I want to find the local maximum, local minimum, or saddle points of this function:
$$f(x,y)=y(e^x-1)=ye^x-y$$
$f_{x}=e^xy$
$f_{y}=e^x-1$
$f_{xx}=e^xy$
$f_{xy}=e^x$
$f_{yy}=0$
$$f_{y}=0=e^x-1\rightarrow x=0$$
$$f_{x}=0 = e^xy\rightarrow y=0$$
This means there is a critical point at (0,0).
Using the second derivative test to check whether this is a local minimum, local maximum, or neither:
$$D(0,0)=f_{xx}(0,0)\cdot f_{yy}(0,0)-[f_{xy}(0,0)]^2$$
$$D(0,0)=0\cdot 0-1 = -1$$
Since $D(0,0) < 0$, the point $(0,0)$ is neither a local maximum or local minimum, so it is a saddle point.
Have I done this correctly?
Best Answer
It is correct. For |x|<<1 and |y|<<1, your function behaves just like f(x,y)=x*y which definitively has a (the classical!) saddle point S(0,0). The saddle though would not be too comfortable to sit on, due to its flattening out for values x < 0.