How can I plot
Sqrt[x - x^3]/(x^2 + 1)
in all its domain?
Domain = FunctionDomain[{ Sqrt[x - x^3]/(x^2 + 1)}, x]
Domain is : x <= -1 || 0 <= x <= 1
I tried
Plot[{Sqrt[x - x^3]/(x^2 + 1)}, {x, -100, -1}]
but I really dont know how!
$\begingroup$
$\endgroup$
$\begingroup$
$\endgroup$
Here's one approach:
Plot[Sqrt[x - x^3]/(x^2 + 1),
Element[x, ImplicitRegion[FunctionDomain[{Sqrt[x - x^3]/(x^2 + 1)}, x], x]]]
answered Mar 29 '16 at 19:26
-
$\begingroup$ Interesting! I wasn't aware you can use region specification directly on
Plotlike that. $\endgroup$ – kirma Mar 30 '16 at 4:36 -
$\begingroup$ BTW, I believe the pedantically correct way would be to write
Element[{x}, ...]. Otherwisexis actually treated as an one-dimensional vector, and it might not always work as it should as a scalar. $\endgroup$ – kirma Mar 30 '16 at 4:40 -
$\begingroup$ There's probably some difference between
xand{x}inPlot, but I don't know of any off the top of my head. If you find any, let me know. $\endgroup$ – Brett Champion Mar 30 '16 at 15:57 -
$\begingroup$ I suspect
Plotdoes some magic with its argument, since it knows that it cannot be multidimensional vector, anyway. Normally there's a difference betweenxand{x}in these cases (and documentation indicates that{x}should be used), but now it seems impossible to spot from insides ofPlot. $\endgroup$ – kirma Mar 30 '16 at 17:10 -
$\begingroup$ I know
Plotdoes some magic with its argument. :-) $\endgroup$ – Brett Champion Mar 30 '16 at 19:25
$\begingroup$
$\endgroup$
I believe that you are looking for something like the following. It uses the Show function to combine the two plots.
Show[Plot[{Sqrt[x - x^3]/(x^2 + 1)}, {x, -10, -1}],
Plot[{Sqrt[x - x^3]/(x^2 + 1)}, {x, 0, 1}], PlotRange -> All]
Plot[{Sqrt[x - x^3]/(x^2 + 1)}, {x, -10, 1}]? $\endgroup$ – garej Mar 29 '16 at 19:12