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

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?

• That's a circle: the function goes to $\pm\infty$ near it. Mar 17 '19 at 22:39
• The derivatives are incorrect. Their denominators should be squares. Mar 17 '19 at 23:25

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.
• What do you mean by the notation $r\to1+(0)$? Mar 17 '19 at 23:26
• We approach $r = 1$ from the above (symbolized by adding a positive $0$, which you can think of like a very very small positive quantity) to $1$. Mar 17 '19 at 23:27
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: