Answer
1. $f\left( {x,y} \right) = {x^2} + {y^2}$
The critical point at $\left( {0,0} \right)$ is a minimum.
2. $g\left( {x,y} \right) = {x^2} - {y^2}$
The critical point at $\left( {0,0} \right)$ is a saddle point.
Work Step by Step
We are given the facts that the functions $f\left( {x,y} \right) = {x^2} + {y^2}$ and $g\left( {x,y} \right) = {x^2} - {y^2}$ both have a critical point at $\left( {0,0} \right)$. However, they behave differently at the critical point at $\left( {0,0} \right)$.
1. Consider $f\left( {x,y} \right) = {x^2} + {y^2}$
Since the right-hand side is non-negative, the critical point at $\left( {0,0} \right)$ is a minimum.
2. Consider $g\left( {x,y} \right) = {x^2} - {y^2}$
If we intersect the graph of $g$ with the $xz$-plane by setting $y=0$, we get $g\left( {x,0} \right) = {x^2}$. Since ${x^2} \ge 0$, the critical point at $\left( {0,0} \right)$ is a minimum in this plane. However, if we intersect the graph with the $yz$-plane by setting $x=0$, we get $g\left( {0,y} \right) = - {y^2}$. Clearly, the critical point at $\left( {0,0} \right)$ is a maximum in this plane. Therefore, we conclude that $\left( {0,0} \right)$ is a saddle point, because it depends on the direction from which takes us uphill or downhill.
