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MathGroup Archive 2004

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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




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