## Calculus (3rd Edition)

The function $f(x)=|x^2-1|$ is not differentailble at $x=\pm1$.
Since $$|x^2-1|=0\Longrightarrow x^2-1=0\Longrightarrow (x+1)(x-1)=0.$$ The function $f(x)=|x^2-1|$ has corners at $x=\pm 1$, hence it is not differentailble at $x=\pm1$.