Answer
Maximum: $f(0.170,-1.215) \approx 3.197$,
Minimum: $f(-1.301,0.549) -\approx 3.145$, $f(1.131,0.549) \approx -0.701$
Saddle points at $(-1.301,-1.215),(0.170,0.549),(1.131,-1.215)$
No Highest or Lowest points
Work Step by Step
Second derivative test: Some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$.
1. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
2.If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point.
Critical points are: $f(0.170,-1.215),(-1.301,0.549), (1.131,0.549),(-1.301,-1.215),(0.170,0.549),(1.131,-1.215)$
For $(x,y)=f(0.170,-1.215)$
$D \gt 0$ and $f_{xx} \lt 0$
For $(x,y)=(-1.301,0.549)$
$D \gt 0$ and $f_{xx}= \gt 0$
For $(x,y)=(1.131,0.549) $
$D \gt 0$ and $f_{xx}= \gt 0$
Thus, when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
For $(x,y)=(-1.301,-1.215),(0.170,0.549),(1.131,-1.215)$
$D \lt 0$ ; saddle points.
Hence,
Maximum: $f(0.170,-1.215) \approx 3.197$,
Minimum: $f(-1.301,0.549) -\approx 3.145$, $f(1.131,0.549) \approx -0.701$
Saddle points at $(-1.301,-1.215),(0.170,0.549),(1.131,-1.215)$
No Highest or Lowest points
