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Should find the inverse of:

$$ f(x) = \frac{1}{2} ( e^x - e^{-x} ) $$

I tried a lot. But I don't know how to proceed on $$ 2x = \frac{(e^y)^2 - 1}{e^y} $$

Writing $ e^{-y} $ as $ \frac{1}{e^y} $ is right? I know somewhere I need to use the $ ln $ but I dont know when

loop
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  • $\operatorname{arsinh} (x) = \ln \left(x + \sqrt{x^{2} + 1} \right)$ – Henry Sep 12 '13 at 13:50
  • @Henry That is wonderfully enlightening. – Did Sep 12 '13 at 13:53
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    @Did: It provides a check at the end. It also suggests the negative root of the quadratic needs a little thought – Henry Sep 12 '13 at 13:57
  • @Henry "Suggests"?? How? I see no suggestion of this sort, just a flat formula. – Did Sep 12 '13 at 14:05
  • @Did: as in Silver Blaze. Sherlock Holmes: "... the curious incident of the dog in the night-time." Inspector Gregory: "The dog did nothing in the night-time." Sherlock Holmes: "That was the curious incident." – Henry Sep 12 '13 at 14:15
  • The Maple code $$with(Student[Calculus1]): InverseTutor(1/2*(exp(x)-exp(-x)), x = -10 .. 10); $$ produces the output. – user64494 Sep 12 '13 at 16:05

2 Answers2

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Let $e^x=z$. Thus, you have:

$$y=\frac{1}{2} (z-\frac{1}{z})$$

Solve the resulting quadratic for $z$ and then set $x = ln(z)$.

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The quantity $z=\mathrm e^y$ solves the quadratic $z^2-2xz-1=0$. Surely you can write down the roots of the quadratic and from there, deduce $y$ (just remember that $z\gt0$).

Did
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  • what is $ z^2 - 2xy - 1 = 0 $ and how would you come up with that? – loop Sep 12 '13 at 13:49
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    You know that $2x=((\mathrm e^y)^2-1)/\mathrm e^y$ and you decide to rename $\mathrm e^y$ as $z$. Hence $2x=(z^2-1)/z$, right? Hence... – Did Sep 12 '13 at 13:52
  • ah. now i know where u want to "direct" me. the typical root formula. thanks! – loop Sep 12 '13 at 13:54
  • Indeed. Note the word "quadratic". Either you know the formula for the roots of any quadratic, or you note that the equation to solve is also $z^2-2xz+x^2-x^2-1=0$, that is, $(z-x)^2=$ $____$ hence... – Did Sep 12 '13 at 13:55
  • know i just take the formula i know from highschool to solving roots. pretty simple and i get $ z_1 = 2 + \sqrt{x^2 + 1} $ and $ z_2 = 2 - \sqrt{x^2+1} $ – loop Sep 12 '13 at 13:59
  • Please have a fresh look at "the formula [you] know from highschool", the one you used is wrong. – Did Sep 12 '13 at 14:00
  • oh yeah. miscalculated! you're right! but if i have 2 roots then, so is this proof for not having a true inversional function? or do i have 2 functions for negative and positve x? – loop Sep 12 '13 at 14:03
  • You do have two roots but you know that $z\gt0$ since $z$ is a real exponential hence you know which root is "your" $z$. (This was explained at the very end of my answer, no?) – Did Sep 12 '13 at 14:04
  • z is > 0 because e^x never gets below 0? or why? – loop Sep 12 '13 at 14:05
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    Yes (not "or why"). – Did Sep 12 '13 at 14:07