If we have a function f(x,y) for which we wish to find the stationary points, then there is one set of rules for classifying those points.(adsbygoogle = window.adsbygoogle || []).push({});

Let there be an extremum of f(x,y) at (x,y) = (a,b), and D = fxx(a,b)*fyy(a,b) - {fxy(a,b)}², then

1) D > 0 and fxx > 0 => min

2) D > 0 and fxx < 0 => max

3) D < 0 => saddle point

4) D = 0 - indeterminate - needs deeper analysis.

Now I can never remember this lot of rules so I made up my own little set of rules - as follows.

In 2-D maths, fxx > 0 gives a min, and fxx < 0 gives a max, so I extended this to 3-D maths, as below.**

My rules

1) if fxx and fyy > 0 then a min

2) if fxx and fyy < 0 then a max

3) if fxx and fyy of opposite sign then a saddle

Now, this has always worked well for me; I find it easy to transfer the 2-D rules to a 3-D situation, and it saves me having to work out fxy.

The only thing it doesn't cover is; if fxx and fyy are the same sign (which by my rules should be either a max or a min) is it possible for D to be < 0, which would mean the TP was a saddle?

So, my questions are:

1) can anyone find an error in my rules?

2) does anyone know of a situation where fxx and fyy were both of the same sign, but the TP was a saddle?

**: here, fxx should really be d²y/dx²

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# Classification of extrema

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