$f(3t^3+2,et^2)=(3,6)$ for all $t\in\mathbb{R}$. Prove: $D_f(2,1)$ not invertible

Question

Let $f\in C^1[\mathbb{R}^2 , \mathbb{R}^2]$

satisfying: $f(3t^3+2,e^{t^2})=(3,6)$ for all $t\in\mathbb{R}$.

Prove: $D_f(2,1)$ not invertible.

My try:

we define $g(t)=(3t^3+2,e^{t^2})$. Then, $g\circ f=(3,6)$.

Using chain rule - $D_f(g(t))\cdot g'(t) = (0,0)$.

For $t=0$:

$D_f(g(0))\cdot g'(0) = D_f((2,1))\cdot (0,0)=(0,0)$

What can I do now?

Thank you for the help.

Answer 1

There is a way to do it without computations (I always prefer those, personally). If your matrix were invertible, by the inverse function theorem, you would have a well defined inverse function from a neighborhood of $(3,6)$ to a neighborhood of $(2,1)$. But this is clearly not the case because to $(3,6)$ there correspond all the points on the intersection of your curve with any neighborhood of $(1,2)$.

Answer 2

Construct the $2\times 2$ matrix $(Df)_{ij} = \frac{\partial f_i}{\partial x_j}$. Now find the determinant of this matrix evaluated at the point $(2,1)$.

$Df = D(f(g(t)) = Df_{g(t)} \times D g(t) = (0,0)\times ( 9t^2, 2e t) = 0$ (i.e. the Jacobian is the zero matrix), hence it has zero determinant.