plot level set of the crescent function

max(f,g) is not possible. use (f+g+abs(f-g))/2

your equation is only a parabol and two points. try another

min(f,g) is (f+g-abs(f-g))/2

tOh yes, thats true. I did not care of this one, it was just one of the a try and go. Nevertheless, this enhancement would be nice.

Actually it is not a parabol, whats this one? did you mean a parabolid? This one surface with 2 dimensions? It looks almost like a crescent, when I plot this with GeoGebra.

It's not clear what you mean I think. Maybe you can post a picture of what you are trying to plot?

I want to visualize the set

L := {(x,y) in R^2 | max(x^2+(y-1)^2+y-1, -x^2-(y-1)^2+y+1)=1 }.

As mathmagic noticed, max(f,g) is equal to (f+g+abs(f-g))/2. So I put this in GeoGebra (see the file), and got sth. that looks (to me) nearly like a crescent. My aim was not to find a formula to plot a crescent, but to visualize this set.

So I just wanted to say, that if you put

max(x^2+(y-1)^2+y-1, -x^2-(y-1)^2+y+1)=1 

it is enough with

x² + (y - 1)² < 5 ∧ x² + (y - 3)² > 5 then line thickness to 0 ?

your graph is not a function because the point over some values of x are not unique

oh yes, it is not a function but a level set of a function. As i said, my aim was not to find a formula to plot a crescent, but to visualize this level set. Actually this function is called crescent function. I just wanted to visualize a problem from my optimization homework, e.g. with GeoGebra.

But this was not my intention for this topic. This should be just a "feature request".

entonces tú querías decir curvas en ecuaciones implicitas a partir de la funcion maximo de dos funciones

"entonces tú querías decir curvas en ecuaciones implicitas a partir de la funcion maximo de dos funciones"

Yes, exactly, thank you for the file, it looks amazing! But really I wanted just say, that it would be nice, if you could just write an equation like `max((x^2+(y-1)^2+y-1, -x^2-(y-1)^2+y+1))=1` and get a plot. It would be better readable, imho.

Oh but actually it should be `c(x,y)=(a(x, y) + b(x, y) + abs(a(x, y) - b(x, y))) * 0.5`, right?

(a+b+abs(a-b))/2 is enough