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?

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    $\begingroup$ If $f$ takes the values $f(x_0)$ and $f(x_1)\geq f(x_0)$ and $f(x_2)\geq f(x_0)$, with $x_1<x_0$ and $x_2>x_0$, then it takes the value $\frac{f(x_0)+\min(f(x_1),f(x_2))}{2}$, both in the interval $[x_1,x_0]$ and in $[x_0,x_2]$. $\endgroup$ – Alamos Apr 25 '15 at 22:27

You are right, a function cannot have extremes, global or local, and be one-to-one. The only case where $\frac {d}{dx}f(x_0)=0$ does not mess up the function's being one-to-one is when $f(x_0)$ is an inflection point, where concave up meets concave down, i.e. the second derivative crosses the x-axis. Your approach of checking with horizontal lines is good in the abstract, but if microscopic changes or variance in near-horizontal segments of a graph are possible, it's not reliable. Comparing $f(x-\varepsilon)$ with $f(x+\varepsilon)$ is very good.


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