Re: Not all points plot on my graph...

*To*: mathgroup at smc.vnet.net*Subject*: [mg105192] Re: [mg105158] Not all points plot on my graph...*From*: Tomas Garza <tgarza10 at msn.com>*Date*: Tue, 24 Nov 2009 05:46:53 -0500 (EST)*References*: <200911231149.GAA21721@smc.vnet.net>

Well, f is complex-valued in the range {-2, -3}. E.g., In[13]:= Table[f[j],{j, -2, -3, -0.1}] Out[13]= {0.,-0.107722+0.18658 I,-0.170998+0.296177 I,-0.22407+0.388101 I,-0.271442+0.470151 I,-0.31498+0.545562 I,-0.355689+0.616072 I,-0.394187+0.682751 I,-0.430887+0.746318 I,-0.466085+0.807283 I,-0.5+0.866025 I} No way to graph it. Also, probably the point (-3, 1) is too small and almost invisible. BTW, I would use SetDelayed (:=) instead of Set (=) in the definition of f. Tomas > Date: Mon, 23 Nov 2009 06:49:43 -0500 > From: davidfrick2003 at yahoo.com > Subject: [mg105158] Not all points plot on my graph... > To: mathgroup at smc.vnet.net > > 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? > >

**References**:**Not all points plot on my graph...***From:*davef <davidfrick2003@yahoo.com>