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Selecting cubic roots in functional form
- To: mathgroup at smc.vnet.net
- Subject: [mg50282] Selecting cubic roots in functional form
- From: suomesta at yahoo.com (Carol Ting)
- Date: Tue, 24 Aug 2004 06:22:27 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
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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