I am currently studying analysis and I am facing a problem with understanding how to determine the local extreme values of a function. I just started studying this, but our prof is already giving us mandatory homeworks with extremely hard examples and I am really struggling, if anyone could please help me it would be very helpful.
The function is:
I would really appreciate the help, I really need it.
$\frac{\partial f}{\partial x}=3x^2-12=0$ $\Rightarrow x=2 $ or $x=-2$
$\frac{\partial f}{\partial y}=3y^2-3=0 \Rightarrow y=1$ or $y=-1$
$(2,1), (2,-1), (-2,1) ,(-2,-1)$ are extreme points
$\frac{\partial^2 f }{\partial x^2}\frac{\partial^2 f }{\partial y^2}-(\frac{\partial^2 f }{\partial x \partial y})^2=36xy$
at point $(2,1) $ $\frac{\partial^2 f }{\partial x^2} >0$ and $\frac{\partial^2 f }{\partial x^2}\frac{\partial^2 f }{\partial y^2}-(\frac{\partial^2 f }{\partial x \partial y})^2=36xy>0$ which means in this point function reaches a local minimum
at point $(-2,-1)$ $\frac{\partial^2 f }{\partial x^2}< 0 \frac{\partial^2 f }{\partial x^2}\frac{\partial^2 f }{\partial y^2}-(\frac{\partial^2 f }{\partial x \partial y})^2=36xy>0 $ that means in this point function reaches a local maximum
and the others are saddle points because $\frac{\partial^2 f }{\partial x^2}\frac{\partial^2 f }{\partial y^2}-(\frac{\partial^2 f }{\partial x \partial y})^2=36xy<0$