RegionPlot[ImplicitRegion[(2 x - 1)/(x - 1) < 3/2, {x, y}],
PlotRange -> {{-9, 9}, {-9, 9}}]
Just checking whether anyone else can replicate this. Can anybody confirm such behavior and, maybe, confirm this as a bug?
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Sign up to join this communityRegionPlot[ImplicitRegion[(2 x - 1)/(x - 1) < 3/2, {x, y}],
PlotRange -> {{-9, 9}, {-9, 9}}]
Just checking whether anyone else can replicate this. Can anybody confirm such behavior and, maybe, confirm this as a bug?
Another thing to do is just cut ImplicitRegion
out of the loop altogether. RegionPlot
is naturally designed to take a predicate of inequalities.
RegionPlot[(2 x - 1)/(x - 1) < 3/2, {x, -9, 9}, {y, -9, 9}]
There have been other examples of RegionPlot
having trouble with ImplicitRegion
, and usually the answer is just to give the inequalities to RegionPlot
directly.
This can worked around by using Reduce
,e.g:
p = Plot[(2 x - 1)/(x - 1), {x, -5, 5},
GridLines -> {{-1, 1}, {3/2, {2, Red}}}, PlotRange -> {-5, 5},
Frame -> True];
rp = RegionPlot[
ImplicitRegion[Reduce[(2 x - 1)/(x - 1) < 3/2, x], {x, y}],
PlotRange -> {{-9, 9}, {-9, 9}}, PlotStyle -> {Pink, Opacity[0.2]}];
Show[p, rp]
10.4.1
too. The following inequalities also do not work:1/x < 1
and1/x <-1
. $\endgroup$ – Anjan Kumar Feb 8 '17 at 8:5011.1.0
$\endgroup$ – user58955 Mar 23 '17 at 17:50