Answer to Question #209278 in Calculus for Sarita bartwal

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

Given function f: R² $\to$ R²

such that f(x,y) = (xy³+1, x²+y²)

we have to show that f is not invertible in R²

Let u = xy³+1,v = x²+y²

Df(x,y) = $\begin{vmatrix} \frac{\delta u}{\delta x} & \frac{\delta u}{\delta y} \\ \frac{\delta v}{\delta x} & \frac{\delta v}{\delta y} \end{vmatrix}$

Df(x,y) = $\begin{vmatrix} \frac{\delta }{\delta x}(xy^3+1) & \frac{\delta }{\delta y}(xy^3+1) \\ \frac{\delta }{\delta x}(x^2+y^2) & \frac{\delta }{\delta y}(x^2+y^2) \end{vmatrix}$

Df(x,y) = $\begin{vmatrix} y^3 & 3xy^2 \\ 2x & 2y \end{vmatrix}$

Df(x,y) = 2y⁴$-$ 6x²y²

Df(x,y) = 2y²(y²$-$ 3x²)

here we see that, Df(x,y) = 0 if 2y²(y²$-$ 3x²) = 0 $\implies$ either y = 0 or (y²$-$ 3x²) = 0 $\implies$ either y = 0 or y = $\pm$ $\sqrt{3}x$

hence,

Df(x,y) is not invertble at (a,0) $\in$ R² for any a $\in$ R also Df(x,y) is not invertble at (x, $\pm\sqrt{3}x$) $\in$ R²

$\implies$ f(x,y) is not invertible.

at point (2,1)

Df(2,1) = 2(1)²[(1)² $-$ 3(2)²]

Df(2,1) = 2(1 $-$ 12)

Df(2,1) = -22 $\neq$ 0

hence, f(x,y) is locally invertible at (2,1)