Re: Not all points plot on my graph...
- To: mathgroup at smc.vnet.net
- Subject: [mg105203] Re: [mg105158] Not all points plot on my graph...
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Tue, 24 Nov 2009 05:49:03 -0500 (EST)
- Reply-to: hanlonr at cox.net
Your assumption that (-1)^(2/3) = 1 is bad.
Clear[f, x]
f[x_] = (x + 2)^(2/3);
f[-3.]
-0.5+0.866025 I
(-1)^(2/3) // N
-0.5+0.866025 I
(-1)^(1/3) // N
0.5+0.866025 I
%^2
-0.5+0.866025 I
% == %%% == %%%%
True
Bob Hanlon
---- davef <davidfrick2003 at yahoo.com> wrote:
=============
Why is it when I write this...
Clear[f, x]
f[x_] = (x + 2)^(2/3)
Plot[f[x], {x, -3, 0}, PlotRange -> {{-3, 0}, {-4, 4}}]
f[-3]
.. I get no values show in the graph for x<-2 even though f[-3]
evaluates to (-1)^(2/3).
(-1)^(2/3) evaluates as f[-3]=1 so why wouldn't the point (-3,1) plot
on my graph?
--
Bob Hanlon