|}=\begin{cases} \sqrt {-x},\ x\lt0 \\ \sqrt x, \ x\geq0\end{cases}$, we have $y'=\begin{cases} \frac{-1}{2\sqrt {-x}}=-\frac{1}{2}(-x)^{-1/2},\ x\lt0 \\\frac{1}{2\sqrt x}=\frac{1}{2}(x)^{-1/2}, \ x\geq0\end{cases}$ and $y'=\begin{cases} -\frac{1}{4}(-x)^{-3/2},\ x\lt0 \\ -\frac{1}{4}(x)^{-3/2}, \ x\geq0\end{cases}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=0$ as a critical points with $y(0)=0$
Step 3. Examine the sign change of $y'$ across the critical points: $..(-)..(0)..(+)..$; we can identify $y(0)=0$ as a local minimum and there are no local maxima. The absolute minimum is at $y(0)=0$ and there is no absolute maximum.
Step 4. The inflection points can be found when $y''=0$ or when it does not exist. We do not have possible inflection points within the domain as $y'$ is not continuous at $x=0$.
Step 5. 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 $..(-)..(0)..(-)..$,
Step 6. The function is concave down on $(-\infty,0)$ and $(0,\infty)$.
Step 7. See graph.
