Need some help on this one.
Let f(x)= x^2 e^x^2, x in R.
Show f^-1 exists and is differentiable on (0, infinity)
Thanks.
Theorem: Let $\displaystyle f$ be a one-to-one function on an open interval $\displaystyle I$, and let $\displaystyle J = f(I)$. If $\displaystyle f$ is differentiable at $\displaystyle x_0 \in I$ and if $\displaystyle f'(x_0) \ne 0$, then $\displaystyle f^{-1}$ is differentiable at $\displaystyle y_0 = f(x_0)$ and $\displaystyle \left( f^{-1}\right)'(y_0) = \frac 1{f'(y_0)}$
are those conditions fulfilled here?
now actually showing that the inverse function exists is another story. we have to cross that bridge before we use the above theorem