# Inverse of $e^{-x^2/2} + 2x$

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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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 $y=f^{-1}(1)=0$.