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# Classify extreme point of $f(x,y)$

2018-03-12 20:20:47

I'm currently studying the first course of multivariable calculus and I have a question.

Suppose $f'_x (a,b) = f'_y (a,b) = 0$, i.e., we have an extreme point at $(x,y)=(a,b)$. To identify the type of this point I would like to use the same method as we use in one variable calculus, but twice. My solution would then look like the following:

if $f''_{xx} (a,b) < 0$ and $f''_{yy} (a,b) < 0$ we have a local maximum point.

if $f''_{xx} (a,b) > 0$ and $f''_{yy}(a,b) > 0$ we have a local minimum point.

if $f''_{xx} (a,b)$ and $f''_{yy} (a,b)$ have different signs, we have a saddle point.

The book suggests that you compute $(f''_{xy} (a,b))^2 -f''_{xx} (a,b) \cdot f''_{yy} (a,b)$, etc.

Doesn't my method work? If it doesn't, why not? Thanks.

Your "method" is used to determine the max/min in any case.

The second derivative test is $f_{xx}f_{yy} - f^2_{xy} <0$

and

$f_{xx}f_{yy} - f^2_{xy} >0$

$$f_{xx}f_{yy} - f^2_{xy} <0 \quad \text{,is a saddle point}$$

$$f_{x • Your "method" is used to determine the max/min in any case. The second derivative test is f_{xx}f_{yy} - f^2_{xy} <0 and f_{xx}f_{yy} - f^2_{xy} >0$$f_{xx}f_{yy} - f^2_{xy} <0 \quad \text{,is a saddle point}f_{xx}f_{yy} - f^2_{xy} >0 \quad \text{,is a min or a max}

If $f_{xx}< 0$ and $f_{yy} <0$ then you have a max

If $f_{xx}> 0$ and $f_{yy} >0$ then you have a min

EDIT: By Example- Suppose $f(x,y)=-x^2-y^2$ , then

$f_{xx}(0,0) = -2 <0$ and

$f_{yy}(0,0) = -2 <0$

and your method works since we have a max at $(0,0)$

Suppose now however that I have the function $f(x,y) = -x^4-y^2-xy$, then your first derivative tests are satisfied at $(0,0)$ but

$f_{xx}(0,0) = 0$ and

$f_{yy}(0,0) = -2 <0$

Here we have that $f_{xx} = 0$ and your method fails to describe what happens at this point. Might be a poor example, but long story short is that your method would be ignoring points.

2018-03-12 21:17:42