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So iI was going through my analysis book and I saw this exercise that needed me to find the inversion of a function called $\tanh(x)$ .

$\tanh(x)=\frac{\sinh(x)}{\cosh(x)}$

I was wondering if I am allowed to just find the inversion of $\sinh(x)$ and $\cosh(x)$ and just say that this is the result of $\tanh(x)$

What I mean is if this is correct or not :

$\tanh^{-1}(x)=\frac{\sinh^{-1}(x)}{\cosh^{-1}(x)}$

Bernard
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Nion
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  • Did you check to see if it works as the inverse? – Randall Oct 31 '19 at 16:12
  • To tell you the truth i didn't. I just wanted to see first if i am allowed to do this or not, and i couldnt find it in my book. – Nion Oct 31 '19 at 16:13
  • You cannot inverse an even function (symmetric). – Andronicus Oct 31 '19 at 16:13
  • As @Randall says, you can convince yourself that your solution is indeed wrong. Moreover, think about the definition of $\tanh(x)$ using $\sinh(x)$ and $\cosh(x)$. From there, try to isolate $x$ (standard approach to find the inverse). However, make sure you are aware that you can invert an even function just adapting its domain, as Andronicus says. – Fede1 Oct 31 '19 at 16:14
  • Welcome @Nion ! No, you are not allowed to do so. Moreover, the hyperbolic tangent is standardly written as $\tanh(x)$ and its inverse $artanh(x)$. – Fede1 Oct 31 '19 at 16:16
  • @Fede1 Thank you , ill try it the way you told me so – Nion Oct 31 '19 at 16:17

2 Answers2

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No we are not allowed in general to make this kind of operations to find the inverse, that is given

$$f(x)=\frac{g(x)}{h(x)} \not \Rightarrow f^{-1}(x)=\frac{g^{-1}(x)}{h^{-1}(x)}$$

think also to the simpler case

$$\tan(x)=\frac{\sin x}{\cos x} \not \Rightarrow \arctan x=\frac{\arcsin x}{\arccos x}$$

For the proper expressions refer to

user
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Here's a sketch of what you should do:

$$y=\operatorname{argtanh}x \stackrel{\text{def}}{\iff} \tanh y=x \iff\frac{\mathrm e^{2y}-1}{\mathrm e^{2y}+1}=x$$ Set $u=\mathrm e^{2x}$ and solve the equation in $u$: $\;\dfrac{u-1}{u+1}=x$ first, then solve in $x$.

Bernard
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