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Re: Plot of (2 x^2 - x^3)^(1/3)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg112132] Re: Plot of (2 x^2 - x^3)^(1/3)
  • From: chungyuandye <chungyuandye at gmail.com>
  • Date: Wed, 1 Sep 2010 06:27:58 -0400 (EDT)
  • References: <i5aqbk$eru$1@smc.vnet.net> <i5ied4$kbl$1@smc.vnet.net>

On 8=E6=9C=8831=E6=97=A5, =E4=B8=8B=E5=8D=884=E6=99=8227=E5=88=86, Peter <i=
nsomnia.ber... at gmail.com> wrote:
> On 28 Aug., 13:02, Bernard <bernard.vuilleum... at edu.ge.ch> wrote:
>
> > I see in Calcul Diff=E9rentiel et int=E9gral, N. Piskounov, Edition=
s
> > MIR, Moscou 1970, on p. 210 the graph of (2 x^2 - x^3)^(1/3) with
> > negative values for x > 2, something like :
>
> > Plot[Piecewise[{{(2 x^2 - x^3)^(1/3), x <= 2}, {-Re[(2 x^2 -
> > x^3)^(1/3)], =C2 x > 2}}], {x, -1, 3}]
>
> > When I plot this function with :
>
> > Plot[(2 x^2 - x^3)^(1/3) // Re, {x, -1, 3}]
>
> > I obtain positives values for x > 3. I don't understand why.
> > Thank you very much for your help !
>
> Hi Bernard,
>
> maybe you want
>
> Plot[Root[#^3-(2 x^2-x^3)&,1]//Re,{x,-1,4}]
>
> ?
>
> Peter

Unprotect[Power];
Power[x_?Negative, Rational[p_, q_?OddQ]] := (-(-x)^(1/q))^p;
Protect[Power];
Plot[(2 x^2 - x^3)^(1/3), {x, -1, 3}]


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