0
$\begingroup$

This question already has an answer here:

I have been trying to understand hyperbolic functions for some time now. I have a problem concerning the expressions of inverse hyperbolic functions. The text( G.B. Thomas ) mentions nothing about their expressions.

While plotting the Hyperbolic Sine was easy, plotting its inverse was not. By definition we have, $$\sinh(x) = \frac{e^x - e^{-x}}{2}$$

I plotted its graph below-

enter image description here

The dotted lines are the function's asymptotes. I then looked at the graph of inverse hyperbolic Sine and decided to obtain its expression. I tried by finding the inverses of the asymptotes because I knew that the inverse function would get close to them eventually. I plotted a graph.

enter image description here

I was happy. But not for long. It turned out that I cannot have an algebraic combination of these two asymptotes. This was my second approach, to be honest. A holy approach to finding inverses is, express $x$ in terms of $y$. I got stuck at the very beginning and could never proceed. Is there a way around this? If there indeed is a way to express $x$ in terms of $y$, could anyone give me a few hints to do it? If there isn't, how do we obtain an expression for the inverse hyperbolic Sine?

$\endgroup$

marked as duplicate by Hans Lundmark, Arthur, kingW3, Glorfindel, Namaste Aug 15 '17 at 20:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I will make sure to search for answers before posting a question. This link guides me well. $\endgroup$ – R004 Aug 15 '17 at 10:55
  • 1
    $\begingroup$ A little search tip: use Google instead of the site's own search function. Like this: google.com/… $\endgroup$ – Hans Lundmark Aug 15 '17 at 10:58
1
$\begingroup$

Let $y\in R $

we look for $x\in R $ such that

$$e^x-e^{-x}-2y=0$$ or $$e^{2x}-2ye^x-1=0$$ put $t=e^x>0$ then $$t^2-2yt-1=0$$ hence $$t=e^x=y+\sqrt {1+y^2} $$

and finally $$x=\ln (y+\sqrt {y^2+1} )$$

$\endgroup$
  • $\begingroup$ I appreciate your answer. $\endgroup$ – R004 Aug 15 '17 at 14:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.