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
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