Function:$$x = {- \sqrt{x^2+y^2}}.$$(a conical surface) To determine whether it has a stationary point or not, 2 condition must be met:
- function must have partial derivatives at point p0, and
- function must have local maxima/minima at point p0.
For this function you can calculate partial derivatives and the gradient is equal 0 when derivative with respect to x variable is equal 0: for x=0; when derivative with respect to y variable is equal 0: for y=0. Also, the cone has a MAXimum in this same point.
HOWEVER: http://www.wolframalpha.com/input/?i=-(x%5E2%2By%5E2)%5E(1%2F2)+stationary+point Wolphram hasn't found any Stationary Points, which means in point 0,0 there is no stationary point.
Doesn't the function have partial derivatives at this point? If so, why?
//Edit:
I tried to calculate (when I asked the question):
$$\lim\limits_{x \to 0^+, y \to 0^+} - \sqrt{x^2+y^2} = 0$$
and
$$\lim\limits_{x \to 0^-, y \to 0^-} - \sqrt{x^2+y^2} = 0$$
.. so I was sure the side limits are the same and equal 0. Now when I know the answer, I assume what I had done is incorrect.
The partial derivatives don’t exist at the cone’s vertex: $f(x,0) = -(x^2)^{1/2}=-|x|$. This function doesn’t have a derivative at $x=0$, which I suspect you know. Working through the limit calculation, anyway, for $h\gt0$, the difference quotient at the origin for differentiation with respect to $x$ is $${f(h,0)-f(0,0)\over h} = -{|h|\over h}=-\frac hh=-1$$ but for $h\lt0$ it is $$-\frac{-h}h = 1,$$ therefore the limit—the partial derivative of $f$ with respect to $x$—clearly doesn’t exist. By symmetry, neither does $f_y$.