4
$\begingroup$

Consider the following code:

f[x_] = Which[x <= 1, x^3 + 3 x^2 - 2, x > 1, 4]

Show[
  Plot[f[x], {x, -4, 4}],
  ListPlot[{{1, f[1]}}], 
  PlotRange -> {{-4, 4}, {-4, 6}}, 
  AspectRatio -> 1,
  AxesLabel -> {x, y}, 
  Epilog -> {PointSize[0.02], Point[{{1, f[1]}}]}]

I get a plot of the discontinuous function $f(x)$.

I would like to ask you if there is a way to show the point of discontinuity on the plot of f, without it being displayed as a black disk. I would like it displayed as a circle.

|improve this question
$\endgroup$
10
$\begingroup$

I recommend plotting the function like this:

f[x_] = Which[x <= 1, x^3 + 3 x^2 - 2, x > 1, 4]

Show[Plot[f[x], {x, -4, 4}], 
 ListPlot[{{1, Limit[f[x], x -> 1, Direction -> "FromAbove"]}}, 
  PlotMarkers -> "OpenMarkers"], 
 PlotRange -> {{-4, 4}, {-4, 6}},
 AspectRatio -> 1, AxesLabel -> {x, y}]

The open marker signifies the left end point is not plotted on the second interval. The Limit function is used to calculate the point.

enter image description here

|improve this answer
$\endgroup$
  • 1
    $\begingroup$ +1 I like the plot marker more than using the disk, so removing my answer. $\endgroup$ – Nasser Nov 17 at 5:53
  • 2
    $\begingroup$ @LouisB's code is elegant and plots the function correctly, but in your original plot the point is highlighted when x == 1. If you want an open circle at that location, just change the limit direction to "FromBelow". $\endgroup$ – Lee Nov 17 at 16:14
  • $\begingroup$ Thanks a lot for your answer, but when I copy and paste your code on the notebook, only the empty circle appears.... and the graph of $f$. What is wrong there? $\endgroup$ – dmtri Nov 18 at 3:08
  • 1
    $\begingroup$ @dmtri My first thought is you should start with a new notebook and a fresh kernel, or Clear your variables. Can you be more explicit about what is missing? Maybe edit your question and add a new paragraph with your latest code and and a picture of the graph. $\endgroup$ – LouisB Nov 18 at 3:25
3
$\begingroup$

Here's how to do it using the Disk graphics primitive. It is similar to Nasser's deleted answer, but with the difference that I use ColorData[97] to make the color of the marker match the curve, and I use Offset to fix the aspect ratio issue. (No matter what aspect ratio you're using, the disk will stay circular.)

f[x_] := Which[x <= 1, x^3 + 3 x^2 - 2, x > 1, 4]

Plot[
 f[x], {x, -4, 4},
 PlotRange -> {{-4, 4}, {-4, 6}},
 AspectRatio -> 1/GoldenRatio,
 AxesLabel -> {x, y},
 Epilog -> {
   EdgeForm[{Thick, ColorData[97, 1]}], FaceForm[White],
   Disk[{1, f[1]}, Offset[{4, 4}]]
   }
 ]

Mathematica graphics

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.