Answer
$y(0)=1$ is a local maximum and $ y(-1)= y(1)=0$ is local minima;
absolute minimum is at $y(-1)=y(1)=0$ and there is no absolute maximum;
concave up on $(-\infty,-1)$ and $(1,\infty)$,
concave down on $(-1,1)$.
See graph.
Work Step by Step
Step 1. Given the function $y=|x^2-1|=\begin{cases} x^2-1,\ x\geq1\ or \ x\leq-1 \\ 1-x^2, \ -1\lt x\lt1\end{cases}$, we have $y'=\begin{cases} 2x,\ x\geq1\ or \ x\leq-1 \\ -2x, \ -1\lt x\lt1\end{cases}$ and $y''=\begin{cases} 2,\ x\geq1\ or \ x\leq-1 \\ -2, \ -1\lt x\lt1\end{cases}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=0,\pm1$ as critical points with $ y(-1)=0, y(0)=1, y(1)=0$
Step 3. Examine the sign change of $y'$ across the critical points: $..(-)..(-1)..(+)..(0)..(-)..(1)..(+)..$; we can identify $y(0)=1$ as a local maximum and $ y(-1)= y(1)=0$ as local minima. The absolute minimum is at $y(-1)=y(1)=0$ 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 $x=1$.
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 $..(+)..(-1)..(-)..(1)..(+)..$,
Step 6. The function is concave up on $(-\infty,-1),(1,\infty)$ and concave down on $(-1,1)$.
Step 7. See graph.
