Re: a question
- To: mathgroup at smc.vnet.net
- Subject: [mg94384] Re: a question
- From: "Jean-Marc Gulliet" <jeanmarc.gulliet at gmail.com>
- Date: Thu, 11 Dec 2008 07:26:48 -0500 (EST)
- References: <ghnfjh$nlv$1@smc.vnet.net> <493F7CCF.2030903@gmail.com>
alberto gonzalez wrote:
> i tried a lot of ways to plot y=4x-6/3-2x and i cant get to make it noticed in the plot the hole of discontinuity that the sketch presents a when the function turns to P(3/2,-2).the best aproximation ive gotten is
> Plot[((4 x-6)/(3-2 x)),{x,1,2},PlotPloints->2]
Hi Alberto,
What is happening here is that the Plot command does some symbolic
pre-processing before plotting the resulting function. In our case, (4
x - 6)/(3 - 2 x) simplifies to -2 (see In/Out[1]), and this is the
constant function y == -2 that is plotting indeed. So there exists no
discontinuity in the plot (even when forcing an explicit exclusion at
x = 3/2, as you can see below (In[2]))!
In[1]:= Simplify[(4 x - 6)/(3 - 2 x)]
Out[1]= -2
In[2]:= Plot[(4 x - 6)/(3 - 2 x), {x, 1, 2}, AxesOrigin -> {0, 0},
PlotStyle -> Thick, Exclusions -> 3/2]
Now, doing some manual massaging, I believe that the following plot is
close to what you were looking for (the point x = 3/2 is "excluded"
from the x-axis and the curve shows the discontinuity at (3/2, -2) by
being drawn in white (matching the background color) on a small
interval around this point):
In[3]:= Plot[(4 x - 6)/(3 - 2 x), {x, 1, 2}, AxesOrigin -> {0, 0},
PlotStyle -> Thick,
Epilog -> {PointSize[Large], Circle[{3/2, 0}, 0.03], White,
Point[{3/2, -2}]}]
Regards,
-- Jean-Marc