# Understanding plot of Root[] object

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

Edit: here is the screenshot. • In Mathematica v12.2 the solution you're interested seems to be Solve[Eq, r][] , but this version doesn't show the gap in your plot. Sep 20 at 7:56
• Thanks for your help. I'm a bit confused: The plot in my initial post shows solution []. I've tried to implement your procedure and it doesn't work for [] due to complex values. However, it does work for solution [] and the plot appears to correspond to the one in my initial post. However, if I just plot solution [] 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. Sep 20 at 8:35
• Plot[Evaluate[ r /. Solve[Eq , r ]], {v, 0, 70} ] shows all real solutions. Sep 20 at 9:08
• 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}]. Sep 20 at 11:25
• Try Show[{Table[ Plot[Evaluate[r /. sol[[i]]], {v, 0, 70}], {i, 1, Length[sol] }]}, PlotRange -> {-10, 10}] which should give the same plot! Sep 20 at 12:28

Try

sol = Solve[Eq , r][] ;


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][]

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