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.
>