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 $f$.
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
|
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 $y=f^{-1}(1)=0$. |
|||||||||||
|
