$\,f \colon \Bbb R^2 \to \Bbb R $ be continuous map such that $\,f(x)=0\,$

Question

I am stuck on the following problem :

Let $\,f \colon \Bbb R^2 \to \Bbb R $ be continuous map such that $\,f(x)=0\,$ for only finitely many values of $x$. Then which of the following options is correct?

1. either $f(x)\le 0 \forall x \,\text{or}\, f(x) \ge 0 \forall x$

2. the map $f$ is one-to-one

3. the map $\,f$ is onto

4. none of the above

My Attempt: Since $\,f(x)=0\,$ for only finitely many values of $x$, $\exists x_1=(a_1,b_1),x_2=(a_2,b_2) \in \Bbb R^2$ such that $f(x_1)=0=f(x_2)$ whereas $x_1 \neq x_2$. So, $f\,$ is not one-to-one. Also,it is a many-to-one mapping and so can not be onto. I am confused about options 1 and 4.

Can someone point me in the right direction with some explanation? Thanks and regards to all.

Answer 1

I guess this is answer one. My idea is the following :assume there exist $x_0$ and $x_1$ such that $f(x_0)\cdot f(x_1)<0$ then take infinitely different curves $\gamma:[0,1]\rightarrow\mathbb{R^2}$ for which $\gamma(0)=x_0$ and $\gamma(1)=x_1$ and that does not have any common points ( an easy exercice to build them with portion of line). then you can apply intermediate value theorem to show that the map $t\mapsto f(\gamma(t))$ vanishes somewhere. Finally you obtain infinitely many zeroes. Then of course $f$ is not "onto"