# 1617-257/TUT-R-7

Suppose ${\displaystyle F:\mathbb {R} ^{2}\to \mathbb {R} ^{2}}$ is ${\displaystyle C^{1}}$ with ${\displaystyle F(t^{4},e^{t^{2}})=(4,5)}$ for all ${\displaystyle t\in \mathbb {R} }$. Prove that ${\displaystyle DF(0,1)}$ is not invertible.
Let ${\displaystyle \gamma (t):=(t^{4},e^{t^{2}})}$. At the end of the tutorial, a couple of students pointed out that it's also true that ${\displaystyle DF(\gamma (t))}$ is not invertible for any ${\displaystyle t\in \mathbb {R} }$. It's easy to check that ${\displaystyle \gamma '(t)\neq 0}$ when ${\displaystyle t\neq 0}$ and that ${\displaystyle \gamma '(t)}$ is in the kernel of ${\displaystyle DF(\gamma (t))}$.
It takes some more care to prove what the problem is asking because one can't immediately deduce that ${\displaystyle DF(\gamma (t))}$ has non-trivial kernel with ${\displaystyle \gamma '(0)=0}$. A key observation here is that since ${\displaystyle F}$ is ${\displaystyle C^{1}}$, invertibility of ${\displaystyle DF}$ at ${\displaystyle \gamma (0)}$ implies invertibility of ${\displaystyle DF}$ at a neighborhood of ${\displaystyle \gamma (0)}$.