Answer
a) Local minimum
b) Saddle point
Work Step by Step
a) $f_{xx}(1,1) = 4$, $f_{xy}(1,1)=1$, $f_{yy}(1,1)=2$
using the formula
$D(a,b) = f_{xx}(a,b)*f_{yy}(a,b)-[f_{xy}(a,b)]^2$ gives
$8-1=7>0$ so it is either a max or a min.
Given $f_{xx}(a,b)=4$ and $4>0$ it is a min.
b) $f_{xx}(1,1) = 4$, $f_{xy}(1,1)=3$, $f_{yy}(1,1)=2$
Using the same formula gives $D(a,b) = 4*2-(3^2) = -1$
so it is a saddle point.
