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The question asks me to find and classify all the critical points of: $$f(x,y)=\text{ln}(1+x^2y)$$ My solution: $$f'_x = \frac{2xy}{1+x^2y}; f'_y=\frac{x^2}{(1+x^2y)}; f''_{xx}=\frac{(2y)(1+x^2y)-(2xy)(2xy)}{(1+x^2y)^2}; f''_{yy}=\frac{-x^4}{-(1+x^2y)^2}\\ f''_{xy}=f''_{yx}= \frac{2x}{(1+x^2y)^2} $$ Using the fact that critical points occur when both first derivatives equal zero, I have that any point of the form $(0,a)$ must be a critical point. To classify these particular critical points, I set up the Hessian matrix as follows: $$H=\begin{bmatrix}0&0\\0&0\end{bmatrix}$$ Thus, the second derivative test is inconclusive. I thought that to classify the point, I could look at the sign of the function - if we fix $x$ then $f(x,y)\ge0$ if $y>0$ and $f(x,y)<0$ if $y<0$, then each point $(0,a)$ must be a saddle point.

My questions:

Is my reasoning correct? If so, what does this mean about the line x=0?

Am I right in thinking that a saddle point is analogous to an inflection point for a function in $\mathbb{R}^2$? That is, a point where the function changes concavity.

Is there any intuition behind why a multivariable function can have infinitely many stationary points, but I have never come across a function in one variable that has this? Would it be possible to have infinitely many local minima/maxima?

Thanks for your help.

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Your computations are not correct. Note that $\frac{\partial^2f}{\partial x^2}(0,a)=2a$ and that therefore the Hessian matrix isn't always the null matrix at the critical points. But you are right when you say that the approach through the Hessian matrix is inconclusive.

Now, suppose that $a>0$. Then, if $(x,y)$ is close to $(0,a)$, then $x^2y\geqslant0$ and so $\ln(1+x^2y)\geqslant0$. Therefore $f$ has a local minimum at $(0,a)$. By the same argument, if $a<0$, $f$ has a local maximum at $(0,a)$.

Finally, if $a=0$, then, near $(0,0)$, the number $x^2y$ can be positive or negative and therefore so can the number $\ln(1+x^2y)$. So, $(0,0)$ is a saddle point.

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