For a function y=f(x), inflection points/stationary points are when (df/dx)=0. For a function z=f(x, y), can you find these points through a similar method? Is it something like (∂f/∂x)=0 and (∂f/∂y)=0?


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Question Number 2548 by Filup last updated on 22/Nov/15

$For a function y = f (x),$ $inflection points / stationary points are$ $when \frac{d f}{d x} = 0 .$ $For a function z = f (x, y), can you find$ $these points through a similar method ?$ $Is it something like \frac{\partial f}{\partial x} = 0 and \frac{\partial f}{\partial y} = 0 ?$
	Answered by Yozzi last updated on 22/Nov/15

	$L e t f b e a f u n c t i o n o f t w o v a r i a b l e s$ $w h o s e f i r s t a n d s e c o n d p a r t i a l$ $d e r i v a t i v e s a r e c o n t i n o u s o n s o m e o p e n$ $d i s c B . S u p p o s e f u r t h e r t h a t a t t h e p o i n t$ $(a, b)$ $f_{x} = \frac{\partial f}{\partial x} = 0 a n d f_{y} = \frac{\partial f}{\partial y} = 0 .$ $L e t p = \frac{\partial^{2} f}{\partial x^{2}} = f_{x x}, q = \frac{\partial^{2} f}{\partial y^{2}} = f_{y y} a n d r = \frac{\partial^{2} f}{\partial y \partial x} = f_{x y}$ $(a, b) i s a l o c a l m i n i m u m i f p q - r^{2} > 0 a n d p > 0 (o r q > 0),$ $(a, b) i s a l o c a l m a x i m u m i f p q - r^{2} > 0 a n d p < 0 (o r q < 0)$ $(a, b) i s a s a d d l e p o i n t i f p q - r^{2} < 0$ $p q - r^{2} = 0 i s i n c o n c l u s i v e .$
	Commented by Filup last updated on 22/Nov/15

	$My question is$ $what is an open disc and what is f_{x y} ?$ $is \frac{\partial^{2} f}{\partial x \partial y} differentiation with two variables ?$
	Commented by Yozzi last updated on 22/Nov/15

	$f_{x} i s p a r t i a l d i f f e r e n t i a t i o n o f f w . r . t x$ $f_{x y} = \frac{\partial}{\partial y} (f_{x}) = \frac{\partial}{\partial y} (\frac{\partial f}{\partial x}) = \frac{\partial^{2} f}{\partial y \partial x} (m i x e d 2 n d p a r t i a l d e r i v a t i v e)$ $A n n - d i m e n s i o n a l o p e n d i s c o f$ $r a d i u s r i s t h e c o l l e c t i o n o f p o i n t s l e s s$ $t h a n a d i s t a n c e o f r a w a y f r o m a f i x e d$ $p o i n t i n E u c l i d e a n n - s p a c e . (W o l f r a m A l p h a)$ $I f n = 1 f o r e x a m p l e, t h i s i s a n o p e n$ $i n t e r v a l (s, t) . S o, f o r n = 2 t h i s$ $d e f i n e s a c o l l e c t i o n o f p o i n t s i n a c i r c l e c o n t a i n i n g$ $t h e p o i n t (a, b) w h e r e b o t h f_{x} a n d f_{y}$ $a r e z e r o . I n t h i s c a s e, o n c e t h e 1 s t$ $a n d 2 n d p a r t i a l d e r i v a t i v e s o f f e x i s t$ $a n d t h e b o t h f_{x y} a n d f_{y x} a r e$ $c o n t i n u o u s f o r t h e s e t o f p o i n t s (x, y)$ $i n t h e d i s c, t h e s e t w o m i x e d p a r t i a l$ $2 n d d e r i v a t i v e s a r e e q u a l, i . e f_{x y} = f_{y x} .$
	Commented by Filup last updated on 22/Nov/15

	$I see . Very interesting . I must go learn$ $multi - variable calculus!$
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