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I'm trying plot an inverse function

InverseFunction[2 A ArcTanh[(# A)/Sqrt[-1 + #^2 B]] + A Log[1 + #^2 (A^2 - B)] - 2 Log[# B + Sqrt[-1 + #^2 B]] &][x]

Plot[Abs[%] /. {A -> 0.2, B -> 0.3}, {x, 0, 2}, PlotRange -> All]

But I can't get any result??!!

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    – Michael E2
    Dec 30, 2020 at 3:44

1 Answer 1

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We can draw y=f[x] by

ParametricPlot[{x, f[x]}, {x, 0, 2}]

and draw its inverse x=f[y] by

ParametricPlot[{f[y], y}, {y, 0, 2}]

f[x_] := 2 A ArcTanh[(# A)/Sqrt[-1 + #^2 B]] + 
       A Log[1 + #^2 (A^2 - B)] - 2 Log[# B + Sqrt[-1 + #^2 B]] &[
     x] /. {A -> 0.2, B -> 0.3} // Abs;
ParametricPlot[{{x, f[x]}, {f[x], x}}, {x, 0, 2}]

enter image description here

We can also use ContourPlot

f[x_] := 2 A ArcTanh[(# A)/Sqrt[-1 + #^2 B]] + 
       A Log[1 + #^2 (A^2 - B)] - 2 Log[# B + Sqrt[-1 + #^2 B]] &[
     x] /. {A -> 0.2, B -> 0.3} // Abs;
ContourPlot[{y == f[x], x == f[y]}, {x, 0, 2}, {y, 0, 2}]
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  • $\begingroup$ You just plot f(x) but as you see I used InverseFunction[f(x)] it means I want to plot f^-1(x) $\endgroup$
    – Armin Sharafi
    Dec 30, 2020 at 2:00
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    $\begingroup$ @ArminSharafi Here we draw y=f[x] and x=f^-1[y] at the same picture. $\endgroup$
    – cvgmt
    Dec 30, 2020 at 2:08

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