Hi everyone I am new in Mathematica and I am trying to find the inverse of a function which contains hyperbolic tangents and I hope someone can help me.

The function is

y=(a)tanh(d(x-g))+ (a/2)(tanh(d(m+g))-tanh(d(m-g)))

Thank you in advance

  • 1
    $\begingroup$ Please post with actual and correct Mathematica code, e.g., use Tanh[d (x - g)] rather than $tanh(d(x - g))$. Also, what are your $d$, $m$, and $g$ — particular constants or arbitrary constants (perhaps with some conditions placed upon them)? $\endgroup$ – murray Aug 10 '15 at 14:53
  • $\begingroup$ Will you be happy with an inverse function involving ArcTan? If so, try InverseFunction. $\endgroup$ – murray Aug 10 '15 at 14:58

The command

InverseFunction[a Tanh[d (# - g)] + a/2 (Tanh[d (m + g)] - Tanh[d (m - g)]) &][y]

gives the output

$$ \frac{\tanh ^{-1}\left(\frac{a \tanh (d (m-g))-a \tanh (d (g+m))+2 y}{2 a}\right)+d g}{d}$$

together with the warning that $\tanh^{-1}$ is a multivalued function.

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