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I used this code on Mathematica but it's not showing any result?!

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

please consider that A, B, and C are positive.

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  • $\begingroup$ Is that all the code you used? What do you expect it to show as a result? You really need to improve your question or it is likely it will get closed and deleted. $\endgroup$
    – Bill
    Nov 6, 2021 at 19:13
  • $\begingroup$ I want to know with the Table of A, B, and C, how should be inverse of this function! $\endgroup$
    – Mathecis
    Nov 6, 2021 at 20:28
  • $\begingroup$ The symbol C is a built-in symbol in the WL. Avoid the use of symbol names that start with a capital letter. $\endgroup$ Nov 6, 2021 at 20:31

1 Answer 1

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Symbolic inverse functions don't exist in general:

f = InverseFunction[Sin[#] + ArcSin[#] &]
(* InverseFunction[Sin[#1] + ArcSin[#1] &] *)

However, Mathematica can evaluate such things numerically:

f[1.5]
(* 0.736689 *)
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  • $\begingroup$ How can I plot this inverse? for example plot inverse f'(x)/inverse f(x) $\endgroup$
    – Mathecis
    Nov 8, 2021 at 16:12
  • $\begingroup$ f[#1]:=AArcTanh[(#1*A)/Sqrt[#1^2*B - C]] - Sqrt[B]*ArcTanh[(#1*Sqrt[B])/Sqrt[#1^2*B - C]])/(A^2 - B) + (ALog[#1^2*(A^2 - B) + C])/(2*(A^2 - B)) $\endgroup$
    – Mathecis
    Nov 8, 2021 at 16:13

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