Answer to Question #195218 in Calculus for Ichigo

Question

Check the local inevitability of the function f defined by f(x,y)=(x²-y²,2xy) at (1,-1). Find a domain for the function f in which f is invertible. 

Answer 1

Given, $f(x,y)=(x^2-y^2,2xy)$

Here, $u=x^2-y^2,v=2xy$

$J|f(x,y)|=\begin{vmatrix} \dfrac{du}{dx} & \dfrac{du}{dy}\\\\\dfrac{dv}{dx}&\dfrac{dv}{dy}\end{vmatrix} =\begin{vmatrix} 2x& -2y\\2y&2x\end{vmatrix}=4x^2+4y^2$

$\Rightarrow J|f(1,-1)|=4(1)^2+4(-1)^2=4+4=8$

$Hence, J|f(1,-1)|\neq 0,$ So F is invertible at (1,-1).

As $4(x^2+y^2)>0,$ So The domain for which f is invertible is all Real numbers R.