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From: J. Batista on 3 Jul 2010 08:20 You may have several options, but here is a simple solution: change your plot range from {x,-5,5} to {x,1,5} and add AxesOrigin->{0,0} in the plot command. If you don't include the command AxesOrigin->{0,0}, your plot may look misleading. Please also note that at x = 1, you may have a singular point. Regards, J. Batista On Fri, Jul 2, 2010 at 2:57 AM, agua <auguaylupo(a)gmail.com> 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. > >
From: Bob Hanlon on 3 Jul 2010 08:21 The function is real for x > 1 or x <= -1 If you want x to start at 1, specify 1 rather than -5. Plot[((2 x + 1) Sqrt[x + 1])/Sqrt[x - 1], {x, 1, 5}] Bob Hanlon ---- agua <auguaylupo(a)gmail.com> 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.
From: Murray Eisenberg on 4 Jul 2010 03:09 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 murray(a)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
From: agua on 5 Jul 2010 21:15
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...(a)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...(a)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 |