Points of Inflection Does $f(x)=|x^2−1|$ have an inflection point at $x=(-1)$ and $x=1$? It seems like the concavity is changing, but I just wanted to make sure.
No. A well-defined derivative does not exist at $x=\pm 1$. It has a local minimum at these points.
Yes, the concavity is changing, so you could say that. But be careful, since the first derivative at those points does not exist (hence, the second neither), so you can't test the signs of $f''(-1)$ and $f''(1)$. Here I asked a question about this same function, and you can see the graph of the function to convince yourself.