I guess this has already be answered.
Plot[1/(x^3 - 2 x^2), {x, -2, 4}, Exclusions -> True]
Plot[1/(x^3 - 2 x^2), {x, -2, 4}, Exclusions -> 2]
I cannot understand why I have to specify the position of discontinuity (2nd plot) in order to get rid of the vertical line at x=2 (which exists in the first plot). Exclusions->True is not for detect the position of discontinuities? The function is rather simple.
1/(x-2)
. - dimitris, Here is the advice on how to do it from the docs (Plot
> "Options" > "Exclusions"):Plot[1/(x^3 - x + 1), {x, -2, 2}, Exclusions -> {x^3 - x + 1 == 0}]
$\endgroup$x == 0
is not excluded per se -- the plot just goes beyond thePlotRange
and re-enters on the same side. It would be nice ifExclusions -> All
would callNSolve
or do something extra. It does not seem to. $\endgroup$Exclusions
, the settingsAll
andTrue
are synonymous withAutomatic
. In the trace,foo = Trace[Plot[1/(x^3 - 2 x^2), {x, -2, 4}, Exclusions -> True], TraceInternal -> True];
, this happens:If[Graphics`PerformanceTuningDump`exclusions === All || Graphics`PerformanceTuningDump`exclusions === True, Graphics`PerformanceTuningDump`exclusions = Automatic]
. $\endgroup$