Re: Re: Log(ln) Function + 2 Parameters + Greater

*To*: mathgroup at smc.vnet.net*Subject*: [mg89320] Re: [mg89297] Re: Log(ln) Function + 2 Parameters + Greater*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Thu, 5 Jun 2008 00:45:03 -0400 (EDT)*Reply-to*: hanlonr at cox.net

Need a space or * between a and x f[a_, x_] := (x/E^(a*x)); Plot[Table[f[a, x], {a, 0, 2}], {x, 0, 3}] Bob Hanlon ---- Felipe Mannshardt <vexie.infamous at googlemail.com> wrote: > dh wrote: > > Hi Felipe, > > > > why are you not simply using an If[] statement: > > > > If[a>0, Plot[...]] > > > > hope this helps, Daniel > > > > > > > > Felipe Mannshardt wrote: > > > >> Hello, > > > > > >> i am having problem telling Mathematica to: > > > > > >> First Try, > > > > > >> Clear[x, f] > > > >> f[x_] := x (a + Log[x])^2 > > > > > >> a > 0 > > > >> Plot[f[x], {x, -5, 5}] > > > > > >> Second Try, > > > > > >> f[x_] := x (a + Log[x])^2 > > > > > >> Plot[f[x], {x, -5, 5}] /.a-> a>0 > > > > > > > >> What i want is to tell Mathematica to draw a Graph of f[x] when a>0 > > > > > >> Thanks for the help. > > > > > > > > > > > > > > Thanks for all replies, worked for my last function, but the suggestions > you guys made so far did work with my last function, but now i have a > new one, and i am here just playing around and i am unable to get it > drawn :( > > First to the first Function, i believe the best suggestion was the use > of "if", > > But, i can not get it to work :/ > > Clear[x, f, a] > f[a_][x_] := x (a + Log[x])^2 > If[a > 0, Plot[f[x], {x, 0, 2}]] > > And all suggestions also did not work at all for my new Function, > > f[a_][x_] := (x/E^(ax)); > Plot[Table[f[a, x], {a, 0, 2}], {x, 0, 3}] > > What am i doing wrong ? > > I have already drawn this function by hand, > and i have two types of function, > for a>0 and a<0 > > But i can not figure out how to draw it in Mathematica :( > > Sorry, that i ask so "noob/dumb" questions, but i have already tried > looking in the help and it did not help me further in this problem, so i > believe the best is asking, right ? > > > Thanks a Lot. >