Answer to Question #192517 in Calculus for Nikhil Rawat

Question

Show that the function f: R^2→R^2 given by

f(x,y) = (xy^3+1, x^2+y^2) is not invertible. Futher

check whether it is locally invertible at the point (2,1)

Answer 1

According to the Inverse function theorem, a function is invertible in a neighborhood of a point in its domain: namely, that its Jacobian is continuous and non-zero at the point. 

Let us build the Jacobian:

$J= \begin{pmatrix} f_{1x} & f_{1y} \\ f_{2x} & f_{2y} \end{pmatrix} = \begin{pmatrix} y^3 & 3xy^2 \\ 2x & 2y \end{pmatrix}$

and the Jacobian determinant is

$\det J=2y^4-6x^2y^2=2y^2(y^2-3x^2)$

$\det J=0 \iff y=0;\quad or \quad y=\pm\sqrt{3}x$

Therefore this function is not invertible.

At the point (2,1)

$\det J(2,1)=2*1^2(1^2-3*2^2)= 2∗1 2 (1 2 −3∗2 2 )=-22\ne 0$

Therefore this function is locally invertible at (2,1)