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I would like to find the inverse of $$f(x) = e^{-x^2/2} + 2x.$$ The reason I need to find the inverse is to evaluate the derivative of the inverse at $x=1$. I realise I can do that without actually finding the inverse of $f$ since $f(0)= 1$ if and only if $f^{-1}(1) =0$. Yet I'd appreciate it if someone could help me find the inverse of $f$.

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up vote 6 down vote accepted

Recall that $$ (f\circ f^{-1})(x)=x $$ implies, by chain rule, $$ f'(f^{-1}(x))(f^{-1})'(x)=1. $$ So $$ f'(f^{-1}(1))(f^{-1})'(1)=f'(0)(f^{-1})'(1)=1. $$ Hence $$ (f^{-1})'(1)=\frac{1}{f'(0)}. $$

Now I'm sure you can compute $f'(0)$.

Edit: if you want to avoid the formula like you say, you can differentiate $$ e^{-y^2/2}+2y=x $$ with respect to $x$. This yields $$ 2\frac{dy}{dx}-y\frac{dy}{dx}e^{-y^2/2}=1\quad\mbox{hence}\quad \frac{dy}{dx}=\frac{1}{2-ye^{-y^2/2}}. $$ Now make $y=f^{-1}(1)=0$.

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I'm aware of the formula you've quoted. I wanted to avoid using it. Thanks though. – Paul Feb 26 '13 at 14:55
@Paul Why would you want to avoid it? Is this required by the problem you are trying to solve? This is certainly the most natural way to do it. And any way around that would be artificially reproving the formula implicitly. – 1015 Feb 26 '13 at 14:57
Maybe, 'avoid' is the wrong word. I already did it using that approach. Perhaps I should have been clear in asking the question. Thanks again. – Paul Feb 26 '13 at 15:05
@Paul I've tried another approach by implicit differentiation. – 1015 Feb 26 '13 at 15:07
Thanks very much Julien. – Paul Feb 26 '13 at 15:31

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