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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



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