Extreme values of $\frac{1}{x^2 + y^2 -1 }$

Question

Let $f(x,y)= \frac{1}{x^2 + y^2 -1 }$ . I want to find its extreme values. Its first partial derivatives are $f_x(x,y) = \frac{-2x}{(x^2 + y^2 -1)^2}$ and $f_y(x,y) = \frac{-2y}{(x^2 + y^2 -1)^2}$ respectively. For the values that makes $f_x$ and $f_y$ equal to $0$ simultaneously, I can apply the second derivative test. However, what can we say about the singular points of $f$?

I mean the set $\{(x,y) | x^2 + y^2 -1 = 0\}$? Can they be extreme?

Answer 1

Hint: Use $x=r\cos \alpha$ and $y=r\sin \alpha$ to reduce the problem to a univariate problem

$$f(r,\alpha)=f(r)=\dfrac{1}{r^2-1}.$$

We can restrict the analysis to $r\geq 0$. Clearly $r= 1$ is a problem. If we approach $r \to 1 + (0)$ the function will diverge to infinity. If we approach $r \to 1 - (0)$ the function will diverge to minus infinity. For $r=0$ we have $f(0)=-1$. For $r\to \infty$ we have $f=0$. You can go ahead and calculate the derivative and set it equal to zero. You will see that this will lead to $r=0$ and $f(0)=-1$ as a local mimimum.

Answer 2

The set $\{(x,y) | x^2 + y^2 -1 = 0\}$ isn't even included in the domain of the function, so can't contain any extreme point. In fact $f_x(x,y)=f_y(x,y)=0$ leads to the unique solution $x=y=0$ and by 2nd order differentiation, we conclude it is a local maximum. Here is the sketch of the function: