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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?

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

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  • $\begingroup$ What do you mean by the notation $r\to1+(0)$? $\endgroup$
    – Allawonder
    Mar 17 '19 at 23:26
  • $\begingroup$ 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 $1$. $\endgroup$ Mar 17 '19 at 23:27
  • $\begingroup$ I see. By the way I like your method. $\endgroup$
    – Allawonder
    Mar 17 '19 at 23:36
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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: enter image description here

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