Re: questions about delayed expression.

*To*: mathgroup at smc.vnet.net*Subject*: [mg19889] Re: [mg19862] questions about delayed expression.*From*: BobHanlon at aol.com*Date*: Sun, 19 Sep 1999 18:47:36 -0400*Sender*: owner-wri-mathgroup at wolfram.com

a[x_] := x + 4; b[x_] := -3x + 8; sol[func_, c_] := Module[{x}, x /. Flatten[Solve[func[x] == c, x]]]; inter[c_] := Interval[{sol[a, c], sol[b, c]}]; inter[c] Interval[{-4 + c, (8 - c)/3}] inter[.3] Interval[{-3.7, 2.56667}] inter[.4] Interval[{-3.6, 2.53333}] Alternate approach: Needs["Algebra`InequalitySolve`"] intvl[c_] := Module[{x, temp}, temp = InequalitySolve[a[x] >= c && b[x] >= c, x]; Interval[{First[temp], Last[temp]}]] intvl[c] Interval[{-4 + c, 8/3 - c/3}] intvl[.3] Interval[{-3.7, 2.56667}] intvl[.4] Interval[{-3.6, 2.53333}] Bob Hanlon In a message dated 9/19/1999 5:26:07 AM, d8442803 at student.nsysu.edu.tw writes: > The following is the process that I run in my notebook. > >In[1]:= >a[x_] := x + 4 >b[x_] := -3 x + 8 > >In[2]:= >sola = x /. Solve[{a[x] == c}, x]; >solb = x /. Solve[{b[x] == c}, x]; > >In[3]:= >Inter[c_] := Interval[{sola[[1]], solb[[1]]}] > >In[4]:= >Inter[.3] >Inter[.4] > >Out[4]= >\!\(Interval[{\(-4\) + c, \(8 - c\)\/3}]\) > >Out[5]= >\!\(Interval[{\(-4\) + c, \(8 - c\)\/3}]\) > >My questions are: >1. Why Inter[.3] and Inter[.4] cannot be evaluated? Their results should > >not be the same. This is not my intention. > >2. I don't know if there is any better way to extract the 'root(s)' from > >the output of 'Solve' command. The output form is {{x->root1}, {x- >>root2}, ...{}}. If I use 'ReplaceAll'(/.) command, it will remain a list > >of solutions of x. It seems I can only use element operation to extract > >the root(s) from the solution list? In my case, I use sola[[1]] and >solb[[1]]. > >Please give me suggestions. Thanks for your help. >