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I didn't manage to obtain the inverse function for this frightening function:

t[a_] := Sqrt[3/(8 Pi GG)]*2/3 (1 + (a^(3 + 3 α) A)/const)^(1/(2 + 2α))(A + a^(-3 (1 + α))const)^(-(1/(2 + 2 α)))Hypergeometric2F1[1/(2 + 2 α), 1/(2 + 2 α),1 + 1/(2 + 2 α), -((a^(3 + 3 α) A)/const)]

I'm a beginner with Mathematica so I don't understand how to apply ParametricPlot described here The graph of inverse function $f^{-1}(y)$

And failed attempt:

Assuming[A > 0 && α > 1 && const > 0 ,{Plot[InverseFunction[t][y], {y,t[1. - 10^-9], t[0. + 10^-9]}]}]

The important assumptions are GG=6.67*10^(-11), A>0, const>0, Alpha>1.

I would be happy to get any help.

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    $\begingroup$ Probably you want:GG = 6.67*10^(-11); α = 2; A = 1; const = 1; f[y_] := Sqrt[3/(8 Pi GG)]*2/ 3 (1 + (y^(3 + 3 α) A)/const)^(1/(2 + 2 α)) (A + y^(-3 (1 + α)) const)^(-(1/(2 + 2 α))) Hypergeometric2F1[1/(2 + 2 α), 1/(2 + 2 α), 1 + 1/(2 + 2 α), -((y^(3 + 3 α) A)/ const)]; ParametricPlot[{f[y], y}, {y, 10^-9, 1}, AspectRatio -> 1/2] $\endgroup$ – Mariusz Iwaniuk Apr 22 '18 at 17:44
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    $\begingroup$ Note that your Plot[InverseFunction[t][y],...] will work only when you assign numeric values to the parameters (e.g. A, const, etc.). This restriction has to do with Plot requiring numeric coordinates more than with InverseFunction: for instance, Plot[a x, {x, 0, 1}] produces a blank plot if a is undefined, but InverseFunction[a # &] works (tacitly assuming a is nonzero). $\endgroup$ – Michael E2 Apr 22 '18 at 18:15
  • $\begingroup$ The ParametricPlot trick is simple enough, you just ParametricPlot[{t[a],a},{a,fromthere,tothere}] which will simply exchange the horizontal and vertical axis for you - but is it what you want? $\endgroup$ – მამუკა ჯიბლაძე Apr 22 '18 at 18:32

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