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Consider a function $f(x)$ such as $x\mapsto 2e^x-\frac1{e^x}$. How do you find $f^{-1}(x)$?

I have tried, logarithms, squaring, substitution, but I wasn't able to isolate $x$. The correct answer, according to Wolfram Alpha is $f^{-1}(x) = \log{\left(\frac14\left(x+\sqrt{x^2+8}\right)\right)}$.

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If you start with

$$x=2e^y-e^{-y}$$

and multiply both sides by $e^y$, you get

$$x e^y=2e^{2y}-1$$

which you can treat as a quadratic equation in $e^y$. Use the quadratic formula to solve for $e^y$ (be careful in choosing roots!), undo the exponential with the logarithm, and you have your needed result.

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    Right, I actually tried that! But this solved to $y=0$ (choosing the right root, like you said), hence eliminating all $y$-terms. What am I doing wrong?2011-05-04
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    BTW, you can figure out why the minus sign in your original question is wrong...2011-05-04
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    BTW, what do you mean??2011-05-04
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    As another way of thinking about it: is $x$ always bigger than $\sqrt{x^2+8}$? Having thought about it, which of $+$ or $-$ should you choose in the quadratic equation?2011-05-04
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    Right, it should be $\pm$. Is that what you mean?2011-05-04
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    @Milosz, plugging in $x = 0$, you get $f(0) = 1$. Therefore, $f^{-1}(1)$ should be $0$.2011-05-04
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    Nonono, $\pm$ means you have two signs to choose from. To know which one to choose, consider my previous comment.2011-05-04
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    @DJC: yes, that's another good sanity check, but I wanted Milosz to think about why he shouldn't be subtracting within the logarithm in this case...2011-05-04
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    Oh. My apologies. I can delete the comment if you like.2011-05-04
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    @DJC, indeed. In my first attempt to a solution I tried factorization, which gave me $(a-1)(a+\frac{1}{2})=0$ where $a=e^{y}$. Hence solving for $y$, gives $y=0$. This however, wouldn't allow me to isolate $y$. I didn't consider the quadratic formula which clearly was the way to go. @J.M. thanks, I see what you mean. The argument of the logarithm cannot be negative!2011-05-04
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    @DJC: No need, and no worries; it helps to have more than one way to see that your solution's hunky-dory. :)2011-05-04