# Marking points of intersection between two curves

I'm trying to illustrate the solutions numerically and graphically for an equation such as Tan[x] == x. I think I did everything ok except I wanted to mark each intersection between Tan[x] and x.

Does anyone know how such a thing can be done?

• Comment by the OP (migrated from the question) -> ** I'm sorry if I'm not 'commenting' properly. I'll figure it out when I get more time here. I wanted to say THANKS to EVERYONE that posted. This is exactly what I was looking for. I will use all the input and make sure I learn from what was given. What a goldmine this site is for learning something like this. Thanks again, it's greatly appreciated!! Sep 12, 2012 at 19:23
• See also the solution using RootsInRange. This solution is more general as it will work when either the exact intersections are not known, or NSolve fails. Mar 26, 2014 at 1:49

Edited to make it a function. For the strange Exclusions specification I use below, see my answer here. Thanks to @Oleksandr and @JM for their great comments.

plInters[{f1_, f2_}, {min_, max_}] :=
Module[{sol, x},
sol = x /. NSolve[f1[x] == f2[x] && min < x < max, x];
Framed@Show[
ListPlot[{#, f1[#]} & /@ sol, PlotStyle -> PointSize[Large]],
Plot[{f1[x], f2[x]}, {x, min, max}, Exclusions -> {True, f2[x] == 10, f1[x] == 10}]
]
]

GraphicsRow[plInters[#, {-10, 10}] & /@ {{# &, Tan}, {Tan, Coth}, {Sin, 1/# &}}] • Damn, yours is better.
– kale
Sep 12, 2012 at 1:00
• @kale That is one of the good things in this site. You can always find other ways. Sep 12, 2012 at 1:01
• Anyway definite +1 from me.
– kale
Sep 12, 2012 at 1:06
• I don't really know why Plot can sometimes determine exclusions properly and sometimes not. Still, this'll look a bit better if you set Exclusions -> Pi/2 Range[-5, 5, 2] for Plot (otherwise, it may be harder for students to see that these aren't really solutions). Sep 12, 2012 at 1:45
• @Alexey I tried to explain why here mathematica.stackexchange.com/a/10502/193. It gets the plot refinement routine to work to your advantage. Sep 13, 2012 at 1:52

Update 3: Using GraphicsMeshFindIntersections to get the intersection points (see also):

showIntersections = Show[#, Graphics @{Red, PointSize[Large],
Point @ GraphicsMeshFindIntersections @ #}] &;


Using the two examples in the original answer:

Row[showIntersections /@ {Plot[{Cos[x], x Sin[x]}, {x, -3 Pi, 3 Pi},
ImageSize -> 400],
Plot[{Tan[x], x Sin[x]}, {x, -3 Pi, 3 Pi}, ImageSize -> 400,
Exclusions -> Range[-5 Pi/2, 5 Pi/2, Pi]]}] You can also use MeshFunctions:

  Plot[{Cos[x], x Sin[x]}, {x, -3 Pi, 3 Pi},
MeshFunctions -> {(Cos[#] - # Sin[#]) &}, Mesh -> {{0}},
MeshStyle -> Directive[Red, PointSize[Large]]] Update: Dealing with Tan[x] using Exclusions

Plot[{Tan[x], x Sin[x]}, {x, -3 Pi, 3 Pi},
MeshFunctions -> {(Tan[#] - # Sin[#]) &}, Mesh -> {{0}},
MeshStyle -> Directive[Red, PointSize[Large]],
Exclusions -> Range[-5 Pi/2, 5 Pi/2, Pi]]
(* or Exclusions -> (Cos[x] == 0) *) Update 2: Using just Mesh and MeshStyle:

points = NSolve[Tan[x] == x Sin[x] && -3 Pi < x < 3 Pi, x][[All, 1, 2]];
Plot[{Tan[x], x Sin[x]}, {x, -3 Pi, 3 Pi},
Mesh -> {points},
MeshStyle -> {Directive[Red, PointSize[Large]]},
Exclusions -> Range[-5 Pi/2, 5 Pi/2, Pi]]
(* same picture as above *)

• very nice! :) 
– rm -rf
Sep 12, 2012 at 1:45
• This has some problems with the exclusions :( Sep 12, 2012 at 5:03
• @R.M thank you. Belisarius, right... without excluding the vertical segments of the Tan function (using Exclusions or RegionFunction) MeshFunctions do not work.
– kglr
Sep 12, 2012 at 5:43
• But that is not always easy > mathematica.stackexchange.com/q/10501/193 Sep 12, 2012 at 13:28