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Given the function

$$ f(x) = e^x + \arctan(x) = y\;, $$

what is the inverse $f^{-1}(y)=\dots\;$, and how can I find it? I’m looking for solutions including all steps and possible explanations along with each.

To give some wider context, I bumped into this problem as part of a bigger question asking me to prove that $f(x)$ is bijective, $(f^{-1})'(y)$ exists for all $y>-\pi/2$ and to calculate $(f^{-1})'(y)$.

I have proven that $f$ is bijective and the rest of the properties follow from the applicability of the inversion theorem for derivatives, but I have a hard time calculating $(f^{-1})'(y)$ now because I can't find $(f^{-1})(y)$

Thanks for your help!

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    I didn’t change it in the edit, but I think that you mean $f^{-1}(y)=\dots\;$.2012-02-25
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    What is the domain and codomain of your function?2012-02-25
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    The fact that Wolfram Alpha only give a [numerical solution for $f(x)=0$](http://www.wolframalpha.com/input/?i=exp%28x%29+%2B+atan%28x%29+%3D+0) suggests that there will not be a simple inverse2012-02-25
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    Added some more context, hope this helps2012-02-25
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    There is no elementary inverse function. Why do you need to find it?2012-02-25

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To find $(f^{-1})'(y)$ you don't need the inverse function. What you need is to use the inverse function theorem: if $f$ is injective and $f(x)=y$ then $(f^{-1})'(y)=\frac{1}{f'(x)}$.

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    [Inverse function theorem](http://en.wikipedia.org/wiki/Inverse_function_theorem).2012-02-25
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    Ouch, thanks a lot!2012-02-25