 Solution

The critical points of $f^{\prime \prime }$ occur where $f^{\prime \prime \prime }=0$, namely $-1,0,1$. Its sign on the four intervals separated by these points is, in order, positive, negative, positive, negative. Thus $f^{\prime \prime }$ has local maxima at $\pm 1$ and a local minimum at zero. Their values areSince $f^{\prime \prime }(x)\to 0$ at both ends of the real line, these are also the global maximum $1/2$ and the global minimum $-2$. The sign analysis establishes extremality; merely solving $f^{\prime \prime \prime }=0$ would not.