Re: Selecting cubic roots in functional form
- To: mathgroup at smc.vnet.net
- Subject: [mg50286] Re: [mg50282] Selecting cubic roots in functional form
- From: "Florian Jaccard" <florian.jaccard at eiaj.ch>
- Date: Wed, 25 Aug 2004 03:35:51 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Carol,
Is this what you want ?
f[c_] = Which @@ Flatten[({#1[[1]], #1[[2,2]]} & ) /@
N[List @@ Reduce[2000*c*(5*c + 7*q) - 4*(c + 8*q)*(12*q + 5*c)^2 == 0
&& c > 0 && q > 0, q]]];
Plot[f[c], {c, 0, 100}]
Regards,
F.Jaccard
-----Message d'origine-----
De : Carol Ting [mailto:suomesta at yahoo.com]
Envoyé : mardi, 24. août 2004 12:22
À : mathgroup at smc.vnet.net
Objet : [mg50282] Selecting cubic roots in functional form
Hello, list,
I have an equation
2000c*(5c+7q)-4(c+8q)*(12q+5c)^2==0
and I need to find q as a function of c, where both c and q are
positive quantities.
I use InequalitySolve and find that when c<3.11665 the 3rd root gives
the positive function I need, and when c>3.11665 I should pick the
first root instead. So I need to pick different roots in different
regions of c and combine them into a piecewise continuous function.
Trouble is, I have a whole series of equations generated by another
function and it is impossible for me to pick the roots one by one. Is
there a way to make Mathematica do this automatically?
Thanks in advance!
Carol