Let $f$ be a continuously differentiable function defined $f : \mathbb R \to \mathbb R$ such that $f(x)$ is defined for for all $x$. Suppose $x_0$ is a local minimizer for $f$. Is $f$ one-to-one?
I was thinking this must be false because at $x_0$, you can probably find two points $p_1$ and $p_2$ in $[x_0 + \epsilon, x_0 - \epsilon]$ with $f(p_1) = f(p_2)$. I think if you drew a horizontal line across the y-axis slightly above $f(x_0)$ it would cross the graph $f(x)$ at least twice which would mean it's not 1-1. However, I am just not sure how to formalize this idea (Though if my idea is wrong too, please correct me).
Any tips for how to approach this?