Re: plot

*To*: mathgroup at smc.vnet.net*Subject*: [mg110783] Re: plot*From*: agua <auguaylupo at gmail.com>*Date*: Mon, 5 Jul 2010 21:15:58 -0400 (EDT)*References*: <i0pc44$jt4$1@smc.vnet.net>

In this case, (2x +1) Sqrt [x +1] and Sqrt [x-1] come from different functions, so the new function ((2x +1) Sqrt [x +1]) / Sqrt [x -1] is real for x> 1. Therefore, we expect a graph only for (x> 1 & & x> -1). Thanks for your comments. On 4 jul, 02:09, Murray Eisenberg <mur... at math.umass.edu> wrote: > Why only for x > 1? Your quotient (which has a redundant set of > parentheses around its numerator) evaluates to a negative real when x < 1. > > Take, e.g., x = -3: > > Sqrt[x - 1] /. x -> -3 // InputForm > 2*I > > (2 x + 1) Sqrt[x + 1] // InputForm > (-5*I)*Sqrt[2] > > (2 x + 1) Sqrt[x + 1)/Sqrt[x - 1] /. x -> -3 // InputForm > -5/Sqrt[2] > > And consider: > > (2 x + 1) Sqrt[x + 1]/Sqrt[x - 1] // Simplify // InputForm > (1 + 2*x)/Sqrt[(-1 + x)/(1 + x)] > > So if you want the graph to exclude negative x, you'll have to do it > explicitly. > > On 7/2/2010 2:57 AM, agua wrote: > > > Hi > > With Plot[( (2x+1) Sqrt[x+1] ) / Sqrt[x-1] ,{x,-5,5}] > > hoped to obtain a graph only for x>1. > > > What happened? > > regards. > > -- > Murray Eisenberg mur... at math.umass.edu > Mathematics & Statistics Dept. > Lederle Graduate Research Tower phone 413 549-1020 (H) > University of Massachusetts 413 545-2859 (W) > 710 North Pleasant Street fax 413 545-1801 > Amherst, MA 01003-9305