My apologies, by the way, I should have specified that I was working for d in [0,2[ (d is actually the sum of 2 duty cycles, each one between 0 and 1). It is such an "obvious" hypothesis for me, because I am so focused on the problem, that it became kind of implicit. So you are right, it wouldn't work out of this interval.
You are also right that, in this example, the min in the x function is not necesary. In the application, I do use x with d in the whole range of d, so I just mechanically put the min whith application in mind. Once again, hypothesis that are implicit considering the application, but are absolutely not obvious when asking for help.
No problem. If you stay with d in the range [0;2[ you would also not need the max() function either as 1/(2-d) would always be larger than 1 for d>1. But I already guessed that your original problem would be a bit more complex.