Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be defined as
$$f(x,y)=\begin{pmatrix} x^3+xy+1\\x+y+y^3+1 \end{pmatrix}$$
Show there is an open neighborhood $U$ of $(1,1)^T$, which is mapped bijectively to a neighborhood $V$ of $(3,4)^T$ by $f$. Find the derivative of the inverse $f^{-1}: V \rightarrow U$ of f in the point $(3,4)^T$.
Explicitly calculating the inverse function is a mess.
My second attempt: define $g: \mathbb{R}^2\times \mathbb{R}^2 \rightarrow\mathbb{R}^2$ such that $g((1,1),(3,4)) = 0$. Then the implicit function theorem gives us that there exists a mapping $\phi$ from a neighborhood of $(1,1)$ to a neighborhood of $(3,4)$, and it tells us that map is continuously differentiable. Using this approach I don't know how to show $f=\phi$, let alone that $\phi$ is bijective.
Is my approach wrong? If not how do I continue?
Best Answer
Show that $f'(1,1)$ is invertible. This shows the existence of the sets $U $ and $V $. The derivative of the inverse at $(3,4)^T $ is given by $f'(1,1)^{-1}$.