How to plot 8 upper-corner-points ?
Thanks in advance.
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As you can see the poles at +/-sqrt 2 and 0 are no valid solutions to your problem this time. This is because they are poles of second order (the square in d(x)) and so there is no change of signs at those positions. A sign change only occurs at poles of odd order.
Many thanks, Werner. And how to keep the same "procedure" and eliminate 3 points are not 6 lower-intersection-points ? ( maybe just edit ? ). ( And how to plot local minimum-points of the f(x) ? )
Many thanks, Werner. And how to keep the same "procedure" and eliminate 3 points are not 6 lower-intersection-points ? ( maybe just edit ? ).
You may think beforehand and don't ask for the zeros of the denominator which are of even order. In your example obviously all are of even order and so you would not ask for d(x)=0 at all.
As an alternative you may consider writing a function which eliminates all points of zeros (nominator and denominator separately) which occur an even number of times. This routine also must consider removable discontinuities like when nominator and denominator have a common zero. Good luck!
Luc, after I have got all of your hints, I try to plot all lower-intersection-points of logical-plot, but I don't know it's true for all of cases :
F.M. and Werner, with your "procedures" above, It seems that I just need "input" a new function ! :
Many thanks for all your time and help.