4.3 Finding stationary points

Solution. We have $\frac{\partial f}{\partial x}=\,{3x^{2}-3,}$ and $\frac{\partial f}{\partial y}=\,{3y^{2}-12.}$ Thus the stationary points are given by $x^{2}=\,{1}$ and $y^{2}=\,{4.}$ We therefore have four stationary points: $(x,y)=\,{(1,2),\;(1,-2),\;(-1,2),\;(-1,-2)}$ with corresponding values of $f$:

In a few pages we will see how to determine whether any of these are local maxima or minima.