Stationary points of $\ln(1+x²y)$ The stationary points of the function lie along with $(0,y)$ for all $y$. The second derivative test is inconclusive.
I understand that the origin is the saddle point. But how do we categorize the other points without looking at the graph? In general, what is the approach to take in case of inconclusive second derivative tests?
 A: Generally speaking, if the second derivative test is inconclusive one needs to assume an epsilon ball around the point of interest, and if the function changes sign within the ball, no matter how small, then the function is a saddle point
From the first derivative test, we know that all points of the form $(0,y)$ are stationary points. Let us look at the function behaviour centered at a point $(0,y_0)$
For $(x,y)$ such that $$||(x,y)-(0,y_0)|| < \epsilon$$
$$f(x,y) = \ln(1 + x^2y)$$
$$f(0,y_0) = 0$$
Now, if I move in any direction other than along $y$ axis - $x^2y >0$ for $y>0$ and $x^2y <0$ for $y<0$
Hence, If $y>0$ - then $$f(x,y) \geq 0 = f(0,y_0)$$
If $y<0$ then
$$f(x,y) \leq0 = f(0,y_0)$$
Can you tell which point is a minima and which is a maxima now?
A: The value of the function at any point $(0,c)$ on Y-axis is $f(0,c)=0$.
Now observe the value of function at the points falling in the $\epsilon$-neighborhood of $(0,c)$ by sliding left or right keeping $c$ fixed, so that $f(0-\epsilon,c)=f(0+\epsilon,c)=In(1+\epsilon^2. c)>f(0,c)$ for $c>0$$\implies $ maxima for all points on positive Y-axis.
and $f(0-\epsilon,c)=f(0+\epsilon,c)=In(1+\epsilon^2. c)<f(0,c)$ for $c<0$$\implies $ minima for all points on negative Y-axis.
