Answer:
A(-1,0) is a local maximum point.
B(-1,0) is a saddle point
C(3,0) is a saddle point
D(3,2) is a local minimum point.
Step-by-step explanation:
The given function is
![f(x,y)=x^3+y^3-3x^2-3y^2-9x](https://tex.z-dn.net/?f=f%28x%2Cy%29%3Dx%5E3%2By%5E3-3x%5E2-3y%5E2-9x)
The first partial derivative with respect to x is
![f_x=3x^2-6x-9](https://tex.z-dn.net/?f=f_x%3D3x%5E2-6x-9)
The first partial derivative with respect to y is
![f_y=3y^2-6y](https://tex.z-dn.net/?f=f_y%3D3y%5E2-6y)
We now set each equation to zero to obtain the system of equations;
![3x^2-6x-9=0](https://tex.z-dn.net/?f=3x%5E2-6x-9%3D0)
![3y^2-6y=0](https://tex.z-dn.net/?f=3y%5E2-6y%3D0)
Solving the two equations simultaneously, gives;
and ![y=0,y=2](https://tex.z-dn.net/?f=y%3D0%2Cy%3D2)
The critical points are
A(-1,0), B(-1,2),C(3,0),and D(3,2).
Now, we need to calculate the discriminant,
![D=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2](https://tex.z-dn.net/?f=D%3Df_%7Bxx%7D%28x%2Cy%29f_%7Byy%7D%28x%2Cy%29-%28f_%7Bxy%7D%28x%2Cy%29%29%5E2)
But, we would have to calculate the second partial derivatives first.
![f_{xx}=6x-6](https://tex.z-dn.net/?f=f_%7Bxx%7D%3D6x-6)
![f_{yy}=6y-6](https://tex.z-dn.net/?f=f_%7Byy%7D%3D6y-6)
![f_{xy}=0](https://tex.z-dn.net/?f=f_%7Bxy%7D%3D0)
![\Rightarrow D=(6x-6)(6y-6)-0^2](https://tex.z-dn.net/?f=%5CRightarrow%20D%3D%286x-6%29%286y-6%29-0%5E2)
![\Rightarrow D=(6x-6)(6y-6)](https://tex.z-dn.net/?f=%5CRightarrow%20D%3D%286x-6%29%286y-6%29)
At A(-1,0),
and ![f_{xx}=6(-1)-6=-18\:](https://tex.z-dn.net/?f=f_%7Bxx%7D%3D6%28-1%29-6%3D-18%5C%3A%3C%5C%3A0)
Hence A(-1,0) is a local maximum point.
See graph
At B(-1,2);
Hence, B(-1,0) is neither a local maximum or a local minimum point.
This is a saddle point.
At C(3,0)
Hence, C(3,0) is neither a local minimum or maximum point. It is a saddle point.
At D(3,2),
and ![f_{xx}=6(3)-6=12\:>\:0](https://tex.z-dn.net/?f=f_%7Bxx%7D%3D6%283%29-6%3D12%5C%3A%3E%5C%3A0)
Hence D(3,2) is a local minimum point.
See graph in attachment.