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I'm trying to understand the plot of this Root object

Root[-1536 (64 - v)^(2/3) + 24 (64 - v)^(2/3)v + (1536 (64 - v)^(2/3) - 24 (64 - v)^(2/3) v) #1^2 + (-384 (64 - v)^(1/3) + 6 (64 - v)^(1/3) v) #1^3 + 4 (64 - v)^(1/3) #1^4 + (-384 (64 - v)^(1/3) + 6 (64 - v)^(1/3) v) #1^5 + (192 - 3 v) #1^6 + 2 #1^7 &, 7]

which is one of the solutions of the equation

Eq = (2 r^8 + 4 r^5 (64 - v)^(1/3) - 6 r^4 (64 - v)^(4/3) - 6 r^6 (64 - v)^(4/3) - 24 r (64 - v)^(5/3) + 24 r^3 (64 - v)^(5/3) - 3 r^7 (-64 + v))/((r^3 + 2 (64 - v)^(1/3)) (64 - v)^(1/3)) == 0;

The plot looks like this:

enter image description here

I would like to identify the values $(v,r)$ for the four points which I have encircled in the plot, i.e. the precise value of $r$ at $v=0$ and the points $(v,r)$ where the function starts/stops. I have tried

Solve[Eq, r][[8, 1, 2]] /. {v -> 0} 

but I am not really sure what to do with the output.

enter image description here

How do I find the points $(v,r)$ that I am interested in?

Edit: here is the screenshot.

enter image description here

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  • $\begingroup$ In Mathematica v12.2 the solution you're interested seems to be Solve[Eq, r][[6]] , but this version doesn't show the gap in your plot. $\endgroup$ Sep 20 at 7:56
  • $\begingroup$ Thanks for your help. I'm a bit confused: The plot in my initial post shows solution [[8]]. I've tried to implement your procedure and it doesn't work for [[8]] due to complex values. However, it does work for solution [[6]] and the plot appears to correspond to the one in my initial post. However, if I just plot solution [[6]] the plot is entirely different and does not include the points that were found using your suggested procedure. Any idea what's going on here? I've attached a screenshot to my post to clarify what I mean. $\endgroup$ Sep 20 at 8:35
  • $\begingroup$ Plot[Evaluate[ r /. Solve[Eq , r ]], {v, 0, 70} ] shows all real solutions. $\endgroup$ Sep 20 at 9:08
  • $\begingroup$ Thanks, that helps. Here's what I don't understand: Plot[Evaluate[r /. Solve[Eq, r]], {v, 0, 70}] gives me a plot of all real-valued solutions. Presumably the different colours correspond to different solutions. When I try to plot the solutions individually to figure out which one corresponds to the one I'm interested in by using Plot[Evaluate[r /. Solve[Eq, r]][[i]], {v, 0, 70}] where $i \in \lbrace 1,2,3, ... \rbrace$, I end up with completely different plots, i.e. non of the individual plots corresponds to any of the coloured lines in Plot[Evaluate[r /. Solve[Eq, r]], {v, 0, 70}]. $\endgroup$ Sep 20 at 11:25
  • $\begingroup$ Try Show[{Table[ Plot[Evaluate[r /. sol[[i]]], {v, 0, 70}], {i, 1, Length[sol] }]}, PlotRange -> {-10, 10}] which should give the same plot! $\endgroup$ Sep 20 at 12:28
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Try

sol = Solve[Eq , r][[6]] ;

The two points on the curve follow

p0 = {v, r} /. sol /. v -> 0 // N
p1 = {v, r /. sol } /. NMinimize[{D[r /. sol, v], v > 50}, v][[2]]

Show[{Plot[r /. sol, {v, 0, 64}], Graphics[{Red,Point[{p0, p1}]}]}, PlotRange -> {0, Automatic}, AxesOrigin -> {0, 0}]

enter image description here

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